From ndbecker2 at gmail.com  Mon Oct  2 14:03:07 2017
From: ndbecker2 at gmail.com (Neal Becker)
Date: Mon, 02 Oct 2017 14:03:07 -0400
Subject: [Matplotlib-users] howto plt.subplots (projection='3d')?
Message-ID: <oqtv0r$m5j$1@blaine.gmane.org>

Any way to use plt.subplots for 3d?

This doesn't work:
  fig, axs = plt.subplots(nrows=opt.rcv_paths, ncols=1, sharex=True, 
subplot_kw=dict(projection='3d'))
...
  File "/home/nbecker/.local/lib/python3.6/site-
packages/matplotlib/projections/__init__.py", line 67, in 
get_projection_class
    raise ValueError("Unknown projection '%s'" % projection)
ValueError: Unknown projection '3d'


From ben.v.root at gmail.com  Mon Oct  2 14:13:28 2017
From: ben.v.root at gmail.com (Benjamin Root)
Date: Mon, 2 Oct 2017 14:13:28 -0400
Subject: [Matplotlib-users] howto plt.subplots (projection='3d')?
In-Reply-To: <oqtv0r$m5j$1@blaine.gmane.org>
References: <oqtv0r$m5j$1@blaine.gmane.org>
Message-ID: <CANNq6F=uLk8JVP4zQORJE=WpmOikAbHAHK1Ae40mRe=BrtiGNw@mail.gmail.com>

You have to import the mplot3d toolkit first. That causes the 3d projection
to get registered.

On Mon, Oct 2, 2017 at 2:03 PM, Neal Becker <ndbecker2 at gmail.com> wrote:

> Any way to use plt.subplots for 3d?
>
> This doesn't work:
>   fig, axs = plt.subplots(nrows=opt.rcv_paths, ncols=1, sharex=True,
> subplot_kw=dict(projection='3d'))
> ...
>   File "/home/nbecker/.local/lib/python3.6/site-
> packages/matplotlib/projections/__init__.py", line 67, in
> get_projection_class
>     raise ValueError("Unknown projection '%s'" % projection)
> ValueError: Unknown projection '3d'
>
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org
> https://mail.python.org/mailman/listinfo/matplotlib-users
>
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From daniele at grinta.net  Tue Oct  3 16:59:48 2017
From: daniele at grinta.net (Daniele Nicolodi)
Date: Tue, 3 Oct 2017 14:59:48 -0600
Subject: [Matplotlib-users] How to make \mathdefault{} the default?
Message-ID: <f698d891-4bb5-3915-79fb-9a62ebd3c21b@grinta.net>

Hello,

when adding LaTeX symbols to labels on my figures, most often I use
\mathdefault{} to make Matplotlib use the same fond for the regular text
and for the LaTeX part.  Is there a way to configure Matplotlib to use
the regular text by default?  Looking at the configuration file I
haven't found anything hinting to it.

Thanks. Cheers,
Daniele

From daniele at grinta.net  Tue Oct  3 17:13:56 2017
From: daniele at grinta.net (Daniele Nicolodi)
Date: Tue, 3 Oct 2017 15:13:56 -0600
Subject: [Matplotlib-users] How to make \mathdefault{} the default?
In-Reply-To: <f698d891-4bb5-3915-79fb-9a62ebd3c21b@grinta.net>
References: <f698d891-4bb5-3915-79fb-9a62ebd3c21b@grinta.net>
Message-ID: <db96ebd4-d976-d487-372f-a4ab91ad0bb3@grinta.net>

On 10/3/17 2:59 PM, Daniele Nicolodi wrote:
> Hello,
> 
> when adding LaTeX symbols to labels on my figures, most often I use
> \mathdefault{} to make Matplotlib use the same fond for the regular text
> and for the LaTeX part.  Is there a way to configure Matplotlib to use
> the regular text by default?  Looking at the configuration file I
> haven't found anything hinting to it.

After staring a bit more at the example configuration file, I can answer
my own question.  This does the trick:

matplotlib.rcParams['mathtext.fontset'] = 'custom'

I'm a bit worried by the note next to this setting in the example
configuration file, though:

# The following settings allow you to select the fonts in math mode.
# They map from a TeX font name to a fontconfig font pattern.
# These settings are only used if mathtext.fontset is 'custom'.
# Note that this "custom" mode is unsupported and may go away in the
# future.

Why is it unsupported?  I find this functionality very neat.

Thanks. Cheers,
Daniele

From ajacobo at mail.rockefeller.edu  Wed Oct  4 17:20:33 2017
From: ajacobo at mail.rockefeller.edu (Adrian Jacobo)
Date: Wed, 4 Oct 2017 17:20:33 -0400
Subject: [Matplotlib-users] Finding where an annotation line ends
Message-ID: <4cb57349-3551-ad41-716f-35830da0c1ef@rockefeller.edu>

Hi,

 ?I'm using an annotation for a plot in this way:

ax.annotate("", xy=(x1,y1), xycoords='data',
 ??????????????????????????? xytext=(x2,y2), textcoords='data',
 ??????????????????????????? arrowprops=dict(arrowstyle='-', patchB=None,
 ??????????????????????????????????????????? shrinkA=25),)

Because I'm setting shrinkA to 25 pixels the line doesn't exactly end at 
(x2,y2). Is there an easy way to find out the end point of the line in 
axis coordinates?
An unrelated question, Is there an easy way to use a stump (a circle) as 
arrowstyle?


Many thanks,
Adrian




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From ndbecker2 at gmail.com  Thu Oct  5 07:01:41 2017
From: ndbecker2 at gmail.com (Neal Becker)
Date: Thu, 05 Oct 2017 07:01:41 -0400
Subject: [Matplotlib-users] surface plot with masked elements?
Message-ID: <or53ep$sre$1@blaine.gmane.org>

I need to make a surface plot where some corner elements of my 2D data are 
invalid, and I'd like to omit them from the plot.

I've seen a couple of stackoverflow questions on the subject, but none of 
the answers seemed satisfactory (or I understood).

It's been some years now, what is the current "best" approach to this 
question?


From ndbecker2 at gmail.com  Thu Oct  5 14:55:43 2017
From: ndbecker2 at gmail.com (Neal Becker)
Date: Thu, 05 Oct 2017 14:55:43 -0400
Subject: [Matplotlib-users] surface plot with masked elements?
References: <or53ep$sre$1@blaine.gmane.org>
Message-ID: <or5v77$2j4$1@blaine.gmane.org>

Neal Becker wrote:

> I need to make a surface plot where some corner elements of my 2D data are
> invalid, and I'd like to omit them from the plot.
> 
> I've seen a couple of stackoverflow questions on the subject, but none of
> the answers seemed satisfactory (or I understood).
> 
> It's been some years now, what is the current "best" approach to this
> question?

I'm trying to follow the approach used here
https://stackoverflow.com/questions/11020160/plotting-a-masked-surface-plot-using-python-numpy-and-matplotlib

My code looks like:

lev = np.linspace(np.min(abs(H)),np.max(abs(H)),10)
import matplotlib.colors as colors
norml = colors.BoundaryNorm(lev, 256)
ax.plot_surface(grid_x, grid_y, H_2d, cmap=cm.Blues, norm=norml)

The numbers '256' and '10' in the above seem to have no particular effect.  
I'm thinking plot_surface is re-quantizing the colormap?  


From padiarushi3012 at gmail.com  Sat Oct  7 15:57:22 2017
From: padiarushi3012 at gmail.com (Rushikesh Padia)
Date: Sun, 8 Oct 2017 01:27:22 +0530
Subject: [Matplotlib-users] Need guidance to start
Message-ID: <CAKZ2tfHZFyLMbG1gaRCBmQQsS_iAkR+xEKTtvJBAwjnGqi+u6w@mail.gmail.com>

I am computer science undergraduate. I have basic understanding of c, c++,
python .I would like to do some contribution to this project. This would be
my first open source project. Can anyone please guide me which part I can
work and how to start.
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From tcaswell at gmail.com  Sat Oct  7 16:44:51 2017
From: tcaswell at gmail.com (Thomas Caswell)
Date: Sat, 07 Oct 2017 20:44:51 +0000
Subject: [Matplotlib-users] Need guidance to start
In-Reply-To: <CAKZ2tfHZFyLMbG1gaRCBmQQsS_iAkR+xEKTtvJBAwjnGqi+u6w@mail.gmail.com>
References: <CAKZ2tfHZFyLMbG1gaRCBmQQsS_iAkR+xEKTtvJBAwjnGqi+u6w@mail.gmail.com>
Message-ID: <CAA48SF_H=rw-Qu+hZALmaEYrfxFeZbVQqhgN24+y=G13gK2h7A@mail.gmail.com>

Rushikesh,

The best place to start is
http://matplotlib.org/devdocs/devel/contributing.html which covers how we
use git / github for development.

We have a 'new contributor friendly' tag
https://github.com/matplotlib/matplotlib/issues?q=is%3Aopen+is%3Aissue+label%3Anew-contributor-friendly
of
issues we think are good entry points for new contributors.

Please ask any questions either here or on
https://gitter.im/matplotlib/matplotlib

Tom

On Sat, Oct 7, 2017 at 3:57 PM Rushikesh Padia <padiarushi3012 at gmail.com>
wrote:

> I am computer science undergraduate. I have basic understanding of c, c++,
> python .I would like to do some contribution to this project. This would be
> my first open source project. Can anyone please guide me which part I can
> work and how to start.
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org
> https://mail.python.org/mailman/listinfo/matplotlib-users
>
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From tcaswell at gmail.com  Sat Oct  7 19:30:11 2017
From: tcaswell at gmail.com (Thomas Caswell)
Date: Sat, 07 Oct 2017 23:30:11 +0000
Subject: [Matplotlib-users] [ANN] Matplotlib 2.1 released
Message-ID: <CAA48SF_7ekHVS36EmWa_+om3f3_Y7LotVyvv6qYd2MBZe5LyOA@mail.gmail.com>

We are happy to announce the release of Matplotlib 2.1. This is the second
minor release in the Matplotlib 2.x series and the first release with major
new features since 1.5.

This release contains approximately 2 years worth of work by 275
contributors
across over 950 pull requests.  Highlights from this release include:

 - support for string categorical values
 - export of animations to interactive javascript widgets
 - major overhaul of polar plots
 - reproducible output for ps/eps, pdf, and svg backends
 - performance improvements in drawing lines and images
 - GUIs show a busy cursor while rendering the plot

along with many other enhancements and bug fixes.

The gallery, examples, and tutorials have been overhauled and consolidated:

  Examples: http://matplotlib.org/gallery/index.html
  Tutorials: http://matplotlib.org/tutorials/index.html

A big thank you to everyone who contributed to this release!

Wheels are available on pypi for win/mac/manylinux and for conda via
conda-forge.

Full whats new:
http://matplotlib.org/users/whats_new.html#new-in-matplotlib-2-1
Full API changes:
http://matplotlib.org/api/api_changes.html#api-changes-in-2-1-0
github stats: http://matplotlib.org/users/github_stats.html

Tom
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From maurobio at gmail.com  Sun Oct  8 14:33:57 2017
From: maurobio at gmail.com (Mauro Cavalcanti)
Date: Sun, 8 Oct 2017 15:33:57 -0300
Subject: [Matplotlib-users] Problem using imshow with Matplotlib/Basemap
Message-ID: <CAC1JhZZeVg76gwauk_D6RWFNau=WF5c+SexqqgDGX4g-PeXi-Q@mail.gmail.com>

Dear ALL,

I have a simple dataset of longitudes/latitudes (see the attached csv
file).

>From such data, I want to generate a grid like this:

0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0 0 0 0
0 0 0 0 0 0 1 1 2 0 0 0 0
0 0 0 0 0 1 1 1 1 0 0 0 0
0 0 0 1 0 1 0 0 0 0 0 0 0
0 0 0 2 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 1 3 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0

which gives the number of data records in each cell of the grid, using one
of the variables in the dataset ("spp") as a categorical (grouping) factor.

>From this grid, I then want to create a heat map, superimposed on a
Matplotlib/Basemap.

I wrote some code which does what I want (see the attachments).

It (mostly) works, but te problem is that the grid image is not being
displayed correctly: as shown in the attached figure, it appears too small,
and in the lower left corner of the map, instead of where it should be (the
West coast of Africa, along the Gulf of Guinea).

Thanks in advance for any assistance you can provide.

Best regards,

-- 
Dr. Mauro J. Cavalcanti
E-mail: maurobio at gmail.com
Web: http://sites.google.com/site/maurobio
"Life is complex. It consists of real and imaginary parts."
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import csv
import numpy as np
from mpl_toolkits.basemap import Basemap
from matplotlib import cm as cmap
import matplotlib.pyplot as plt

import warnings
warnings.filterwarnings("ignore")

#read input data
with open('testdata.csv', 'r') as csvfile:
    reader = csv.reader(csvfile, delimiter=',')
    headers = reader.next()
    input_data = list(reader)

#grid dimensions with one-degree resolution
lat_inf, lon_inf = 0, 0
lat_sup, lon_sup = 90, 360
resolution = 1

latitude, longitude = [], []
latitude = range(lat_inf, lat_sup, resolution)
longitude = range(lon_inf, lon_sup, resolution)

#create output grid
output_grid = []
for i in latitude:
   output_grid.append([])
   for j in longitude:
       output_grid[i].append(0)

#traverse the input_data evaluating the lat, lon coordinates
#summing +1 in the output_grid[latitude][longitude].
for row in input_data:
   lat = int(row[2]) 
   lon = int(row[3]) 
   #sp = row[1]

   #check its indexes
   i_lat = latitude.index(lat)
   i_lon = longitude.index(lon)

   #increase counter
   output_grid[i_lat][i_lon] += 1

output_grid = np.array(output_grid, np.int16)

#create map
m = Basemap()
m.drawcoastlines(linewidth=0.25)

#display image
im = m.imshow(output_grid.transpose(), cmap='summer', origin='lower', aspect='auto', interpolation='none')
m.colorbar(im)
plt.show()
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From ben.v.root at gmail.com  Mon Oct  9 10:29:31 2017
From: ben.v.root at gmail.com (Benjamin Root)
Date: Mon, 9 Oct 2017 10:29:31 -0400
Subject: [Matplotlib-users] Problem using imshow with Matplotlib/Basemap
In-Reply-To: <CAC1JhZZeVg76gwauk_D6RWFNau=WF5c+SexqqgDGX4g-PeXi-Q@mail.gmail.com>
References: <CAC1JhZZeVg76gwauk_D6RWFNau=WF5c+SexqqgDGX4g-PeXi-Q@mail.gmail.com>
Message-ID: <CANNq6FnS3Oze=E82axb8bwAhPdSM=5MR_EpJcW-RNpzdXmAEdg@mail.gmail.com>

First, you shouldn't need to transpose your image... that'll effectively
rotate the data by 90 degrees. Second, you didn't specify the extents of
your image, so Basemap is putting everything starting at coordinate 0,0 in
the default projection.

If you specify the extent keyword argument to imshow as the (lon1, lat1,
lon2, lat2) tuple for the lower-left and upper right corners, you won't
even need the origin='lower', and you definitely won't need the transpose.

Cheers!
Ben Root


On Sun, Oct 8, 2017 at 2:33 PM, Mauro Cavalcanti <maurobio at gmail.com> wrote:

> Dear ALL,
>
> I have a simple dataset of longitudes/latitudes (see the attached csv
> file).
>
> From such data, I want to generate a grid like this:
>
> 0 0 0 0 0 0 0 0 0 0 0 0 0
> 0 0 0 0 0 0 0 0 0 0 0 0 0
> 0 0 0 0 0 0 0 0 0 0 0 0 0
> 0 0 0 0 0 0 1 0 0 0 0 0 0
> 0 0 0 0 0 0 1 1 2 0 0 0 0
> 0 0 0 0 0 1 1 1 1 0 0 0 0
> 0 0 0 1 0 1 0 0 0 0 0 0 0
> 0 0 0 2 0 0 0 0 0 0 0 0 0
> 0 0 0 0 0 0 0 0 0 1 3 0 0
> 0 0 0 0 0 0 0 0 0 0 0 0 0
> 0 0 0 0 0 0 0 0 0 1 0 0 0
> 0 0 0 0 0 0 0 0 0 1 0 0 0
> 0 0 0 0 0 0 0 0 0 0 0 0 0
> 0 0 0 0 0 0 0 0 0 0 1 0 0
> 0 0 0 0 0 0 0 0 0 0 1 0 0
> 0 0 0 0 0 0 0 0 0 0 0 0 0
> 0 0 0 0 0 0 0 0 0 0 0 0 0
>
> which gives the number of data records in each cell of the grid, using one
> of the variables in the dataset ("spp") as a categorical (grouping) factor.
>
> From this grid, I then want to create a heat map, superimposed on a
> Matplotlib/Basemap.
>
> I wrote some code which does what I want (see the attachments).
>
> It (mostly) works, but te problem is that the grid image is not being
> displayed correctly: as shown in the attached figure, it appears too small,
> and in the lower left corner of the map, instead of where it should be (the
> West coast of Africa, along the Gulf of Guinea).
>
> Thanks in advance for any assistance you can provide.
>
> Best regards,
>
> --
> Dr. Mauro J. Cavalcanti
> E-mail: maurobio at gmail.com
> Web: http://sites.google.com/site/maurobio
> "Life is complex. It consists of real and imaginary parts."
>
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org
> https://mail.python.org/mailman/listinfo/matplotlib-users
>
>
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From strozzi2 at llnl.gov  Mon Oct  9 12:23:16 2017
From: strozzi2 at llnl.gov (Strozzi, David J.)
Date: Mon, 9 Oct 2017 16:23:16 +0000
Subject: [Matplotlib-users] Windows7: figure window "not responding"
In-Reply-To: <BL2PR09MB1073E3D3287D9CE8AA654113E99E0@BL2PR09MB1073.namprd09.prod.outlook.com>
References: <SN1PR09MB1088D46F9EB6671B67E61753E9A30@SN1PR09MB1088.namprd09.prod.outlook.com>
 <CAA48SF-GQi7t0cqUr2ck3TCO7M+xTEvf9-9w17=bDK_NpfxRcw@mail.gmail.com>
 <SN1PR09MB1088197E28477B0005E3545FE9810@SN1PR09MB1088.namprd09.prod.outlook.com>
 <CAA48SF_F=Nq8stppqS8VO5jhOESWaD0q1YPYMznpiyvH-J6AwQ@mail.gmail.com>
 <SN1PR09MB10886B644F25551A6E3070F5E9810@SN1PR09MB1088.namprd09.prod.outlook.com>
 <CAA48SF-8pdjL7XD--C2+uNc7ANdmsoVV6kA6zK7F_W=55_w6Jw@mail.gmail.com>
 <BL2PR09MB1073E3D3287D9CE8AA654113E99E0@BL2PR09MB1073.namprd09.prod.outlook.com>
Message-ID: <SN1PR09MB108875F7B6F571FB6EB220B3E9740@SN1PR09MB1088.namprd09.prod.outlook.com>

Hope this appears in the thread.  I still haven?t solved this issue, trying to bump this to the top of someone?s stack.  No one responded to this message.

Any help appreciated.  Thanks.

Dave

From: Matplotlib-users [mailto:matplotlib-users-bounces+strozzi2=llnl.gov at python.org] On Behalf Of Strozzi, David J.
Sent: Monday, August 28, 2017 1:48 PM
To: Thomas Caswell <tcaswell at gmail.com>; matplotlib-users at python.org
Subject: Re: [Matplotlib-users] Windows7: figure window "not responding"

Sorry for the delay.  Life, work, etc. intervene.  I tried tk backend, failed but in a different way.  Seems to get ?farther? in that I can get a plot to render.

* Versions (all very recent from Anaconda): Python: 3.6.2  ipython: 6.1.0  matplotlib: 2.0.2

* My emacs startup has (elpy-use-ipython).

* With backend ?TkAgg?:

plt.ion()
figure(1)  --> window appears, with buttons on bottom (doesn?t happen with ?Qt5Agg? backend.  ipyton prompt works.  But, if I try interacting with the figure window, I get ?not responding? in title, and windows? spinning shell.  If I close the figure window by clicking on the X in upper right, windows says ?app not responding.?  Dialog box w/ End Process or Cancel.  End process kills figure and ipython session.

plt.ioff() instead:
figure(1)
plot(1,1)
show()
I get a figure window with a plot, but if I interact with it I get ?not responding.?  ipython prompt seems fine.  I can close window from prompt close(1) and ipython fine.

Help appreciated!  I hopefully won?t wait so long to respond.

Thanks,
Dave


From: Thomas Caswell [mailto:tcaswell at gmail.com]
Sent: Saturday, August 19, 2017 2:32 PM
To: Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>>; matplotlib-users at python.org<mailto:matplotlib-users at python.org>
Subject: Re: [Matplotlib-users] Windows7: figure window "not responding"

That is at least all consistent!  The issue as that for what ever reason the input hook is not being set up correctly (which is what lets the GUI event loop spin and process events while the terminal waits for you to type).  With `plt.ion()` the terminal is blocking the process waiting for you to type (which is why the window is not responding), with `plt.ioff()` when you do `plt.show()` the GUI event loop is blocking the process (which is why it works, but you don't get the next prompt back until you close the plot window).  Again, see https://github.com/matplotlib/matplotlib/pull/4779 for a longer pass at explaining how this all works (and I would appreciate feed back about where it does not make sense!).

Try using `tkagg` as the backend, the integration between python and tk in a bit tighter (because it is done in the cpython source).

Which version of IPython and matplotlib are you using (and are you sure that elpy is using the versions you think it is?)?

are you using `(elpy-use-ipython)` or just hacking the python executable path?

You are siting at a very interesting intersection of technology here (emacs + python + GUI + windows).  It might make sense to ask either on the elpy mailing list or the ipython mailing list.

Tom

On Sat, Aug 19, 2017 at 3:28 PM Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>> wrote:
Hi,

* Just subscribed to list.

* Backend: mpl.get_backend() = ?Qt5Agg?.  I am not committed to this if another one will work.

* ?If you get up a window:? I start ipython within emacs (the GNU win32 build, not cygwin), and do figure(1), a figure window appears, but nothing renders, and window title bar says ?Not responding.?  If I plot something, it doesn?t appear.

fig is not defined, by default.  I can do fig=gcf().  Then fig.canvas.flush_events() has an interesting effect ? the plot appears, but then window title still says ?not responding? and when I hover the mouse over the figure window it becomes Windows? spinning blue circle (the busy symbol, like Mac?s spinning beachball).

?control? is returned to ipython.  I can continue to execute commands there.

* At this point, plt.ioff() doesn?t seem to do anything.  plt.show() does ? ?not responding? disappears, the figure window works like normal.  Buttons on it work.  But the ipython window is ?frozen? no In[] prompt.

OK so this seems like a lot of information.  Maybe we can make progress?

Dave


From: Thomas Caswell [mailto:tcaswell at gmail.com<mailto:tcaswell at gmail.com>]
Sent: Saturday, August 19, 2017 10:54 AM

To: Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>>; matplotlib-users at python.org<mailto:matplotlib-users at python.org>
Subject: Re: [Matplotlib-users] Windows7: figure window "not responding"

David,

Please subscribe to the mailing list so you can post un-moderated.

Which backend are you using?

If you get a window up and then do `fig.canvas.flush_events()` will it update?

If you do `plt.ioff()` and then `plt.show()` does it work?

Tom


On Sat, Aug 19, 2017 at 12:40 PM Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>> wrote:
Thanks for the respone.  Sadly, I can?t get anything useful out of the doc.  And %matplotlib doesn?t help, same effect.

This is very frustrating.  I don?t know how to debug this.  It?s just a ?black box?, I do what I?m ?supposed to?, doesn?t work.

Help appreciated!
Dave

From: Thomas Caswell [mailto:tcaswell at gmail.com<mailto:tcaswell at gmail.com>]
Sent: Saturday, August 05, 2017 10:36 AM
To: Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>>; matplotlib-users at python.org<mailto:matplotlib-users at python.org>
Subject: Re: [Matplotlib-users] Windows7: figure window "not responding"

I do exactly this as well (but on linux)!

The symptoms your are describing sounds like the GUI event loop is not being spun behind the scenes (see https://github.com/matplotlib/matplotlib/pull/4779 for some draft text at explaining how all of that works).  Have you tried doing `%matplotlib` is IPython?

Tom

On Sat, Aug 5, 2017 at 12:39 PM Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>> wrote:

This problem has been driving me nuts, posted it elsewhere, so far no one's been able to help.  I am running Python 3.6.1 from Anaconda (all packages updated to latest and greatest) on Windows 7.  ipython and mpl figure windows work fine from a console, or spyder.  The problem is when I try running ipython within emacs using elpy.



To back up, the workflow I'm shooting for is using emacs as my IDE, and running python within emacs.  I've done this for many years with the Yorick interpreter, and it's quite addictive.  I get the same problem whether I use the GNU windows build of emacs, or emacs -nw (terminal, no X) from cygwin (ver 25.2 both cases).  elpy seems to be a good Python package for emacs, and nominally supports using ipython.



OK, so - I can use ipython + emacs + elpy.  The only problem is matplotlib figures.  When I do figure(), a figure window appears, but is not fully rendered (e.g. no buttons), and the window title says "Not responding".  If I do a plot(), nothing is plotted.  Playing with ion() and show() doesn't help.



I've posted this on the elpy github, various stackexchange forums, no one has had any ideas yet.



Any help is greatly appreciated!

Dave
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From thibault.marin.us at ieee.org  Mon Oct  9 12:53:56 2017
From: thibault.marin.us at ieee.org (Thibault Marin)
Date: Mon, 09 Oct 2017 11:53:56 -0500
Subject: [Matplotlib-users] Windows7: figure window "not responding"
In-Reply-To: <SN1PR09MB108875F7B6F571FB6EB220B3E9740@SN1PR09MB1088.namprd09.prod.outlook.com>
References: <SN1PR09MB1088D46F9EB6671B67E61753E9A30@SN1PR09MB1088.namprd09.prod.outlook.com>
 <CAA48SF-GQi7t0cqUr2ck3TCO7M+xTEvf9-9w17=bDK_NpfxRcw@mail.gmail.com>
 <SN1PR09MB1088197E28477B0005E3545FE9810@SN1PR09MB1088.namprd09.prod.outlook.com>
 <CAA48SF_F=Nq8stppqS8VO5jhOESWaD0q1YPYMznpiyvH-J6AwQ@mail.gmail.com>
 <SN1PR09MB10886B644F25551A6E3070F5E9810@SN1PR09MB1088.namprd09.prod.outlook.com>
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 <BL2PR09MB1073E3D3287D9CE8AA654113E99E0@BL2PR09MB1073.namprd09.prod.outlook.com>
 <SN1PR09MB108875F7B6F571FB6EB220B3E9740@SN1PR09MB1088.namprd09.prod.outlook.com>
Message-ID: <87tvz8b6u3.fsf@dell-desktop.WORKGROUP>

Hi, I thought I replied to that, but I probably failed for some
reason.  Here is my (slightly edited) reply, I hope it helps.

----

Hi,

Sorry for breaking the thread, I have just subscribed to the list and I
don't know how to respond to a message that predates my subscription.  I
hope this messages makes its way.

I believe I have a similar setup (Windows + emacs + Qt5Agg, python 3.5
and matplotlib 2+).  I also see freezes when starting ipython from
within emacs.

The way I work around that is by using org-mode and ob-ipython
(https://github.com/gregsexton/ob-ipython).

To work with matplotlib, I do the following:

1 - Open my org file, which starts with something like:
,----
| #+PROPERTY: header-args :eval no-export :session Python
|
| #+NAME: py-startup-0
| #+BEGIN_SRC ipython :results silent :exports none
| # %% Standard imports
| # General
| import os
| import sys
| import re
| import matplotlib as mpl
| mpl.use('Qt5Agg')
| #+END_SRC
`----

2 - Execute the source block (move point at the beginning of the
"BEGIN_SRC line" and C-c C-c)

I then get a warning about the interpreter which I ignore.  Note that it
sometimes hangs at that point, in which case I use C-g.

3 - The *Python* buffer then contains the ipython session, in which I
can use matplotlib: I get not freezes and calls do not block the ipython
instance.  Note that I have to run `plt.ion()` once to get matplotlib
figures to show up.  Important note: I often get an error on the first
command I type in the *Python* shell, so I typically go to the *Python*
buffer, press enter and ignore the error I may get (I am guessing some
line return is missing in some previous command sent during the setup).

4 - I usually write most of my code in org-mode anyway, but after
running the source block I can open any .py file and work with it ignoring
the org-mode part.

For completeness, my emacs init file contains the following:
,----
| (when (eq system-type 'windows-nt)
|       (progn
|         (setq python-shell-interpreter-args "--simple-prompt -i")
|         (setq ob-ipython-command "ipython")))
`----

I hope this helps.

thibault

Strozzi, David J. writes:

> Hope this appears in the thread.  I still haven?t solved this issue, trying to bump this to the top of someone?s stack.  No one responded to this message.
>
> Any help appreciated.  Thanks.
>
> Dave
>
> From: Matplotlib-users [mailto:matplotlib-users-bounces+strozzi2=llnl.gov at python.org] On Behalf Of Strozzi, David J.
> Sent: Monday, August 28, 2017 1:48 PM
> To: Thomas Caswell <tcaswell at gmail.com>; matplotlib-users at python.org
> Subject: Re: [Matplotlib-users] Windows7: figure window "not responding"
>
> Sorry for the delay.  Life, work, etc. intervene.  I tried tk backend, failed but in a different way.  Seems to get ?farther? in that I can get a plot to render.
>
> * Versions (all very recent from Anaconda): Python: 3.6.2  ipython: 6.1.0  matplotlib: 2.0.2
>
> * My emacs startup has (elpy-use-ipython).
>
> * With backend ?TkAgg?:
>
> plt.ion()
> figure(1)  --> window appears, with buttons on bottom (doesn?t happen with ?Qt5Agg? backend.  ipyton prompt works.  But, if I try interacting with the figure window, I get ?not responding? in title, and windows? spinning shell.  If I close the figure window by clicking on the X in upper right, windows says ?app not responding.?  Dialog box w/ End Process or Cancel.  End process kills figure and ipython session.
>
> plt.ioff() instead:
> figure(1)
> plot(1,1)
> show()
> I get a figure window with a plot, but if I interact with it I get ?not responding.?  ipython prompt seems fine.  I can close window from prompt close(1) and ipython fine.
>
> Help appreciated!  I hopefully won?t wait so long to respond.
>
> Thanks,
> Dave
>
>
> From: Thomas Caswell [mailto:tcaswell at gmail.com]
> Sent: Saturday, August 19, 2017 2:32 PM
> To: Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>>; matplotlib-users at python.org<mailto:matplotlib-users at python.org>
> Subject: Re: [Matplotlib-users] Windows7: figure window "not responding"
>
> That is at least all consistent!  The issue as that for what ever reason the input hook is not being set up correctly (which is what lets the GUI event loop spin and process events while the terminal waits for you to type).  With `plt.ion()` the terminal is blocking the process waiting for you to type (which is why the window is not responding), with `plt.ioff()` when you do `plt.show()` the GUI event loop is blocking the process (which is why it works, but you don't get the next prompt back until you close the plot window).  Again, see https://github.com/matplotlib/matplotlib/pull/4779 for a longer pass at explaining how this all works (and I would appreciate feed back about where it does not make sense!).
>
> Try using `tkagg` as the backend, the integration between python and tk in a bit tighter (because it is done in the cpython source).
>
> Which version of IPython and matplotlib are you using (and are you sure that elpy is using the versions you think it is?)?
>
> are you using `(elpy-use-ipython)` or just hacking the python executable path?
>
> You are siting at a very interesting intersection of technology here (emacs + python + GUI + windows).  It might make sense to ask either on the elpy mailing list or the ipython mailing list.
>
> Tom
>
> On Sat, Aug 19, 2017 at 3:28 PM Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>> wrote:
> Hi,
>
> * Just subscribed to list.
>
> * Backend: mpl.get_backend() = ?Qt5Agg?.  I am not committed to this if another one will work.
>
> * ?If you get up a window:? I start ipython within emacs (the GNU win32 build, not cygwin), and do figure(1), a figure window appears, but nothing renders, and window title bar says ?Not responding.?  If I plot something, it doesn?t appear.
>
> fig is not defined, by default.  I can do fig=gcf().  Then fig.canvas.flush_events() has an interesting effect ? the plot appears, but then window title still says ?not responding? and when I hover the mouse over the figure window it becomes Windows? spinning blue circle (the busy symbol, like Mac?s spinning beachball).
>
> ?control? is returned to ipython.  I can continue to execute commands there.
>
> * At this point, plt.ioff() doesn?t seem to do anything.  plt.show() does ? ?not responding? disappears, the figure window works like normal.  Buttons on it work.  But the ipython window is ?frozen? no In[] prompt.
>
> OK so this seems like a lot of information.  Maybe we can make progress?
>
> Dave
>
>
> From: Thomas Caswell [mailto:tcaswell at gmail.com<mailto:tcaswell at gmail.com>]
> Sent: Saturday, August 19, 2017 10:54 AM
>
> To: Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>>; matplotlib-users at python.org<mailto:matplotlib-users at python.org>
> Subject: Re: [Matplotlib-users] Windows7: figure window "not responding"
>
> David,
>
> Please subscribe to the mailing list so you can post un-moderated.
>
> Which backend are you using?
>
> If you get a window up and then do `fig.canvas.flush_events()` will it update?
>
> If you do `plt.ioff()` and then `plt.show()` does it work?
>
> Tom
>
>
> On Sat, Aug 19, 2017 at 12:40 PM Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>> wrote:
> Thanks for the respone.  Sadly, I can?t get anything useful out of the doc.  And %matplotlib doesn?t help, same effect.
>
> This is very frustrating.  I don?t know how to debug this.  It?s just a ?black box?, I do what I?m ?supposed to?, doesn?t work.
>
> Help appreciated!
> Dave
>
> From: Thomas Caswell [mailto:tcaswell at gmail.com<mailto:tcaswell at gmail.com>]
> Sent: Saturday, August 05, 2017 10:36 AM
> To: Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>>; matplotlib-users at python.org<mailto:matplotlib-users at python.org>
> Subject: Re: [Matplotlib-users] Windows7: figure window "not responding"
>
> I do exactly this as well (but on linux)!
>
> The symptoms your are describing sounds like the GUI event loop is not being spun behind the scenes (see https://github.com/matplotlib/matplotlib/pull/4779 for some draft text at explaining how all of that works).  Have you tried doing `%matplotlib` is IPython?
>
> Tom
>
> On Sat, Aug 5, 2017 at 12:39 PM Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>> wrote:
>
> This problem has been driving me nuts, posted it elsewhere, so far no one's been able to help.  I am running Python 3.6.1 from Anaconda (all packages updated to latest and greatest) on Windows 7.  ipython and mpl figure windows work fine from a console, or spyder.  The problem is when I try running ipython within emacs using elpy.
>
>
>
> To back up, the workflow I'm shooting for is using emacs as my IDE, and running python within emacs.  I've done this for many years with the Yorick interpreter, and it's quite addictive.  I get the same problem whether I use the GNU windows build of emacs, or emacs -nw (terminal, no X) from cygwin (ver 25.2 both cases).  elpy seems to be a good Python package for emacs, and nominally supports using ipython.
>
>
>
> OK, so - I can use ipython + emacs + elpy.  The only problem is matplotlib figures.  When I do figure(), a figure window appears, but is not fully rendered (e.g. no buttons), and the window title says "Not responding".  If I do a plot(), nothing is plotted.  Playing with ion() and show() doesn't help.
>
>
>
> I've posted this on the elpy github, various stackexchange forums, no one has had any ideas yet.
>
>
>
> Any help is greatly appreciated!
>
> Dave
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org<mailto:Matplotlib-users at python.org>
> https://mail.python.org/mailman/listinfo/matplotlib-users
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org
> https://mail.python.org/mailman/listinfo/matplotlib-users

From strozzi2 at llnl.gov  Mon Oct  9 13:07:40 2017
From: strozzi2 at llnl.gov (Strozzi, David J.)
Date: Mon, 9 Oct 2017 17:07:40 +0000
Subject: [Matplotlib-users] Windows7: figure window "not responding"
In-Reply-To: <87tvz8b6u3.fsf@dell-desktop.WORKGROUP>
References: <SN1PR09MB1088D46F9EB6671B67E61753E9A30@SN1PR09MB1088.namprd09.prod.outlook.com>
 <CAA48SF-GQi7t0cqUr2ck3TCO7M+xTEvf9-9w17=bDK_NpfxRcw@mail.gmail.com>
 <SN1PR09MB1088197E28477B0005E3545FE9810@SN1PR09MB1088.namprd09.prod.outlook.com>
 <CAA48SF_F=Nq8stppqS8VO5jhOESWaD0q1YPYMznpiyvH-J6AwQ@mail.gmail.com>
 <SN1PR09MB10886B644F25551A6E3070F5E9810@SN1PR09MB1088.namprd09.prod.outlook.com>
 <CAA48SF-8pdjL7XD--C2+uNc7ANdmsoVV6kA6zK7F_W=55_w6Jw@mail.gmail.com>
 <BL2PR09MB1073E3D3287D9CE8AA654113E99E0@BL2PR09MB1073.namprd09.prod.outlook.com>
 <SN1PR09MB108875F7B6F571FB6EB220B3E9740@SN1PR09MB1088.namprd09.prod.outlook.com>
 <87tvz8b6u3.fsf@dell-desktop.WORKGROUP>
Message-ID: <SN1PR09MB10887B5DF71D375B5DB9452EE9740@SN1PR09MB1088.namprd09.prod.outlook.com>

Hi,

Thanks, I did get this response, but consider it too 'convoluted' or 'non-standard'.  I think this workflow of running ipython inside emacs should "just work" with some set of standard tools, like elpy + ipython + matplotlib, and not need one to "roll their own".  Of course Windows may be a pretty non-standard misfit but such is life!

Dave

-----Original Message-----
From: Thibault Marin [mailto:thibault.marin.us at ieee.org] 
Sent: Monday, October 09, 2017 9:54 AM
To: Strozzi, David J. <strozzi2 at llnl.gov>
Cc: matplotlib-users at python.org
Subject: Re: [Matplotlib-users] Windows7: figure window "not responding"

Hi, I thought I replied to that, but I probably failed for some reason.  Here is my (slightly edited) reply, I hope it helps.

----

Hi,

Sorry for breaking the thread, I have just subscribed to the list and I don't know how to respond to a message that predates my subscription.  I hope this messages makes its way.

I believe I have a similar setup (Windows + emacs + Qt5Agg, python 3.5 and matplotlib 2+).  I also see freezes when starting ipython from within emacs.

The way I work around that is by using org-mode and ob-ipython (https://github.com/gregsexton/ob-ipython).

To work with matplotlib, I do the following:

1 - Open my org file, which starts with something like:
,----
| #+PROPERTY: header-args :eval no-export :session Python
|
| #+NAME: py-startup-0
| #+BEGIN_SRC ipython :results silent :exports none # %% Standard 
| imports # General import os import sys import re import matplotlib as 
| mpl
| mpl.use('Qt5Agg')
| #+END_SRC
`----

2 - Execute the source block (move point at the beginning of the "BEGIN_SRC line" and C-c C-c)

I then get a warning about the interpreter which I ignore.  Note that it sometimes hangs at that point, in which case I use C-g.

3 - The *Python* buffer then contains the ipython session, in which I can use matplotlib: I get not freezes and calls do not block the ipython instance.  Note that I have to run `plt.ion()` once to get matplotlib figures to show up.  Important note: I often get an error on the first command I type in the *Python* shell, so I typically go to the *Python* buffer, press enter and ignore the error I may get (I am guessing some line return is missing in some previous command sent during the setup).

4 - I usually write most of my code in org-mode anyway, but after running the source block I can open any .py file and work with it ignoring the org-mode part.

For completeness, my emacs init file contains the following:
,----
| (when (eq system-type 'windows-nt)
|       (progn
|         (setq python-shell-interpreter-args "--simple-prompt -i")
|         (setq ob-ipython-command "ipython")))
`----

I hope this helps.

thibault

Strozzi, David J. writes:

> Hope this appears in the thread.  I still haven?t solved this issue, trying to bump this to the top of someone?s stack.  No one responded to this message.
>
> Any help appreciated.  Thanks.
>
> Dave
>
> From: Matplotlib-users [mailto:matplotlib-users-bounces+strozzi2=llnl.gov at python.org] On Behalf Of Strozzi, David J.
> Sent: Monday, August 28, 2017 1:48 PM
> To: Thomas Caswell <tcaswell at gmail.com>; matplotlib-users at python.org
> Subject: Re: [Matplotlib-users] Windows7: figure window "not responding"
>
> Sorry for the delay.  Life, work, etc. intervene.  I tried tk backend, failed but in a different way.  Seems to get ?farther? in that I can get a plot to render.
>
> * Versions (all very recent from Anaconda): Python: 3.6.2  ipython: 
> 6.1.0  matplotlib: 2.0.2
>
> * My emacs startup has (elpy-use-ipython).
>
> * With backend ?TkAgg?:
>
> plt.ion()
> figure(1)  --> window appears, with buttons on bottom (doesn?t happen with ?Qt5Agg? backend.  ipyton prompt works.  But, if I try interacting with the figure window, I get ?not responding? in title, and windows? spinning shell.  If I close the figure window by clicking on the X in upper right, windows says ?app not responding.?  Dialog box w/ End Process or Cancel.  End process kills figure and ipython session.
>
> plt.ioff() instead:
> figure(1)
> plot(1,1)
> show()
> I get a figure window with a plot, but if I interact with it I get ?not responding.?  ipython prompt seems fine.  I can close window from prompt close(1) and ipython fine.
>
> Help appreciated!  I hopefully won?t wait so long to respond.
>
> Thanks,
> Dave
>
>
> From: Thomas Caswell [mailto:tcaswell at gmail.com]
> Sent: Saturday, August 19, 2017 2:32 PM
> To: Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>>; 
> matplotlib-users at python.org<mailto:matplotlib-users at python.org>
> Subject: Re: [Matplotlib-users] Windows7: figure window "not responding"
>
> That is at least all consistent!  The issue as that for what ever reason the input hook is not being set up correctly (which is what lets the GUI event loop spin and process events while the terminal waits for you to type).  With `plt.ion()` the terminal is blocking the process waiting for you to type (which is why the window is not responding), with `plt.ioff()` when you do `plt.show()` the GUI event loop is blocking the process (which is why it works, but you don't get the next prompt back until you close the plot window).  Again, see https://github.com/matplotlib/matplotlib/pull/4779 for a longer pass at explaining how this all works (and I would appreciate feed back about where it does not make sense!).
>
> Try using `tkagg` as the backend, the integration between python and tk in a bit tighter (because it is done in the cpython source).
>
> Which version of IPython and matplotlib are you using (and are you sure that elpy is using the versions you think it is?)?
>
> are you using `(elpy-use-ipython)` or just hacking the python executable path?
>
> You are siting at a very interesting intersection of technology here (emacs + python + GUI + windows).  It might make sense to ask either on the elpy mailing list or the ipython mailing list.
>
> Tom
>
> On Sat, Aug 19, 2017 at 3:28 PM Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>> wrote:
> Hi,
>
> * Just subscribed to list.
>
> * Backend: mpl.get_backend() = ?Qt5Agg?.  I am not committed to this if another one will work.
>
> * ?If you get up a window:? I start ipython within emacs (the GNU win32 build, not cygwin), and do figure(1), a figure window appears, but nothing renders, and window title bar says ?Not responding.?  If I plot something, it doesn?t appear.
>
> fig is not defined, by default.  I can do fig=gcf().  Then fig.canvas.flush_events() has an interesting effect ? the plot appears, but then window title still says ?not responding? and when I hover the mouse over the figure window it becomes Windows? spinning blue circle (the busy symbol, like Mac?s spinning beachball).
>
> ?control? is returned to ipython.  I can continue to execute commands there.
>
> * At this point, plt.ioff() doesn?t seem to do anything.  plt.show() does ? ?not responding? disappears, the figure window works like normal.  Buttons on it work.  But the ipython window is ?frozen? no In[] prompt.
>
> OK so this seems like a lot of information.  Maybe we can make progress?
>
> Dave
>
>
> From: Thomas Caswell 
> [mailto:tcaswell at gmail.com<mailto:tcaswell at gmail.com>]
> Sent: Saturday, August 19, 2017 10:54 AM
>
> To: Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>>; 
> matplotlib-users at python.org<mailto:matplotlib-users at python.org>
> Subject: Re: [Matplotlib-users] Windows7: figure window "not responding"
>
> David,
>
> Please subscribe to the mailing list so you can post un-moderated.
>
> Which backend are you using?
>
> If you get a window up and then do `fig.canvas.flush_events()` will it update?
>
> If you do `plt.ioff()` and then `plt.show()` does it work?
>
> Tom
>
>
> On Sat, Aug 19, 2017 at 12:40 PM Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>> wrote:
> Thanks for the respone.  Sadly, I can?t get anything useful out of the doc.  And %matplotlib doesn?t help, same effect.
>
> This is very frustrating.  I don?t know how to debug this.  It?s just a ?black box?, I do what I?m ?supposed to?, doesn?t work.
>
> Help appreciated!
> Dave
>
> From: Thomas Caswell 
> [mailto:tcaswell at gmail.com<mailto:tcaswell at gmail.com>]
> Sent: Saturday, August 05, 2017 10:36 AM
> To: Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>>; 
> matplotlib-users at python.org<mailto:matplotlib-users at python.org>
> Subject: Re: [Matplotlib-users] Windows7: figure window "not responding"
>
> I do exactly this as well (but on linux)!
>
> The symptoms your are describing sounds like the GUI event loop is not being spun behind the scenes (see https://github.com/matplotlib/matplotlib/pull/4779 for some draft text at explaining how all of that works).  Have you tried doing `%matplotlib` is IPython?
>
> Tom
>
> On Sat, Aug 5, 2017 at 12:39 PM Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>> wrote:
>
> This problem has been driving me nuts, posted it elsewhere, so far no one's been able to help.  I am running Python 3.6.1 from Anaconda (all packages updated to latest and greatest) on Windows 7.  ipython and mpl figure windows work fine from a console, or spyder.  The problem is when I try running ipython within emacs using elpy.
>
>
>
> To back up, the workflow I'm shooting for is using emacs as my IDE, and running python within emacs.  I've done this for many years with the Yorick interpreter, and it's quite addictive.  I get the same problem whether I use the GNU windows build of emacs, or emacs -nw (terminal, no X) from cygwin (ver 25.2 both cases).  elpy seems to be a good Python package for emacs, and nominally supports using ipython.
>
>
>
> OK, so - I can use ipython + emacs + elpy.  The only problem is matplotlib figures.  When I do figure(), a figure window appears, but is not fully rendered (e.g. no buttons), and the window title says "Not responding".  If I do a plot(), nothing is plotted.  Playing with ion() and show() doesn't help.
>
>
>
> I've posted this on the elpy github, various stackexchange forums, no one has had any ideas yet.
>
>
>
> Any help is greatly appreciated!
>
> Dave
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org<mailto:Matplotlib-users at python.org>
> https://mail.python.org/mailman/listinfo/matplotlib-users
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org
> https://mail.python.org/mailman/listinfo/matplotlib-users

From ben.v.root at gmail.com  Mon Oct  9 13:20:30 2017
From: ben.v.root at gmail.com (Benjamin Root)
Date: Mon, 9 Oct 2017 13:20:30 -0400
Subject: [Matplotlib-users] Windows7: figure window "not responding"
In-Reply-To: <SN1PR09MB10887B5DF71D375B5DB9452EE9740@SN1PR09MB1088.namprd09.prod.outlook.com>
References: <SN1PR09MB1088D46F9EB6671B67E61753E9A30@SN1PR09MB1088.namprd09.prod.outlook.com>
 <CAA48SF-GQi7t0cqUr2ck3TCO7M+xTEvf9-9w17=bDK_NpfxRcw@mail.gmail.com>
 <SN1PR09MB1088197E28477B0005E3545FE9810@SN1PR09MB1088.namprd09.prod.outlook.com>
 <CAA48SF_F=Nq8stppqS8VO5jhOESWaD0q1YPYMznpiyvH-J6AwQ@mail.gmail.com>
 <SN1PR09MB10886B644F25551A6E3070F5E9810@SN1PR09MB1088.namprd09.prod.outlook.com>
 <CAA48SF-8pdjL7XD--C2+uNc7ANdmsoVV6kA6zK7F_W=55_w6Jw@mail.gmail.com>
 <BL2PR09MB1073E3D3287D9CE8AA654113E99E0@BL2PR09MB1073.namprd09.prod.outlook.com>
 <SN1PR09MB108875F7B6F571FB6EB220B3E9740@SN1PR09MB1088.namprd09.prod.outlook.com>
 <87tvz8b6u3.fsf@dell-desktop.WORKGROUP>
 <SN1PR09MB10887B5DF71D375B5DB9452EE9740@SN1PR09MB1088.namprd09.prod.outlook.com>
Message-ID: <CANNq6F=UJqbPY0UBORb-SzZFCdmKc95HbktsTDEXmcT=m9BvVg@mail.gmail.com>

Just some thoughts... have you roped in any ipython devs into this
discussion? Can you confirm that doing similar things, but outside of
emacs, works as expected? Can you confirm that just simply using ipython
within emacs, but without matplotlib, also works as expected?

Also, now that I think of it, could you also re-confirm exactly which
versions of matplotlib, emacs, and ipython you are using, and where your
matplotlib and ipython were installed from?

Just trying to narrow down the possible sources of issues.

Cheers!
Ben Root


On Mon, Oct 9, 2017 at 1:07 PM, Strozzi, David J. <strozzi2 at llnl.gov> wrote:

> Hi,
>
> Thanks, I did get this response, but consider it too 'convoluted' or
> 'non-standard'.  I think this workflow of running ipython inside emacs
> should "just work" with some set of standard tools, like elpy + ipython +
> matplotlib, and not need one to "roll their own".  Of course Windows may be
> a pretty non-standard misfit but such is life!
>
> Dave
>
> -----Original Message-----
> From: Thibault Marin [mailto:thibault.marin.us at ieee.org]
> Sent: Monday, October 09, 2017 9:54 AM
> To: Strozzi, David J. <strozzi2 at llnl.gov>
> Cc: matplotlib-users at python.org
> Subject: Re: [Matplotlib-users] Windows7: figure window "not responding"
>
> Hi, I thought I replied to that, but I probably failed for some reason.
> Here is my (slightly edited) reply, I hope it helps.
>
> ----
>
> Hi,
>
> Sorry for breaking the thread, I have just subscribed to the list and I
> don't know how to respond to a message that predates my subscription.  I
> hope this messages makes its way.
>
> I believe I have a similar setup (Windows + emacs + Qt5Agg, python 3.5 and
> matplotlib 2+).  I also see freezes when starting ipython from within emacs.
>
> The way I work around that is by using org-mode and ob-ipython (
> https://github.com/gregsexton/ob-ipython).
>
> To work with matplotlib, I do the following:
>
> 1 - Open my org file, which starts with something like:
> ,----
> | #+PROPERTY: header-args :eval no-export :session Python
> |
> | #+NAME: py-startup-0
> | #+BEGIN_SRC ipython :results silent :exports none # %% Standard
> | imports # General import os import sys import re import matplotlib as
> | mpl
> | mpl.use('Qt5Agg')
> | #+END_SRC
> `----
>
> 2 - Execute the source block (move point at the beginning of the
> "BEGIN_SRC line" and C-c C-c)
>
> I then get a warning about the interpreter which I ignore.  Note that it
> sometimes hangs at that point, in which case I use C-g.
>
> 3 - The *Python* buffer then contains the ipython session, in which I can
> use matplotlib: I get not freezes and calls do not block the ipython
> instance.  Note that I have to run `plt.ion()` once to get matplotlib
> figures to show up.  Important note: I often get an error on the first
> command I type in the *Python* shell, so I typically go to the *Python*
> buffer, press enter and ignore the error I may get (I am guessing some line
> return is missing in some previous command sent during the setup).
>
> 4 - I usually write most of my code in org-mode anyway, but after running
> the source block I can open any .py file and work with it ignoring the
> org-mode part.
>
> For completeness, my emacs init file contains the following:
> ,----
> | (when (eq system-type 'windows-nt)
> |       (progn
> |         (setq python-shell-interpreter-args "--simple-prompt -i")
> |         (setq ob-ipython-command "ipython")))
> `----
>
> I hope this helps.
>
> thibault
>
> Strozzi, David J. writes:
>
> > Hope this appears in the thread.  I still haven?t solved this issue,
> trying to bump this to the top of someone?s stack.  No one responded to
> this message.
> >
> > Any help appreciated.  Thanks.
> >
> > Dave
> >
> > From: Matplotlib-users [mailto:matplotlib-users-bounces+strozzi2=
> llnl.gov at python.org] On Behalf Of Strozzi, David J.
> > Sent: Monday, August 28, 2017 1:48 PM
> > To: Thomas Caswell <tcaswell at gmail.com>; matplotlib-users at python.org
> > Subject: Re: [Matplotlib-users] Windows7: figure window "not responding"
> >
> > Sorry for the delay.  Life, work, etc. intervene.  I tried tk backend,
> failed but in a different way.  Seems to get ?farther? in that I can get a
> plot to render.
> >
> > * Versions (all very recent from Anaconda): Python: 3.6.2  ipython:
> > 6.1.0  matplotlib: 2.0.2
> >
> > * My emacs startup has (elpy-use-ipython).
> >
> > * With backend ?TkAgg?:
> >
> > plt.ion()
> > figure(1)  --> window appears, with buttons on bottom (doesn?t happen
> with ?Qt5Agg? backend.  ipyton prompt works.  But, if I try interacting
> with the figure window, I get ?not responding? in title, and windows?
> spinning shell.  If I close the figure window by clicking on the X in upper
> right, windows says ?app not responding.?  Dialog box w/ End Process or
> Cancel.  End process kills figure and ipython session.
> >
> > plt.ioff() instead:
> > figure(1)
> > plot(1,1)
> > show()
> > I get a figure window with a plot, but if I interact with it I get ?not
> responding.?  ipython prompt seems fine.  I can close window from prompt
> close(1) and ipython fine.
> >
> > Help appreciated!  I hopefully won?t wait so long to respond.
> >
> > Thanks,
> > Dave
> >
> >
> > From: Thomas Caswell [mailto:tcaswell at gmail.com]
> > Sent: Saturday, August 19, 2017 2:32 PM
> > To: Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>>;
> > matplotlib-users at python.org<mailto:matplotlib-users at python.org>
> > Subject: Re: [Matplotlib-users] Windows7: figure window "not responding"
> >
> > That is at least all consistent!  The issue as that for what ever reason
> the input hook is not being set up correctly (which is what lets the GUI
> event loop spin and process events while the terminal waits for you to
> type).  With `plt.ion()` the terminal is blocking the process waiting for
> you to type (which is why the window is not responding), with `plt.ioff()`
> when you do `plt.show()` the GUI event loop is blocking the process (which
> is why it works, but you don't get the next prompt back until you close the
> plot window).  Again, see https://github.com/matplotlib/
> matplotlib/pull/4779 for a longer pass at explaining how this all works
> (and I would appreciate feed back about where it does not make sense!).
> >
> > Try using `tkagg` as the backend, the integration between python and tk
> in a bit tighter (because it is done in the cpython source).
> >
> > Which version of IPython and matplotlib are you using (and are you sure
> that elpy is using the versions you think it is?)?
> >
> > are you using `(elpy-use-ipython)` or just hacking the python executable
> path?
> >
> > You are siting at a very interesting intersection of technology here
> (emacs + python + GUI + windows).  It might make sense to ask either on the
> elpy mailing list or the ipython mailing list.
> >
> > Tom
> >
> > On Sat, Aug 19, 2017 at 3:28 PM Strozzi, David J. <strozzi2 at llnl.gov
> <mailto:strozzi2 at llnl.gov>> wrote:
> > Hi,
> >
> > * Just subscribed to list.
> >
> > * Backend: mpl.get_backend() = ?Qt5Agg?.  I am not committed to this if
> another one will work.
> >
> > * ?If you get up a window:? I start ipython within emacs (the GNU win32
> build, not cygwin), and do figure(1), a figure window appears, but nothing
> renders, and window title bar says ?Not responding.?  If I plot something,
> it doesn?t appear.
> >
> > fig is not defined, by default.  I can do fig=gcf().  Then
> fig.canvas.flush_events() has an interesting effect ? the plot appears, but
> then window title still says ?not responding? and when I hover the mouse
> over the figure window it becomes Windows? spinning blue circle (the busy
> symbol, like Mac?s spinning beachball).
> >
> > ?control? is returned to ipython.  I can continue to execute commands
> there.
> >
> > * At this point, plt.ioff() doesn?t seem to do anything.  plt.show()
> does ? ?not responding? disappears, the figure window works like normal.
> Buttons on it work.  But the ipython window is ?frozen? no In[] prompt.
> >
> > OK so this seems like a lot of information.  Maybe we can make progress?
> >
> > Dave
> >
> >
> > From: Thomas Caswell
> > [mailto:tcaswell at gmail.com<mailto:tcaswell at gmail.com>]
> > Sent: Saturday, August 19, 2017 10:54 AM
> >
> > To: Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>>;
> > matplotlib-users at python.org<mailto:matplotlib-users at python.org>
> > Subject: Re: [Matplotlib-users] Windows7: figure window "not responding"
> >
> > David,
> >
> > Please subscribe to the mailing list so you can post un-moderated.
> >
> > Which backend are you using?
> >
> > If you get a window up and then do `fig.canvas.flush_events()` will it
> update?
> >
> > If you do `plt.ioff()` and then `plt.show()` does it work?
> >
> > Tom
> >
> >
> > On Sat, Aug 19, 2017 at 12:40 PM Strozzi, David J. <strozzi2 at llnl.gov
> <mailto:strozzi2 at llnl.gov>> wrote:
> > Thanks for the respone.  Sadly, I can?t get anything useful out of the
> doc.  And %matplotlib doesn?t help, same effect.
> >
> > This is very frustrating.  I don?t know how to debug this.  It?s just a
> ?black box?, I do what I?m ?supposed to?, doesn?t work.
> >
> > Help appreciated!
> > Dave
> >
> > From: Thomas Caswell
> > [mailto:tcaswell at gmail.com<mailto:tcaswell at gmail.com>]
> > Sent: Saturday, August 05, 2017 10:36 AM
> > To: Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>>;
> > matplotlib-users at python.org<mailto:matplotlib-users at python.org>
> > Subject: Re: [Matplotlib-users] Windows7: figure window "not responding"
> >
> > I do exactly this as well (but on linux)!
> >
> > The symptoms your are describing sounds like the GUI event loop is not
> being spun behind the scenes (see https://github.com/matplotlib/
> matplotlib/pull/4779 for some draft text at explaining how all of that
> works).  Have you tried doing `%matplotlib` is IPython?
> >
> > Tom
> >
> > On Sat, Aug 5, 2017 at 12:39 PM Strozzi, David J. <strozzi2 at llnl.gov
> <mailto:strozzi2 at llnl.gov>> wrote:
> >
> > This problem has been driving me nuts, posted it elsewhere, so far no
> one's been able to help.  I am running Python 3.6.1 from Anaconda (all
> packages updated to latest and greatest) on Windows 7.  ipython and mpl
> figure windows work fine from a console, or spyder.  The problem is when I
> try running ipython within emacs using elpy.
> >
> >
> >
> > To back up, the workflow I'm shooting for is using emacs as my IDE, and
> running python within emacs.  I've done this for many years with the Yorick
> interpreter, and it's quite addictive.  I get the same problem whether I
> use the GNU windows build of emacs, or emacs -nw (terminal, no X) from
> cygwin (ver 25.2 both cases).  elpy seems to be a good Python package for
> emacs, and nominally supports using ipython.
> >
> >
> >
> > OK, so - I can use ipython + emacs + elpy.  The only problem is
> matplotlib figures.  When I do figure(), a figure window appears, but is
> not fully rendered (e.g. no buttons), and the window title says "Not
> responding".  If I do a plot(), nothing is plotted.  Playing with ion() and
> show() doesn't help.
> >
> >
> >
> > I've posted this on the elpy github, various stackexchange forums, no
> one has had any ideas yet.
> >
> >
> >
> > Any help is greatly appreciated!
> >
> > Dave
> > _______________________________________________
> > Matplotlib-users mailing list
> > Matplotlib-users at python.org<mailto:Matplotlib-users at python.org>
> > https://mail.python.org/mailman/listinfo/matplotlib-users
> > _______________________________________________
> > Matplotlib-users mailing list
> > Matplotlib-users at python.org
> > https://mail.python.org/mailman/listinfo/matplotlib-users
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org
> https://mail.python.org/mailman/listinfo/matplotlib-users
>
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From strozzi2 at llnl.gov  Mon Oct  9 14:14:21 2017
From: strozzi2 at llnl.gov (Strozzi, David J.)
Date: Mon, 9 Oct 2017 18:14:21 +0000
Subject: [Matplotlib-users] Windows7: figure window "not responding"
In-Reply-To: <CANNq6F=UJqbPY0UBORb-SzZFCdmKc95HbktsTDEXmcT=m9BvVg@mail.gmail.com>
References: <SN1PR09MB1088D46F9EB6671B67E61753E9A30@SN1PR09MB1088.namprd09.prod.outlook.com>
 <CAA48SF-GQi7t0cqUr2ck3TCO7M+xTEvf9-9w17=bDK_NpfxRcw@mail.gmail.com>
 <SN1PR09MB1088197E28477B0005E3545FE9810@SN1PR09MB1088.namprd09.prod.outlook.com>
 <CAA48SF_F=Nq8stppqS8VO5jhOESWaD0q1YPYMznpiyvH-J6AwQ@mail.gmail.com>
 <SN1PR09MB10886B644F25551A6E3070F5E9810@SN1PR09MB1088.namprd09.prod.outlook.com>
 <CAA48SF-8pdjL7XD--C2+uNc7ANdmsoVV6kA6zK7F_W=55_w6Jw@mail.gmail.com>
 <BL2PR09MB1073E3D3287D9CE8AA654113E99E0@BL2PR09MB1073.namprd09.prod.outlook.com>
 <SN1PR09MB108875F7B6F571FB6EB220B3E9740@SN1PR09MB1088.namprd09.prod.outlook.com>
 <87tvz8b6u3.fsf@dell-desktop.WORKGROUP>
 <SN1PR09MB10887B5DF71D375B5DB9452EE9740@SN1PR09MB1088.namprd09.prod.outlook.com>
 <CANNq6F=UJqbPY0UBORb-SzZFCdmKc95HbktsTDEXmcT=m9BvVg@mail.gmail.com>
Message-ID: <SN1PR09MB1088DB6EC4AD7F808F1702EDE9740@SN1PR09MB1088.namprd09.prod.outlook.com>

Haven?t roped in ipython devs yet, maybe I should.

Everything works as expected outside of emacs, either an ipython prompt from a console, or running Spyder GUI.

Ipython works fine inside emacs, as long as I don?t do any matplotlib.

Emacs version:
GNU Emacs 25.2.1 (x86_64-w64-mingw32)
of 2017-04-24

ipython: 6.1.0 with python 3.6.2, Matplotlib 2.0.2, all from Conda.  I updated within the last few weeks.

I am happy to update anything, change backend, etc. if it fixes the problem.

Thanks!

From: Benjamin Root [mailto:ben.v.root at gmail.com]
Sent: Monday, October 09, 2017 10:21 AM
To: Strozzi, David J. <strozzi2 at llnl.gov>
Cc: thibault.marin.us at ieee.org; matplotlib-users at python.org
Subject: Re: [Matplotlib-users] Windows7: figure window "not responding"

Just some thoughts... have you roped in any ipython devs into this discussion? Can you confirm that doing similar things, but outside of emacs, works as expected? Can you confirm that just simply using ipython within emacs, but without matplotlib, also works as expected?
Also, now that I think of it, could you also re-confirm exactly which versions of matplotlib, emacs, and ipython you are using, and where your matplotlib and ipython were installed from?

Just trying to narrow down the possible sources of issues.
Cheers!
Ben Root

On Mon, Oct 9, 2017 at 1:07 PM, Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>> wrote:
Hi,

Thanks, I did get this response, but consider it too 'convoluted' or 'non-standard'.  I think this workflow of running ipython inside emacs should "just work" with some set of standard tools, like elpy + ipython + matplotlib, and not need one to "roll their own".  Of course Windows may be a pretty non-standard misfit but such is life!

Dave

-----Original Message-----
From: Thibault Marin [mailto:thibault.marin.us at ieee.org<mailto:thibault.marin.us at ieee.org>]
Sent: Monday, October 09, 2017 9:54 AM
To: Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>>
Cc: matplotlib-users at python.org<mailto:matplotlib-users at python.org>
Subject: Re: [Matplotlib-users] Windows7: figure window "not responding"

Hi, I thought I replied to that, but I probably failed for some reason.  Here is my (slightly edited) reply, I hope it helps.

----

Hi,

Sorry for breaking the thread, I have just subscribed to the list and I don't know how to respond to a message that predates my subscription.  I hope this messages makes its way.

I believe I have a similar setup (Windows + emacs + Qt5Agg, python 3.5 and matplotlib 2+).  I also see freezes when starting ipython from within emacs.

The way I work around that is by using org-mode and ob-ipython (https://github.com/gregsexton/ob-ipython).

To work with matplotlib, I do the following:

1 - Open my org file, which starts with something like:
,----
| #+PROPERTY: header-args :eval no-export :session Python
|
| #+NAME: py-startup-0
| #+BEGIN_SRC ipython :results silent :exports none # %% Standard
| imports # General import os import sys import re import matplotlib as
| mpl
| mpl.use('Qt5Agg')
| #+END_SRC
`----

2 - Execute the source block (move point at the beginning of the "BEGIN_SRC line" and C-c C-c)

I then get a warning about the interpreter which I ignore.  Note that it sometimes hangs at that point, in which case I use C-g.

3 - The *Python* buffer then contains the ipython session, in which I can use matplotlib: I get not freezes and calls do not block the ipython instance.  Note that I have to run `plt.ion()` once to get matplotlib figures to show up.  Important note: I often get an error on the first command I type in the *Python* shell, so I typically go to the *Python* buffer, press enter and ignore the error I may get (I am guessing some line return is missing in some previous command sent during the setup).

4 - I usually write most of my code in org-mode anyway, but after running the source block I can open any .py file and work with it ignoring the org-mode part.

For completeness, my emacs init file contains the following:
,----
| (when (eq system-type 'windows-nt)
|       (progn
|         (setq python-shell-interpreter-args "--simple-prompt -i")
|         (setq ob-ipython-command "ipython")))
`----

I hope this helps.

thibault

Strozzi, David J. writes:

> Hope this appears in the thread.  I still haven?t solved this issue, trying to bump this to the top of someone?s stack.  No one responded to this message.
>
> Any help appreciated.  Thanks.
>
> Dave
>
> From: Matplotlib-users [mailto:matplotlib-users-bounces+strozzi2<mailto:matplotlib-users-bounces%2Bstrozzi2>=llnl.gov at python.org<mailto:llnl.gov at python.org>] On Behalf Of Strozzi, David J.
> Sent: Monday, August 28, 2017 1:48 PM
> To: Thomas Caswell <tcaswell at gmail.com<mailto:tcaswell at gmail.com>>; matplotlib-users at python.org<mailto:matplotlib-users at python.org>
> Subject: Re: [Matplotlib-users] Windows7: figure window "not responding"
>
> Sorry for the delay.  Life, work, etc. intervene.  I tried tk backend, failed but in a different way.  Seems to get ?farther? in that I can get a plot to render.
>
> * Versions (all very recent from Anaconda): Python: 3.6.2  ipython:
> 6.1.0  matplotlib: 2.0.2
>
> * My emacs startup has (elpy-use-ipython).
>
> * With backend ?TkAgg?:
>
> plt.ion()
> figure(1)  --> window appears, with buttons on bottom (doesn?t happen with ?Qt5Agg? backend.  ipyton prompt works.  But, if I try interacting with the figure window, I get ?not responding? in title, and windows? spinning shell.  If I close the figure window by clicking on the X in upper right, windows says ?app not responding.?  Dialog box w/ End Process or Cancel.  End process kills figure and ipython session.
>
> plt.ioff() instead:
> figure(1)
> plot(1,1)
> show()
> I get a figure window with a plot, but if I interact with it I get ?not responding.?  ipython prompt seems fine.  I can close window from prompt close(1) and ipython fine.
>
> Help appreciated!  I hopefully won?t wait so long to respond.
>
> Thanks,
> Dave
>
>
> From: Thomas Caswell [mailto:tcaswell at gmail.com<mailto:tcaswell at gmail.com>]
> Sent: Saturday, August 19, 2017 2:32 PM
> To: Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov><mailto:strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>>>;
> matplotlib-users at python.org<mailto:matplotlib-users at python.org><mailto:matplotlib-users at python.org<mailto:matplotlib-users at python.org>>
> Subject: Re: [Matplotlib-users] Windows7: figure window "not responding"
>
> That is at least all consistent!  The issue as that for what ever reason the input hook is not being set up correctly (which is what lets the GUI event loop spin and process events while the terminal waits for you to type).  With `plt.ion()` the terminal is blocking the process waiting for you to type (which is why the window is not responding), with `plt.ioff()` when you do `plt.show()` the GUI event loop is blocking the process (which is why it works, but you don't get the next prompt back until you close the plot window).  Again, see https://github.com/matplotlib/matplotlib/pull/4779 for a longer pass at explaining how this all works (and I would appreciate feed back about where it does not make sense!).
>
> Try using `tkagg` as the backend, the integration between python and tk in a bit tighter (because it is done in the cpython source).
>
> Which version of IPython and matplotlib are you using (and are you sure that elpy is using the versions you think it is?)?
>
> are you using `(elpy-use-ipython)` or just hacking the python executable path?
>
> You are siting at a very interesting intersection of technology here (emacs + python + GUI + windows).  It might make sense to ask either on the elpy mailing list or the ipython mailing list.
>
> Tom
>
> On Sat, Aug 19, 2017 at 3:28 PM Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov><mailto:strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>>> wrote:
> Hi,
>
> * Just subscribed to list.
>
> * Backend: mpl.get_backend() = ?Qt5Agg?.  I am not committed to this if another one will work.
>
> * ?If you get up a window:? I start ipython within emacs (the GNU win32 build, not cygwin), and do figure(1), a figure window appears, but nothing renders, and window title bar says ?Not responding.?  If I plot something, it doesn?t appear.
>
> fig is not defined, by default.  I can do fig=gcf().  Then fig.canvas.flush_events() has an interesting effect ? the plot appears, but then window title still says ?not responding? and when I hover the mouse over the figure window it becomes Windows? spinning blue circle (the busy symbol, like Mac?s spinning beachball).
>
> ?control? is returned to ipython.  I can continue to execute commands there.
>
> * At this point, plt.ioff() doesn?t seem to do anything.  plt.show() does ? ?not responding? disappears, the figure window works like normal.  Buttons on it work.  But the ipython window is ?frozen? no In[] prompt.
>
> OK so this seems like a lot of information.  Maybe we can make progress?
>
> Dave
>
>
> From: Thomas Caswell
> [mailto:tcaswell at gmail.com<mailto:tcaswell at gmail.com><mailto:tcaswell at gmail.com<mailto:tcaswell at gmail.com>>]
> Sent: Saturday, August 19, 2017 10:54 AM
>
> To: Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov><mailto:strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>>>;
> matplotlib-users at python.org<mailto:matplotlib-users at python.org><mailto:matplotlib-users at python.org<mailto:matplotlib-users at python.org>>
> Subject: Re: [Matplotlib-users] Windows7: figure window "not responding"
>
> David,
>
> Please subscribe to the mailing list so you can post un-moderated.
>
> Which backend are you using?
>
> If you get a window up and then do `fig.canvas.flush_events()` will it update?
>
> If you do `plt.ioff()` and then `plt.show()` does it work?
>
> Tom
>
>
> On Sat, Aug 19, 2017 at 12:40 PM Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov><mailto:strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>>> wrote:
> Thanks for the respone.  Sadly, I can?t get anything useful out of the doc.  And %matplotlib doesn?t help, same effect.
>
> This is very frustrating.  I don?t know how to debug this.  It?s just a ?black box?, I do what I?m ?supposed to?, doesn?t work.
>
> Help appreciated!
> Dave
>
> From: Thomas Caswell
> [mailto:tcaswell at gmail.com<mailto:tcaswell at gmail.com><mailto:tcaswell at gmail.com<mailto:tcaswell at gmail.com>>]
> Sent: Saturday, August 05, 2017 10:36 AM
> To: Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov><mailto:strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>>>;
> matplotlib-users at python.org<mailto:matplotlib-users at python.org><mailto:matplotlib-users at python.org<mailto:matplotlib-users at python.org>>
> Subject: Re: [Matplotlib-users] Windows7: figure window "not responding"
>
> I do exactly this as well (but on linux)!
>
> The symptoms your are describing sounds like the GUI event loop is not being spun behind the scenes (see https://github.com/matplotlib/matplotlib/pull/4779 for some draft text at explaining how all of that works).  Have you tried doing `%matplotlib` is IPython?
>
> Tom
>
> On Sat, Aug 5, 2017 at 12:39 PM Strozzi, David J. <strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov><mailto:strozzi2 at llnl.gov<mailto:strozzi2 at llnl.gov>>> wrote:
>
> This problem has been driving me nuts, posted it elsewhere, so far no one's been able to help.  I am running Python 3.6.1 from Anaconda (all packages updated to latest and greatest) on Windows 7.  ipython and mpl figure windows work fine from a console, or spyder.  The problem is when I try running ipython within emacs using elpy.
>
>
>
> To back up, the workflow I'm shooting for is using emacs as my IDE, and running python within emacs.  I've done this for many years with the Yorick interpreter, and it's quite addictive.  I get the same problem whether I use the GNU windows build of emacs, or emacs -nw (terminal, no X) from cygwin (ver 25.2 both cases).  elpy seems to be a good Python package for emacs, and nominally supports using ipython.
>
>
>
> OK, so - I can use ipython + emacs + elpy.  The only problem is matplotlib figures.  When I do figure(), a figure window appears, but is not fully rendered (e.g. no buttons), and the window title says "Not responding".  If I do a plot(), nothing is plotted.  Playing with ion() and show() doesn't help.
>
>
>
> I've posted this on the elpy github, various stackexchange forums, no one has had any ideas yet.
>
>
>
> Any help is greatly appreciated!
>
> Dave
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org<mailto:Matplotlib-users at python.org><mailto:Matplotlib-users at python.org<mailto:Matplotlib-users at python.org>>
> https://mail.python.org/mailman/listinfo/matplotlib-users
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org<mailto:Matplotlib-users at python.org>
> https://mail.python.org/mailman/listinfo/matplotlib-users
_______________________________________________
Matplotlib-users mailing list
Matplotlib-users at python.org<mailto:Matplotlib-users at python.org>
https://mail.python.org/mailman/listinfo/matplotlib-users

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From maurobio at gmail.com  Mon Oct  9 19:54:40 2017
From: maurobio at gmail.com (Mauro Cavalcanti)
Date: Mon, 9 Oct 2017 20:54:40 -0300
Subject: [Matplotlib-users] Problem using imshow with Matplotlib/Basemap
In-Reply-To: <CANNq6FnS3Oze=E82axb8bwAhPdSM=5MR_EpJcW-RNpzdXmAEdg@mail.gmail.com>
References: <CAC1JhZZeVg76gwauk_D6RWFNau=WF5c+SexqqgDGX4g-PeXi-Q@mail.gmail.com>
 <CANNq6FnS3Oze=E82axb8bwAhPdSM=5MR_EpJcW-RNpzdXmAEdg@mail.gmail.com>
Message-ID: <CAC1JhZacB18gaM8tb8LWw0Kn0VTPjpP7gNsn7MwaYkakGEXm1A@mail.gmail.com>

Hi,

Thanks for you reply and suggestions.

I changed the imshow call to:

im = m.imshow(grilla_salida, cmap='summer', extent=(lon_inf, lat_inf,
lon_sup, lat_sup), aspect='auto', interpolation='none')

However, the figure is stil wrong (see attachment).

Maybe if instead of imshow, should I use meshgrid/pcolormesh?

Best regards,

2017-10-09 11:29 GMT-03:00 Benjamin Root <ben.v.root at gmail.com>:

> First, you shouldn't need to transpose your image... that'll effectively
> rotate the data by 90 degrees. Second, you didn't specify the extents of
> your image, so Basemap is putting everything starting at coordinate 0,0 in
> the default projection.
>
> If you specify the extent keyword argument to imshow as the (lon1, lat1,
> lon2, lat2) tuple for the lower-left and upper right corners, you won't
> even need the origin='lower', and you definitely won't need the transpose.
>
> Cheers!
> Ben Root
>
>
> On Sun, Oct 8, 2017 at 2:33 PM, Mauro Cavalcanti <maurobio at gmail.com>
> wrote:
>
>> Dear ALL,
>>
>> I have a simple dataset of longitudes/latitudes (see the attached csv
>> file).
>>
>> From such data, I want to generate a grid like this:
>>
>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>> 0 0 0 0 0 0 1 0 0 0 0 0 0
>> 0 0 0 0 0 0 1 1 2 0 0 0 0
>> 0 0 0 0 0 1 1 1 1 0 0 0 0
>> 0 0 0 1 0 1 0 0 0 0 0 0 0
>> 0 0 0 2 0 0 0 0 0 0 0 0 0
>> 0 0 0 0 0 0 0 0 0 1 3 0 0
>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>> 0 0 0 0 0 0 0 0 0 1 0 0 0
>> 0 0 0 0 0 0 0 0 0 1 0 0 0
>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>> 0 0 0 0 0 0 0 0 0 0 1 0 0
>> 0 0 0 0 0 0 0 0 0 0 1 0 0
>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>
>> which gives the number of data records in each cell of the grid, using
>> one of the variables in the dataset ("spp") as a categorical (grouping)
>> factor.
>>
>> From this grid, I then want to create a heat map, superimposed on a
>> Matplotlib/Basemap.
>>
>> I wrote some code which does what I want (see the attachments).
>>
>> It (mostly) works, but te problem is that the grid image is not being
>> displayed correctly: as shown in the attached figure, it appears too small,
>> and in the lower left corner of the map, instead of where it should be (the
>> West coast of Africa, along the Gulf of Guinea).
>>
>> Thanks in advance for any assistance you can provide.
>>
>> Best regards,
>>
>> --
>> Dr. Mauro J. Cavalcanti
>> E-mail: maurobio at gmail.com
>> Web: http://sites.google.com/site/maurobio
>> "Life is complex. It consists of real and imaginary parts."
>>
>> _______________________________________________
>> Matplotlib-users mailing list
>> Matplotlib-users at python.org
>> https://mail.python.org/mailman/listinfo/matplotlib-users
>>
>>
>


-- 
Dr. Mauro J. Cavalcanti
E-mail: maurobio at gmail.com
Web: http://sites.google.com/site/maurobio
"Life is complex. It consists of real and imaginary parts."
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From Stuart_Mentzer at objexx.com  Mon Oct  9 23:32:02 2017
From: Stuart_Mentzer at objexx.com (Stuart Mentzer)
Date: Mon, 9 Oct 2017 20:32:02 -0700 (MST)
Subject: [Matplotlib-users] Unzoom problem in 2.1.0
Message-ID: <1507606322709-0.post@n5.nabble.com>

Hello,

I have a Pyside application (using Qt4Agg backend) that does line and
spectral/contour plots.
After updating to 2.1.0 "unzooming" via the back toolbar button or
right-click won't always
return to the "home" view. It is not consistent but the problem appears
regularly. It is as
though the zoom stack is losing entries off the end.

My attempts to reproduce this in a small (non-Pyside) pyplot example that I
could submit
as a bug haven't been successful yet. I'm wondering if anyone else is seeing
this or has an
idea on how best to isolate the problem.

Thanks,
Stuart



--
Sent from: http://matplotlib.1069221.n5.nabble.com/matplotlib-users-f3.html

From a.h.jaffe at gmail.com  Tue Oct 10 10:40:00 2017
From: a.h.jaffe at gmail.com (Andrew Jaffe)
Date: Tue, 10 Oct 2017 15:40:00 +0100
Subject: [Matplotlib-users] ax.bar edgecolor fails for scalar color?
Message-ID: <orim3o$4fd$1@blaine.gmane.org>

Hi,

I've been working through Ben Root's excellent Anatomy of Matplotlib 
tutorial at https://github.com/matplotlib/AnatomyOfMatplotlib/.

In the course of this (exercise 2.1 in particular), I've noticed that 
the `edgecolor` argument to `ax.bar(...)` doesn't work as advertised:

 > edgecolor : scalar or array-like, optional
 >    the colors of the bar edges

However, it doesn't seem to work as a scalar: in that case it only 
applies to the first bar; instead you need to supply `(color,)*len(x)` 
or similar.

It fails the same way whether or not we use something like 
`plt.style.use('classic')` or `mpl.rcParams['patch.force_edgecolor'] = True`

Is this a bug, something weird in my setup, or am I missing something?

Andrew


From ben.v.root at gmail.com  Tue Oct 10 10:50:15 2017
From: ben.v.root at gmail.com (Benjamin Root)
Date: Tue, 10 Oct 2017 10:50:15 -0400
Subject: [Matplotlib-users] ax.bar edgecolor fails for scalar color?
In-Reply-To: <orim3o$4fd$1@blaine.gmane.org>
References: <orim3o$4fd$1@blaine.gmane.org>
Message-ID: <CANNq6FnWuN0K0Zb7SJ4m7KqvEkSZic-P5utyYOL-dcQhgS+X9g@mail.gmail.com>

Yeah, I think we ran into that during the live tutorial. IIRC, the `color`
keyword argument was overriding the `edgecolor` keyword argument.

On Tue, Oct 10, 2017 at 10:40 AM, Andrew Jaffe <a.h.jaffe at gmail.com> wrote:

> Hi,
>
> I've been working through Ben Root's excellent Anatomy of Matplotlib
> tutorial at https://github.com/matplotlib/AnatomyOfMatplotlib/.
>
> In the course of this (exercise 2.1 in particular), I've noticed that the
> `edgecolor` argument to `ax.bar(...)` doesn't work as advertised:
>
> > edgecolor : scalar or array-like, optional
> >    the colors of the bar edges
>
> However, it doesn't seem to work as a scalar: in that case it only applies
> to the first bar; instead you need to supply `(color,)*len(x)` or similar.
>
> It fails the same way whether or not we use something like
> `plt.style.use('classic')` or `mpl.rcParams['patch.force_edgecolor'] =
> True`
>
> Is this a bug, something weird in my setup, or am I missing something?
>
> Andrew
>
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org
> https://mail.python.org/mailman/listinfo/matplotlib-users
>
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From ben.v.root at gmail.com  Tue Oct 10 10:52:34 2017
From: ben.v.root at gmail.com (Benjamin Root)
Date: Tue, 10 Oct 2017 10:52:34 -0400
Subject: [Matplotlib-users] ax.bar edgecolor fails for scalar color?
In-Reply-To: <CANNq6FnWuN0K0Zb7SJ4m7KqvEkSZic-P5utyYOL-dcQhgS+X9g@mail.gmail.com>
References: <orim3o$4fd$1@blaine.gmane.org>
 <CANNq6FnWuN0K0Zb7SJ4m7KqvEkSZic-P5utyYOL-dcQhgS+X9g@mail.gmail.com>
Message-ID: <CANNq6FmMLH=mpFvi9Q8LcJp9BxB+QNwBPdpO0hSVnMWU6Not5g@mail.gmail.com>

Actually, to be honest, I can't remember what the solution was.  CC'ing
Thomas because I think he figured it out later on during the tutorial.

Ben

On Tue, Oct 10, 2017 at 10:50 AM, Benjamin Root <ben.v.root at gmail.com>
wrote:

> Yeah, I think we ran into that during the live tutorial. IIRC, the `color`
> keyword argument was overriding the `edgecolor` keyword argument.
>
> On Tue, Oct 10, 2017 at 10:40 AM, Andrew Jaffe <a.h.jaffe at gmail.com>
> wrote:
>
>> Hi,
>>
>> I've been working through Ben Root's excellent Anatomy of Matplotlib
>> tutorial at https://github.com/matplotlib/AnatomyOfMatplotlib/.
>>
>> In the course of this (exercise 2.1 in particular), I've noticed that the
>> `edgecolor` argument to `ax.bar(...)` doesn't work as advertised:
>>
>> > edgecolor : scalar or array-like, optional
>> >    the colors of the bar edges
>>
>> However, it doesn't seem to work as a scalar: in that case it only
>> applies to the first bar; instead you need to supply `(color,)*len(x)` or
>> similar.
>>
>> It fails the same way whether or not we use something like
>> `plt.style.use('classic')` or `mpl.rcParams['patch.force_edgecolor'] =
>> True`
>>
>> Is this a bug, something weird in my setup, or am I missing something?
>>
>> Andrew
>>
>> _______________________________________________
>> Matplotlib-users mailing list
>> Matplotlib-users at python.org
>> https://mail.python.org/mailman/listinfo/matplotlib-users
>>
>
>
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From ben.v.root at gmail.com  Tue Oct 10 11:02:32 2017
From: ben.v.root at gmail.com (Benjamin Root)
Date: Tue, 10 Oct 2017 11:02:32 -0400
Subject: [Matplotlib-users] Unzoom problem in 2.1.0
In-Reply-To: <1507606322709-0.post@n5.nabble.com>
References: <1507606322709-0.post@n5.nabble.com>
Message-ID: <CANNq6FkxQJufOe98-najbDY=3v43uv62HqBe2TWJBfhx83o4Mg@mail.gmail.com>

Stuart,

right clicks? Did you add a callback for handling right-clicks? I don't
think matplotlib does anything (unless you are talking about using the
right mouse button to create an "unzoom" box?)

Ben Root

On Mon, Oct 9, 2017 at 11:32 PM, Stuart Mentzer <Stuart_Mentzer at objexx.com>
wrote:

> Hello,
>
> I have a Pyside application (using Qt4Agg backend) that does line and
> spectral/contour plots.
> After updating to 2.1.0 "unzooming" via the back toolbar button or
> right-click won't always
> return to the "home" view. It is not consistent but the problem appears
> regularly. It is as
> though the zoom stack is losing entries off the end.
>
> My attempts to reproduce this in a small (non-Pyside) pyplot example that I
> could submit
> as a bug haven't been successful yet. I'm wondering if anyone else is
> seeing
> this or has an
> idea on how best to isolate the problem.
>
> Thanks,
> Stuart
>
>
>
> --
> Sent from: http://matplotlib.1069221.n5.nabble.com/matplotlib-users-
> f3.html
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org
> https://mail.python.org/mailman/listinfo/matplotlib-users
>
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From ben.v.root at gmail.com  Tue Oct 10 11:20:04 2017
From: ben.v.root at gmail.com (Benjamin Root)
Date: Tue, 10 Oct 2017 11:20:04 -0400
Subject: [Matplotlib-users] Problem using imshow with Matplotlib/Basemap
In-Reply-To: <CAC1JhZacB18gaM8tb8LWw0Kn0VTPjpP7gNsn7MwaYkakGEXm1A@mail.gmail.com>
References: <CAC1JhZZeVg76gwauk_D6RWFNau=WF5c+SexqqgDGX4g-PeXi-Q@mail.gmail.com>
 <CANNq6FnS3Oze=E82axb8bwAhPdSM=5MR_EpJcW-RNpzdXmAEdg@mail.gmail.com>
 <CAC1JhZacB18gaM8tb8LWw0Kn0VTPjpP7gNsn7MwaYkakGEXm1A@mail.gmail.com>
Message-ID: <CANNq6F=bD48rdS5wgVu8L2f6snAWtQ_6PH4HsrWX=o130exHoQ@mail.gmail.com>

Ah, the problem was that by default, the limits for a Basemap goes from
-180 to 180, and -90 to 90. If you do:

m = Basemap(llcrnrlat=lat_inf, llcrnrlon=lon_inf,
            urcrnrlat=lat_sup, urcrnrlon=lon_sup)

Then things line up correctly, and you don't need the origin keyword
argument, the transpose, or even the extent argument.

In fact, there might even be a bug, as I would have expected specifying the
extent should have worked regardless of the original bounds. Might need to
look into that.

Cheers!
Ben Root


On Mon, Oct 9, 2017 at 7:54 PM, Mauro Cavalcanti <maurobio at gmail.com> wrote:

> Hi,
>
> Thanks for you reply and suggestions.
>
> I changed the imshow call to:
>
> im = m.imshow(grilla_salida, cmap='summer', extent=(lon_inf, lat_inf,
> lon_sup, lat_sup), aspect='auto', interpolation='none')
>
> However, the figure is stil wrong (see attachment).
>
> Maybe if instead of imshow, should I use meshgrid/pcolormesh?
>
> Best regards,
>
> 2017-10-09 11:29 GMT-03:00 Benjamin Root <ben.v.root at gmail.com>:
>
>> First, you shouldn't need to transpose your image... that'll effectively
>> rotate the data by 90 degrees. Second, you didn't specify the extents of
>> your image, so Basemap is putting everything starting at coordinate 0,0 in
>> the default projection.
>>
>> If you specify the extent keyword argument to imshow as the (lon1, lat1,
>> lon2, lat2) tuple for the lower-left and upper right corners, you won't
>> even need the origin='lower', and you definitely won't need the transpose.
>>
>> Cheers!
>> Ben Root
>>
>>
>> On Sun, Oct 8, 2017 at 2:33 PM, Mauro Cavalcanti <maurobio at gmail.com>
>> wrote:
>>
>>> Dear ALL,
>>>
>>> I have a simple dataset of longitudes/latitudes (see the attached csv
>>> file).
>>>
>>> From such data, I want to generate a grid like this:
>>>
>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>> 0 0 0 0 0 0 1 0 0 0 0 0 0
>>> 0 0 0 0 0 0 1 1 2 0 0 0 0
>>> 0 0 0 0 0 1 1 1 1 0 0 0 0
>>> 0 0 0 1 0 1 0 0 0 0 0 0 0
>>> 0 0 0 2 0 0 0 0 0 0 0 0 0
>>> 0 0 0 0 0 0 0 0 0 1 3 0 0
>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>> 0 0 0 0 0 0 0 0 0 1 0 0 0
>>> 0 0 0 0 0 0 0 0 0 1 0 0 0
>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>> 0 0 0 0 0 0 0 0 0 0 1 0 0
>>> 0 0 0 0 0 0 0 0 0 0 1 0 0
>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>
>>> which gives the number of data records in each cell of the grid, using
>>> one of the variables in the dataset ("spp") as a categorical (grouping)
>>> factor.
>>>
>>> From this grid, I then want to create a heat map, superimposed on a
>>> Matplotlib/Basemap.
>>>
>>> I wrote some code which does what I want (see the attachments).
>>>
>>> It (mostly) works, but te problem is that the grid image is not being
>>> displayed correctly: as shown in the attached figure, it appears too small,
>>> and in the lower left corner of the map, instead of where it should be (the
>>> West coast of Africa, along the Gulf of Guinea).
>>>
>>> Thanks in advance for any assistance you can provide.
>>>
>>> Best regards,
>>>
>>> --
>>> Dr. Mauro J. Cavalcanti
>>> E-mail: maurobio at gmail.com
>>> Web: http://sites.google.com/site/maurobio
>>> "Life is complex. It consists of real and imaginary parts."
>>>
>>> _______________________________________________
>>> Matplotlib-users mailing list
>>> Matplotlib-users at python.org
>>> https://mail.python.org/mailman/listinfo/matplotlib-users
>>>
>>>
>>
>
>
> --
> Dr. Mauro J. Cavalcanti
> E-mail: maurobio at gmail.com
> Web: http://sites.google.com/site/maurobio
> "Life is complex. It consists of real and imaginary parts."
>
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From michael.francis.corcoran at gmail.com  Tue Oct 10 14:08:50 2017
From: michael.francis.corcoran at gmail.com (mfacorcoran)
Date: Tue, 10 Oct 2017 11:08:50 -0700 (MST)
Subject: [Matplotlib-users] can't import cbook/matplotlib v. 2.1.0
Message-ID: <1507658930741-0.post@n5.nabble.com>

I've recently updated matplotlib to 2.1.0 and now when I try to import
matplotlib/pylab I get the following error that cbook cannot be imported:

% python
Python 2.7.14 (default, Sep 22 2017, 00:05:22)
[GCC 4.2.1 Compatible Apple LLVM 7.3.0 (clang-703.0.31)] on darwin
Type "help", "copyright", "credits" or "license" for more information.

Importing pyfits
Could not import pylab
Traceback (most recent call last):
  File "/Users/corcoran/.pythonrc", line 34, in <module>
    import matplotlib
  File
"/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/matplotlib/__init__.py",
line 124, in <module>
    from . import cbook
ImportError: cannot import name cbook

any ideas what might be going on?

Mike



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From pmhobson at gmail.com  Tue Oct 10 14:12:00 2017
From: pmhobson at gmail.com (Paul Hobson)
Date: Tue, 10 Oct 2017 11:12:00 -0700
Subject: [Matplotlib-users] can't import cbook/matplotlib v. 2.1.0
In-Reply-To: <1507658930741-0.post@n5.nabble.com>
References: <1507658930741-0.post@n5.nabble.com>
Message-ID: <CADT3MEDsSPJCveCA-6D1KnnsrqPK0RQDOX4NTzoxZLoRQ1z5Vg@mail.gmail.com>

Were you in the matplotlib source directory when you launched your python
interpreter?
-p

On Tue, Oct 10, 2017 at 11:08 AM, mfacorcoran <
michael.francis.corcoran at gmail.com> wrote:

> I've recently updated matplotlib to 2.1.0 and now when I try to import
> matplotlib/pylab I get the following error that cbook cannot be imported:
>
> % python
> Python 2.7.14 (default, Sep 22 2017, 00:05:22)
> [GCC 4.2.1 Compatible Apple LLVM 7.3.0 (clang-703.0.31)] on darwin
> Type "help", "copyright", "credits" or "license" for more information.
>
> Importing pyfits
> Could not import pylab
> Traceback (most recent call last):
>   File "/Users/corcoran/.pythonrc", line 34, in <module>
>     import matplotlib
>   File
> "/opt/local/Library/Frameworks/Python.framework/
> Versions/2.7/lib/python2.7/site-packages/matplotlib/__init__.py",
> line 124, in <module>
>     from . import cbook
> ImportError: cannot import name cbook
>
> any ideas what might be going on?
>
> Mike
>
>
>
> --
> Sent from: http://matplotlib.1069221.n5.nabble.com/matplotlib-users-
> f3.html
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org
> https://mail.python.org/mailman/listinfo/matplotlib-users
>
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From michael.francis.corcoran at gmail.com  Tue Oct 10 14:20:48 2017
From: michael.francis.corcoran at gmail.com (mfacorcoran)
Date: Tue, 10 Oct 2017 11:20:48 -0700 (MST)
Subject: [Matplotlib-users] can't import cbook/matplotlib v. 2.1.0
In-Reply-To: <CADT3MEDsSPJCveCA-6D1KnnsrqPK0RQDOX4NTzoxZLoRQ1z5Vg@mail.gmail.com>
References: <1507658930741-0.post@n5.nabble.com>
 <CADT3MEDsSPJCveCA-6D1KnnsrqPK0RQDOX4NTzoxZLoRQ1z5Vg@mail.gmail.com>
Message-ID: <1507659648138-0.post@n5.nabble.com>

I don't think so.  I re-tried it from my home directory and got the same
error...

cheers

Mike




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From ben.v.root at gmail.com  Tue Oct 10 14:41:07 2017
From: ben.v.root at gmail.com (Benjamin Root)
Date: Tue, 10 Oct 2017 14:41:07 -0400
Subject: [Matplotlib-users] can't import cbook/matplotlib v. 2.1.0
In-Reply-To: <1507659648138-0.post@n5.nabble.com>
References: <1507658930741-0.post@n5.nabble.com>
 <CADT3MEDsSPJCveCA-6D1KnnsrqPK0RQDOX4NTzoxZLoRQ1z5Vg@mail.gmail.com>
 <1507659648138-0.post@n5.nabble.com>
Message-ID: <CANNq6F=O9ekAU5ZjodU0WNqVTh46Sza_gTXrvT_asC9uC=fZZg@mail.gmail.com>

How did you install matplotlib?

On Tue, Oct 10, 2017 at 2:20 PM, mfacorcoran <
michael.francis.corcoran at gmail.com> wrote:

> I don't think so.  I re-tried it from my home directory and got the same
> error...
>
> cheers
>
> Mike
>
>
>
>
> --
> Sent from: http://matplotlib.1069221.n5.nabble.com/matplotlib-users-
> f3.html
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org
> https://mail.python.org/mailman/listinfo/matplotlib-users
>
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From michael.francis.corcoran at gmail.com  Tue Oct 10 14:49:49 2017
From: michael.francis.corcoran at gmail.com (mfacorcoran)
Date: Tue, 10 Oct 2017 11:49:49 -0700 (MST)
Subject: [Matplotlib-users] can't import cbook/matplotlib v. 2.1.0
In-Reply-To: <CANNq6F=O9ekAU5ZjodU0WNqVTh46Sza_gTXrvT_asC9uC=fZZg@mail.gmail.com>
References: <1507658930741-0.post@n5.nabble.com>
 <CADT3MEDsSPJCveCA-6D1KnnsrqPK0RQDOX4NTzoxZLoRQ1z5Vg@mail.gmail.com>
 <1507659648138-0.post@n5.nabble.com>
 <CANNq6F=O9ekAU5ZjodU0WNqVTh46Sza_gTXrvT_asC9uC=fZZg@mail.gmail.com>
Message-ID: <1507661389563-0.post@n5.nabble.com>

using macports.  I'm on a macbook pro running 10.11.6.  I've used matplotlib
for the past 2-3 years; this is the first time I had this issue



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From ben.v.root at gmail.com  Tue Oct 10 14:52:26 2017
From: ben.v.root at gmail.com (Benjamin Root)
Date: Tue, 10 Oct 2017 14:52:26 -0400
Subject: [Matplotlib-users] can't import cbook/matplotlib v. 2.1.0
In-Reply-To: <1507661389563-0.post@n5.nabble.com>
References: <1507658930741-0.post@n5.nabble.com>
 <CADT3MEDsSPJCveCA-6D1KnnsrqPK0RQDOX4NTzoxZLoRQ1z5Vg@mail.gmail.com>
 <1507659648138-0.post@n5.nabble.com>
 <CANNq6F=O9ekAU5ZjodU0WNqVTh46Sza_gTXrvT_asC9uC=fZZg@mail.gmail.com>
 <1507661389563-0.post@n5.nabble.com>
Message-ID: <CANNq6FmKg-9sttBW2GkmEY37skwL7rCpXTuKjwa_cffaVfmwEA@mail.gmail.com>

Then exactly which version of the matplotlib package did you install from
MacPorts? Just today, we had another bug report from another user using
MacPorts and python 2.7 that turned out to be a packaging error on their
part.

On Tue, Oct 10, 2017 at 2:49 PM, mfacorcoran <
michael.francis.corcoran at gmail.com> wrote:

> using macports.  I'm on a macbook pro running 10.11.6.  I've used
> matplotlib
> for the past 2-3 years; this is the first time I had this issue
>
>
>
> --
> Sent from: http://matplotlib.1069221.n5.nabble.com/matplotlib-users-
> f3.html
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org
> https://mail.python.org/mailman/listinfo/matplotlib-users
>
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From maurobio at gmail.com  Tue Oct 10 14:57:01 2017
From: maurobio at gmail.com (Mauro Cavalcanti)
Date: Tue, 10 Oct 2017 15:57:01 -0300
Subject: [Matplotlib-users] Problem using imshow with Matplotlib/Basemap
In-Reply-To: <CANNq6F=bD48rdS5wgVu8L2f6snAWtQ_6PH4HsrWX=o130exHoQ@mail.gmail.com>
References: <CAC1JhZZeVg76gwauk_D6RWFNau=WF5c+SexqqgDGX4g-PeXi-Q@mail.gmail.com>
 <CANNq6FnS3Oze=E82axb8bwAhPdSM=5MR_EpJcW-RNpzdXmAEdg@mail.gmail.com>
 <CAC1JhZacB18gaM8tb8LWw0Kn0VTPjpP7gNsn7MwaYkakGEXm1A@mail.gmail.com>
 <CANNq6F=bD48rdS5wgVu8L2f6snAWtQ_6PH4HsrWX=o130exHoQ@mail.gmail.com>
Message-ID: <CAC1JhZZQAuYHh_DeVW1Kv23ariB8-pHk5d=DK8R4qGEyP-8+gg@mail.gmail.com>

Dear Ben,

Thanks a lot again for your help and patience.

After modifying the call to Basemap according to your suggestion:

m = Basemap(llcrnrlat=lat_inf, llcrnrlon=lon_inf, urcrnrlat=lat_sup,
urcrnrlon=lon_sup)

the grid is correctly displayed where it belongs (thr west coast of
Africa). But now I got a map centered at the Pacific basin, with the grid
appearing on the lower left corner of it! In fact, See the attached image.
I would like a conventional map centered around the Atlantic basin (ie.,
with center coordinates at lat_0=0 and long_0=0).

After using the superb Matplotlib for almost a decade (in the beginning, I
even got some help here from the legendary John Hunter!), I nonetheless
feel somewhat ashamed of having found a potential bug in the library...

All the best,

2017-10-10 12:20 GMT-03:00 Benjamin Root <ben.v.root at gmail.com>:

> Ah, the problem was that by default, the limits for a Basemap goes from
> -180 to 180, and -90 to 90. If you do:
>
> m = Basemap(llcrnrlat=lat_inf, llcrnrlon=lon_inf,
>             urcrnrlat=lat_sup, urcrnrlon=lon_sup)
>
> Then things line up correctly, and you don't need the origin keyword
> argument, the transpose, or even the extent argument.
>
> In fact, there might even be a bug, as I would have expected specifying
> the extent should have worked regardless of the original bounds. Might need
> to look into that.
>
> Cheers!
> Ben Root
>
>
> On Mon, Oct 9, 2017 at 7:54 PM, Mauro Cavalcanti <maurobio at gmail.com>
> wrote:
>
>> Hi,
>>
>> Thanks for you reply and suggestions.
>>
>> I changed the imshow call to:
>>
>> im = m.imshow(grilla_salida, cmap='summer', extent=(lon_inf, lat_inf,
>> lon_sup, lat_sup), aspect='auto', interpolation='none')
>>
>> However, the figure is stil wrong (see attachment).
>>
>> Maybe if instead of imshow, should I use meshgrid/pcolormesh?
>>
>> Best regards,
>>
>> 2017-10-09 11:29 GMT-03:00 Benjamin Root <ben.v.root at gmail.com>:
>>
>>> First, you shouldn't need to transpose your image... that'll effectively
>>> rotate the data by 90 degrees. Second, you didn't specify the extents of
>>> your image, so Basemap is putting everything starting at coordinate 0,0 in
>>> the default projection.
>>>
>>> If you specify the extent keyword argument to imshow as the (lon1, lat1,
>>> lon2, lat2) tuple for the lower-left and upper right corners, you won't
>>> even need the origin='lower', and you definitely won't need the transpose.
>>>
>>> Cheers!
>>> Ben Root
>>>
>>>
>>> On Sun, Oct 8, 2017 at 2:33 PM, Mauro Cavalcanti <maurobio at gmail.com>
>>> wrote:
>>>
>>>> Dear ALL,
>>>>
>>>> I have a simple dataset of longitudes/latitudes (see the attached csv
>>>> file).
>>>>
>>>> From such data, I want to generate a grid like this:
>>>>
>>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>> 0 0 0 0 0 0 1 0 0 0 0 0 0
>>>> 0 0 0 0 0 0 1 1 2 0 0 0 0
>>>> 0 0 0 0 0 1 1 1 1 0 0 0 0
>>>> 0 0 0 1 0 1 0 0 0 0 0 0 0
>>>> 0 0 0 2 0 0 0 0 0 0 0 0 0
>>>> 0 0 0 0 0 0 0 0 0 1 3 0 0
>>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>> 0 0 0 0 0 0 0 0 0 1 0 0 0
>>>> 0 0 0 0 0 0 0 0 0 1 0 0 0
>>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>> 0 0 0 0 0 0 0 0 0 0 1 0 0
>>>> 0 0 0 0 0 0 0 0 0 0 1 0 0
>>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>>
>>>> which gives the number of data records in each cell of the grid, using
>>>> one of the variables in the dataset ("spp") as a categorical (grouping)
>>>> factor.
>>>>
>>>> From this grid, I then want to create a heat map, superimposed on a
>>>> Matplotlib/Basemap.
>>>>
>>>> I wrote some code which does what I want (see the attachments).
>>>>
>>>> It (mostly) works, but te problem is that the grid image is not being
>>>> displayed correctly: as shown in the attached figure, it appears too small,
>>>> and in the lower left corner of the map, instead of where it should be (the
>>>> West coast of Africa, along the Gulf of Guinea).
>>>>
>>>> Thanks in advance for any assistance you can provide.
>>>>
>>>> Best regards,
>>>>
>>>> --
>>>> Dr. Mauro J. Cavalcanti
>>>> E-mail: maurobio at gmail.com
>>>> Web: http://sites.google.com/site/maurobio
>>>> "Life is complex. It consists of real and imaginary parts."
>>>>
>>>> _______________________________________________
>>>> Matplotlib-users mailing list
>>>> Matplotlib-users at python.org
>>>> https://mail.python.org/mailman/listinfo/matplotlib-users
>>>>
>>>>
>>>
>>
>>
>> --
>> Dr. Mauro J. Cavalcanti
>> E-mail: maurobio at gmail.com
>> Web: http://sites.google.com/site/maurobio
>> "Life is complex. It consists of real and imaginary parts."
>>
>
>


-- 
Dr. Mauro J. Cavalcanti
E-mail: maurobio at gmail.com
Web: http://sites.google.com/site/maurobio
"Life is complex. It consists of real and imaginary parts."
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From michael.francis.corcoran at gmail.com  Tue Oct 10 15:06:44 2017
From: michael.francis.corcoran at gmail.com (mfacorcoran)
Date: Tue, 10 Oct 2017 12:06:44 -0700 (MST)
Subject: [Matplotlib-users] can't import cbook/matplotlib v. 2.1.0
In-Reply-To: <CANNq6FmKg-9sttBW2GkmEY37skwL7rCpXTuKjwa_cffaVfmwEA@mail.gmail.com>
References: <1507658930741-0.post@n5.nabble.com>
 <CADT3MEDsSPJCveCA-6D1KnnsrqPK0RQDOX4NTzoxZLoRQ1z5Vg@mail.gmail.com>
 <1507659648138-0.post@n5.nabble.com>
 <CANNq6F=O9ekAU5ZjodU0WNqVTh46Sza_gTXrvT_asC9uC=fZZg@mail.gmail.com>
 <1507661389563-0.post@n5.nabble.com>
 <CANNq6FmKg-9sttBW2GkmEY37skwL7rCpXTuKjwa_cffaVfmwEA@mail.gmail.com>
Message-ID: <1507662404822-0.post@n5.nabble.com>

This is the python 2.7 version:  py27-matplotlib 2.1.0

I also checked the python3 version - that seems to work ok (btw)



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From ben.v.root at gmail.com  Tue Oct 10 16:25:12 2017
From: ben.v.root at gmail.com (Benjamin Root)
Date: Tue, 10 Oct 2017 16:25:12 -0400
Subject: [Matplotlib-users] Problem using imshow with Matplotlib/Basemap
In-Reply-To: <CAC1JhZZQAuYHh_DeVW1Kv23ariB8-pHk5d=DK8R4qGEyP-8+gg@mail.gmail.com>
References: <CAC1JhZZeVg76gwauk_D6RWFNau=WF5c+SexqqgDGX4g-PeXi-Q@mail.gmail.com>
 <CANNq6FnS3Oze=E82axb8bwAhPdSM=5MR_EpJcW-RNpzdXmAEdg@mail.gmail.com>
 <CAC1JhZacB18gaM8tb8LWw0Kn0VTPjpP7gNsn7MwaYkakGEXm1A@mail.gmail.com>
 <CANNq6F=bD48rdS5wgVu8L2f6snAWtQ_6PH4HsrWX=o130exHoQ@mail.gmail.com>
 <CAC1JhZZQAuYHh_DeVW1Kv23ariB8-pHk5d=DK8R4qGEyP-8+gg@mail.gmail.com>
Message-ID: <CANNq6Fk8Rtcd63AN6Fo-spHM6110YnvGgTGu0M04SgfQkU1eUA@mail.gmail.com>

So, the problem here is that what you are asking for is in conflict with
the settings. You are giving bounding box that starts at llcrnrlon=lon_inf
and goes to urcrnrlon=lon_sup, and that is the extent of the data you are
providing, too. But you want to view it in a different bounds. It is
possible to get what you want, but it starts to get tricky at this point.

Before venturing further, I should point out to you that Basemap is
effectively deprecated. Its support will be terminated in 2020. The CartoPy
project, also built on top of matplotlib, is a far superior library, and I
think would be much easier for you to use. It doesn't do everything that
basemap does yet, but the things that it does do, it does better. I suggest
checking it out before we go any further here.

Cheers!
Ben Root


On Tue, Oct 10, 2017 at 2:57 PM, Mauro Cavalcanti <maurobio at gmail.com>
wrote:

> Dear Ben,
>
> Thanks a lot again for your help and patience.
>
> After modifying the call to Basemap according to your suggestion:
>
> m = Basemap(llcrnrlat=lat_inf, llcrnrlon=lon_inf, urcrnrlat=lat_sup,
> urcrnrlon=lon_sup)
>
> the grid is correctly displayed where it belongs (thr west coast of
> Africa). But now I got a map centered at the Pacific basin, with the grid
> appearing on the lower left corner of it! In fact, See the attached image.
> I would like a conventional map centered around the Atlantic basin (ie.,
> with center coordinates at lat_0=0 and long_0=0).
>
> After using the superb Matplotlib for almost a decade (in the beginning, I
> even got some help here from the legendary John Hunter!), I nonetheless
> feel somewhat ashamed of having found a potential bug in the library...
>
> All the best,
>
> 2017-10-10 12:20 GMT-03:00 Benjamin Root <ben.v.root at gmail.com>:
>
>> Ah, the problem was that by default, the limits for a Basemap goes from
>> -180 to 180, and -90 to 90. If you do:
>>
>> m = Basemap(llcrnrlat=lat_inf, llcrnrlon=lon_inf,
>>             urcrnrlat=lat_sup, urcrnrlon=lon_sup)
>>
>> Then things line up correctly, and you don't need the origin keyword
>> argument, the transpose, or even the extent argument.
>>
>> In fact, there might even be a bug, as I would have expected specifying
>> the extent should have worked regardless of the original bounds. Might need
>> to look into that.
>>
>> Cheers!
>> Ben Root
>>
>>
>> On Mon, Oct 9, 2017 at 7:54 PM, Mauro Cavalcanti <maurobio at gmail.com>
>> wrote:
>>
>>> Hi,
>>>
>>> Thanks for you reply and suggestions.
>>>
>>> I changed the imshow call to:
>>>
>>> im = m.imshow(grilla_salida, cmap='summer', extent=(lon_inf, lat_inf,
>>> lon_sup, lat_sup), aspect='auto', interpolation='none')
>>>
>>> However, the figure is stil wrong (see attachment).
>>>
>>> Maybe if instead of imshow, should I use meshgrid/pcolormesh?
>>>
>>> Best regards,
>>>
>>> 2017-10-09 11:29 GMT-03:00 Benjamin Root <ben.v.root at gmail.com>:
>>>
>>>> First, you shouldn't need to transpose your image... that'll
>>>> effectively rotate the data by 90 degrees. Second, you didn't specify the
>>>> extents of your image, so Basemap is putting everything starting at
>>>> coordinate 0,0 in the default projection.
>>>>
>>>> If you specify the extent keyword argument to imshow as the (lon1,
>>>> lat1, lon2, lat2) tuple for the lower-left and upper right corners, you
>>>> won't even need the origin='lower', and you definitely won't need the
>>>> transpose.
>>>>
>>>> Cheers!
>>>> Ben Root
>>>>
>>>>
>>>> On Sun, Oct 8, 2017 at 2:33 PM, Mauro Cavalcanti <maurobio at gmail.com>
>>>> wrote:
>>>>
>>>>> Dear ALL,
>>>>>
>>>>> I have a simple dataset of longitudes/latitudes (see the attached csv
>>>>> file).
>>>>>
>>>>> From such data, I want to generate a grid like this:
>>>>>
>>>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>>> 0 0 0 0 0 0 1 0 0 0 0 0 0
>>>>> 0 0 0 0 0 0 1 1 2 0 0 0 0
>>>>> 0 0 0 0 0 1 1 1 1 0 0 0 0
>>>>> 0 0 0 1 0 1 0 0 0 0 0 0 0
>>>>> 0 0 0 2 0 0 0 0 0 0 0 0 0
>>>>> 0 0 0 0 0 0 0 0 0 1 3 0 0
>>>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>>> 0 0 0 0 0 0 0 0 0 1 0 0 0
>>>>> 0 0 0 0 0 0 0 0 0 1 0 0 0
>>>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>>> 0 0 0 0 0 0 0 0 0 0 1 0 0
>>>>> 0 0 0 0 0 0 0 0 0 0 1 0 0
>>>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>>>
>>>>> which gives the number of data records in each cell of the grid, using
>>>>> one of the variables in the dataset ("spp") as a categorical (grouping)
>>>>> factor.
>>>>>
>>>>> From this grid, I then want to create a heat map, superimposed on a
>>>>> Matplotlib/Basemap.
>>>>>
>>>>> I wrote some code which does what I want (see the attachments).
>>>>>
>>>>> It (mostly) works, but te problem is that the grid image is not being
>>>>> displayed correctly: as shown in the attached figure, it appears too small,
>>>>> and in the lower left corner of the map, instead of where it should be (the
>>>>> West coast of Africa, along the Gulf of Guinea).
>>>>>
>>>>> Thanks in advance for any assistance you can provide.
>>>>>
>>>>> Best regards,
>>>>>
>>>>> --
>>>>> Dr. Mauro J. Cavalcanti
>>>>> E-mail: maurobio at gmail.com
>>>>> Web: http://sites.google.com/site/maurobio
>>>>> "Life is complex. It consists of real and imaginary parts."
>>>>>
>>>>> _______________________________________________
>>>>> Matplotlib-users mailing list
>>>>> Matplotlib-users at python.org
>>>>> https://mail.python.org/mailman/listinfo/matplotlib-users
>>>>>
>>>>>
>>>>
>>>
>>>
>>> --
>>> Dr. Mauro J. Cavalcanti
>>> E-mail: maurobio at gmail.com
>>> Web: http://sites.google.com/site/maurobio
>>> "Life is complex. It consists of real and imaginary parts."
>>>
>>
>>
>
>
> --
> Dr. Mauro J. Cavalcanti
> E-mail: maurobio at gmail.com
> Web: http://sites.google.com/site/maurobio
> "Life is complex. It consists of real and imaginary parts."
>
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From ben.v.root at gmail.com  Tue Oct 10 16:28:18 2017
From: ben.v.root at gmail.com (Benjamin Root)
Date: Tue, 10 Oct 2017 16:28:18 -0400
Subject: [Matplotlib-users] can't import cbook/matplotlib v. 2.1.0
In-Reply-To: <1507662404822-0.post@n5.nabble.com>
References: <1507658930741-0.post@n5.nabble.com>
 <CADT3MEDsSPJCveCA-6D1KnnsrqPK0RQDOX4NTzoxZLoRQ1z5Vg@mail.gmail.com>
 <1507659648138-0.post@n5.nabble.com>
 <CANNq6F=O9ekAU5ZjodU0WNqVTh46Sza_gTXrvT_asC9uC=fZZg@mail.gmail.com>
 <1507661389563-0.post@n5.nabble.com>
 <CANNq6FmKg-9sttBW2GkmEY37skwL7rCpXTuKjwa_cffaVfmwEA@mail.gmail.com>
 <1507662404822-0.post@n5.nabble.com>
Message-ID: <CANNq6F=mH+ikajFWVoZT=Su-BS5Y3yALVUZn9bSe=RUWHabB_Q@mail.gmail.com>

Check to see if you have the latest package from macports, the version the
other user had was 2.1.0_1, which indicates that there was at least a
bugfix in their packaging. That _1 release did have another bug, where the
backports package wasn't specified as a dependency in py27.

Ben Root

On Tue, Oct 10, 2017 at 3:06 PM, mfacorcoran <
michael.francis.corcoran at gmail.com> wrote:

> This is the python 2.7 version:  py27-matplotlib 2.1.0
>
> I also checked the python3 version - that seems to work ok (btw)
>
>
>
> --
> Sent from: http://matplotlib.1069221.n5.nabble.com/matplotlib-users-
> f3.html
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org
> https://mail.python.org/mailman/listinfo/matplotlib-users
>
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From michael.francis.corcoran at gmail.com  Tue Oct 10 16:40:15 2017
From: michael.francis.corcoran at gmail.com (mfacorcoran)
Date: Tue, 10 Oct 2017 13:40:15 -0700 (MST)
Subject: [Matplotlib-users] can't import cbook/matplotlib v. 2.1.0
In-Reply-To: <CANNq6F=mH+ikajFWVoZT=Su-BS5Y3yALVUZn9bSe=RUWHabB_Q@mail.gmail.com>
References: <1507658930741-0.post@n5.nabble.com>
 <CADT3MEDsSPJCveCA-6D1KnnsrqPK0RQDOX4NTzoxZLoRQ1z5Vg@mail.gmail.com>
 <1507659648138-0.post@n5.nabble.com>
 <CANNq6F=O9ekAU5ZjodU0WNqVTh46Sza_gTXrvT_asC9uC=fZZg@mail.gmail.com>
 <1507661389563-0.post@n5.nabble.com>
 <CANNq6FmKg-9sttBW2GkmEY37skwL7rCpXTuKjwa_cffaVfmwEA@mail.gmail.com>
 <1507662404822-0.post@n5.nabble.com>
 <CANNq6F=mH+ikajFWVoZT=Su-BS5Y3yALVUZn9bSe=RUWHabB_Q@mail.gmail.com>
Message-ID: <1507668015883-0.post@n5.nabble.com>

I have  py27-matplotlib @2.1.0_1+cairo+tkinter (active) installed

I just went back to the 2.0.2 version via:

% sudo port activate py27-matplotlib @2.0.2_0+cairo+tkinter

and that seems to have resolved the issue.  So it looks like there's
something strange with 
the port of py27-matplotlib @2.1.0_1+cairo+tkinter

Mike



--
Sent from: http://matplotlib.1069221.n5.nabble.com/matplotlib-users-f3.html

From maurobio at gmail.com  Tue Oct 10 19:04:17 2017
From: maurobio at gmail.com (Mauro Cavalcanti)
Date: Tue, 10 Oct 2017 20:04:17 -0300
Subject: [Matplotlib-users] Problem using imshow with Matplotlib/Basemap
In-Reply-To: <CANNq6Fk8Rtcd63AN6Fo-spHM6110YnvGgTGu0M04SgfQkU1eUA@mail.gmail.com>
References: <CAC1JhZZeVg76gwauk_D6RWFNau=WF5c+SexqqgDGX4g-PeXi-Q@mail.gmail.com>
 <CANNq6FnS3Oze=E82axb8bwAhPdSM=5MR_EpJcW-RNpzdXmAEdg@mail.gmail.com>
 <CAC1JhZacB18gaM8tb8LWw0Kn0VTPjpP7gNsn7MwaYkakGEXm1A@mail.gmail.com>
 <CANNq6F=bD48rdS5wgVu8L2f6snAWtQ_6PH4HsrWX=o130exHoQ@mail.gmail.com>
 <CAC1JhZZQAuYHh_DeVW1Kv23ariB8-pHk5d=DK8R4qGEyP-8+gg@mail.gmail.com>
 <CANNq6Fk8Rtcd63AN6Fo-spHM6110YnvGgTGu0M04SgfQkU1eUA@mail.gmail.com>
Message-ID: <CAC1JhZbu=1Uw-S5eOFG7jz95pDuOW1uvx0ZWuO5EfC6mQoLTAQ@mail.gmail.com>

Dear Ben,

Sure, I am aware of Cartopy and have been following its development.
However, I am afraid it is not yet mature enough for the kind of
application I have been working on (a PyQt-based multiplatform desktop
application for mapping biodiversity data). Documentation is also not as
good as that already available for Matplotlib/Basemap. But I surely intend
to take the time to eventually port my application to Cartopy!

That said, I will be very grateful if you can help me with the 'tricky'
part of my problem. I have also another test version using meshgrid instead
of imshow. It looks promising because only the gridded data is displayed on
the map (while imshow covers the whole map). However, as the attached image
show, I have to fiddle with the grid limits when creating the mesh, and
even so the grid is not yet plotted in the correct position.

With warmest regards,

2017-10-10 17:25 GMT-03:00 Benjamin Root <ben.v.root at gmail.com>:

> So, the problem here is that what you are asking for is in conflict with
> the settings. You are giving bounding box that starts at llcrnrlon=lon_inf
> and goes to urcrnrlon=lon_sup, and that is the extent of the data you are
> providing, too. But you want to view it in a different bounds. It is
> possible to get what you want, but it starts to get tricky at this point.
>
> Before venturing further, I should point out to you that Basemap is
> effectively deprecated. Its support will be terminated in 2020. The CartoPy
> project, also built on top of matplotlib, is a far superior library, and I
> think would be much easier for you to use. It doesn't do everything that
> basemap does yet, but the things that it does do, it does better. I suggest
> checking it out before we go any further here.
>
> Cheers!
> Ben Root
>
>
> On Tue, Oct 10, 2017 at 2:57 PM, Mauro Cavalcanti <maurobio at gmail.com>
> wrote:
>
>> Dear Ben,
>>
>> Thanks a lot again for your help and patience.
>>
>> After modifying the call to Basemap according to your suggestion:
>>
>> m = Basemap(llcrnrlat=lat_inf, llcrnrlon=lon_inf, urcrnrlat=lat_sup,
>> urcrnrlon=lon_sup)
>>
>> the grid is correctly displayed where it belongs (thr west coast of
>> Africa). But now I got a map centered at the Pacific basin, with the grid
>> appearing on the lower left corner of it! In fact, See the attached image.
>> I would like a conventional map centered around the Atlantic basin (ie.,
>> with center coordinates at lat_0=0 and long_0=0).
>>
>> After using the superb Matplotlib for almost a decade (in the beginning,
>> I even got some help here from the legendary John Hunter!), I nonetheless
>> feel somewhat ashamed of having found a potential bug in the library...
>>
>> All the best,
>>
>> 2017-10-10 12:20 GMT-03:00 Benjamin Root <ben.v.root at gmail.com>:
>>
>>> Ah, the problem was that by default, the limits for a Basemap goes from
>>> -180 to 180, and -90 to 90. If you do:
>>>
>>> m = Basemap(llcrnrlat=lat_inf, llcrnrlon=lon_inf,
>>>             urcrnrlat=lat_sup, urcrnrlon=lon_sup)
>>>
>>> Then things line up correctly, and you don't need the origin keyword
>>> argument, the transpose, or even the extent argument.
>>>
>>> In fact, there might even be a bug, as I would have expected specifying
>>> the extent should have worked regardless of the original bounds. Might need
>>> to look into that.
>>>
>>> Cheers!
>>> Ben Root
>>>
>>>
>>> On Mon, Oct 9, 2017 at 7:54 PM, Mauro Cavalcanti <maurobio at gmail.com>
>>> wrote:
>>>
>>>> Hi,
>>>>
>>>> Thanks for you reply and suggestions.
>>>>
>>>> I changed the imshow call to:
>>>>
>>>> im = m.imshow(grilla_salida, cmap='summer', extent=(lon_inf, lat_inf,
>>>> lon_sup, lat_sup), aspect='auto', interpolation='none')
>>>>
>>>> However, the figure is stil wrong (see attachment).
>>>>
>>>> Maybe if instead of imshow, should I use meshgrid/pcolormesh?
>>>>
>>>> Best regards,
>>>>
>>>> 2017-10-09 11:29 GMT-03:00 Benjamin Root <ben.v.root at gmail.com>:
>>>>
>>>>> First, you shouldn't need to transpose your image... that'll
>>>>> effectively rotate the data by 90 degrees. Second, you didn't specify the
>>>>> extents of your image, so Basemap is putting everything starting at
>>>>> coordinate 0,0 in the default projection.
>>>>>
>>>>> If you specify the extent keyword argument to imshow as the (lon1,
>>>>> lat1, lon2, lat2) tuple for the lower-left and upper right corners, you
>>>>> won't even need the origin='lower', and you definitely won't need the
>>>>> transpose.
>>>>>
>>>>> Cheers!
>>>>> Ben Root
>>>>>
>>>>>
>>>>> On Sun, Oct 8, 2017 at 2:33 PM, Mauro Cavalcanti <maurobio at gmail.com>
>>>>> wrote:
>>>>>
>>>>>> Dear ALL,
>>>>>>
>>>>>> I have a simple dataset of longitudes/latitudes (see the attached csv
>>>>>> file).
>>>>>>
>>>>>> From such data, I want to generate a grid like this:
>>>>>>
>>>>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>>>> 0 0 0 0 0 0 1 0 0 0 0 0 0
>>>>>> 0 0 0 0 0 0 1 1 2 0 0 0 0
>>>>>> 0 0 0 0 0 1 1 1 1 0 0 0 0
>>>>>> 0 0 0 1 0 1 0 0 0 0 0 0 0
>>>>>> 0 0 0 2 0 0 0 0 0 0 0 0 0
>>>>>> 0 0 0 0 0 0 0 0 0 1 3 0 0
>>>>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>>>> 0 0 0 0 0 0 0 0 0 1 0 0 0
>>>>>> 0 0 0 0 0 0 0 0 0 1 0 0 0
>>>>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>>>> 0 0 0 0 0 0 0 0 0 0 1 0 0
>>>>>> 0 0 0 0 0 0 0 0 0 0 1 0 0
>>>>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>>>>
>>>>>> which gives the number of data records in each cell of the grid,
>>>>>> using one of the variables in the dataset ("spp") as a categorical
>>>>>> (grouping) factor.
>>>>>>
>>>>>> From this grid, I then want to create a heat map, superimposed on a
>>>>>> Matplotlib/Basemap.
>>>>>>
>>>>>> I wrote some code which does what I want (see the attachments).
>>>>>>
>>>>>> It (mostly) works, but te problem is that the grid image is not being
>>>>>> displayed correctly: as shown in the attached figure, it appears too small,
>>>>>> and in the lower left corner of the map, instead of where it should be (the
>>>>>> West coast of Africa, along the Gulf of Guinea).
>>>>>>
>>>>>> Thanks in advance for any assistance you can provide.
>>>>>>
>>>>>> Best regards,
>>>>>>
>>>>>> --
>>>>>> Dr. Mauro J. Cavalcanti
>>>>>> E-mail: maurobio at gmail.com
>>>>>> Web: http://sites.google.com/site/maurobio
>>>>>> "Life is complex. It consists of real and imaginary parts."
>>>>>>
>>>>>> _______________________________________________
>>>>>> Matplotlib-users mailing list
>>>>>> Matplotlib-users at python.org
>>>>>> https://mail.python.org/mailman/listinfo/matplotlib-users
>>>>>>
>>>>>>
>>>>>
>>>>
>>>>
>>>> --
>>>> Dr. Mauro J. Cavalcanti
>>>> E-mail: maurobio at gmail.com
>>>> Web: http://sites.google.com/site/maurobio
>>>> "Life is complex. It consists of real and imaginary parts."
>>>>
>>>
>>>
>>
>>
>> --
>> Dr. Mauro J. Cavalcanti
>> E-mail: maurobio at gmail.com
>> Web: http://sites.google.com/site/maurobio
>> "Life is complex. It consists of real and imaginary parts."
>>
>
>


-- 
Dr. Mauro J. Cavalcanti
E-mail: maurobio at gmail.com
Web: http://sites.google.com/site/maurobio
"Life is complex. It consists of real and imaginary parts."
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import csv
import numpy as np
from mpl_toolkits.basemap import Basemap
from matplotlib import cm as cmap
import matplotlib.pyplot as plt

import warnings
warnings.filterwarnings("ignore")

#read input data
with open('testdata.csv', 'r') as csvfile:
    reader = csv.reader(csvfile, delimiter=',')
    headers = reader.next()
    input_data = list(reader)

#grid dimensions with one-degree resolution
lat_inf, lon_inf = 0, 0
lat_sup, lon_sup = 90, 360
resolution = 1

latitude, longitude = [], []
latitude = range(lat_inf, lat_sup, resolution)
longitude = range(lon_inf, lon_sup, resolution)

#create output grid
output_grid = []
for i in latitude:
   output_grid.append([])
   for j in longitude:
       output_grid[i].append(0)

#traverse the input_data evaluating the lat, lon coordinates
#summing +1 in the output_grid[latitude][longitude].
for row in input_data:
   lat = int(row[2])
   lon = int(row[3])
   #sp = row[1]

   #check its indexes
   i_lat = latitude.index(lat)
   i_lon = longitude.index(lon)

   #increase counter
   output_grid[i_lat][i_lon] += 1

output_grid = np.array(output_grid, np.int16)

lons,lats = np.meshgrid(np.linspace(1, 30, output_grid.shape[1]+1),
            np.linspace(2, 60, output_grid.shape[0]+1))

#create map
m = Basemap()
m.drawcoastlines(linewidth=0.25)

X, Y = m(lons,lats)

#display image
pc = m.pcolormesh(X, Y, output_grid, cmap='summer')
m.colorbar(pc)
plt.show()
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From jklymak at uvic.ca  Tue Oct 10 19:10:46 2017
From: jklymak at uvic.ca (Jody Klymak)
Date: Tue, 10 Oct 2017 16:10:46 -0700
Subject: [Matplotlib-users] Problem using imshow with Matplotlib/Basemap
In-Reply-To: <CAC1JhZbu=1Uw-S5eOFG7jz95pDuOW1uvx0ZWuO5EfC6mQoLTAQ@mail.gmail.com>
References: <CAC1JhZZeVg76gwauk_D6RWFNau=WF5c+SexqqgDGX4g-PeXi-Q@mail.gmail.com>
 <CANNq6FnS3Oze=E82axb8bwAhPdSM=5MR_EpJcW-RNpzdXmAEdg@mail.gmail.com>
 <CAC1JhZacB18gaM8tb8LWw0Kn0VTPjpP7gNsn7MwaYkakGEXm1A@mail.gmail.com>
 <CANNq6F=bD48rdS5wgVu8L2f6snAWtQ_6PH4HsrWX=o130exHoQ@mail.gmail.com>
 <CAC1JhZZQAuYHh_DeVW1Kv23ariB8-pHk5d=DK8R4qGEyP-8+gg@mail.gmail.com>
 <CANNq6Fk8Rtcd63AN6Fo-spHM6110YnvGgTGu0M04SgfQkU1eUA@mail.gmail.com>
 <CAC1JhZbu=1Uw-S5eOFG7jz95pDuOW1uvx0ZWuO5EfC6mQoLTAQ@mail.gmail.com>
Message-ID: <D41C77E7-3E28-4A14-AFCC-4995AD0E09D6@uvic.ca>

I?m not 100% following, but can?t you just `np.roll` your data by N/2 to get it between -180 and +180 instead of 0 and 360, and set llcrnrlon=-180 and urcrnrlon=180?

I didn?t test, so sorry if that is a dumb suggestion...

Cheers,  Jody





> On 10 Oct 2017, at  16:04 PM, Mauro Cavalcanti <maurobio at gmail.com> wrote:
> 
> Dear Ben,
> 
> Sure, I am aware of Cartopy and have been following its development. However, I am afraid it is not yet mature enough for the kind of application I have been working on (a PyQt-based multiplatform desktop application for mapping biodiversity data). Documentation is also not as good as that already available for Matplotlib/Basemap. But I surely intend to take the time to eventually port my application to Cartopy!
> 
> That said, I will be very grateful if you can help me with the 'tricky' part of my problem. I have also another test version using meshgrid instead of imshow. It looks promising because only the gridded data is displayed on the map (while imshow covers the whole map). However, as the attached image show, I have to fiddle with the grid limits when creating the mesh, and even so the grid is not yet plotted in the correct position. 
> 
> With warmest regards,
> 
> 2017-10-10 17:25 GMT-03:00 Benjamin Root <ben.v.root at gmail.com <mailto:ben.v.root at gmail.com>>:
> So, the problem here is that what you are asking for is in conflict with the settings. You are giving bounding box that starts at llcrnrlon=lon_inf and goes to urcrnrlon=lon_sup, and that is the extent of the data you are providing, too. But you want to view it in a different bounds. It is possible to get what you want, but it starts to get tricky at this point.
> 
> Before venturing further, I should point out to you that Basemap is effectively deprecated. Its support will be terminated in 2020. The CartoPy project, also built on top of matplotlib, is a far superior library, and I think would be much easier for you to use. It doesn't do everything that basemap does yet, but the things that it does do, it does better. I suggest checking it out before we go any further here.
> 
> Cheers!
> Ben Root
> 
> 
> On Tue, Oct 10, 2017 at 2:57 PM, Mauro Cavalcanti <maurobio at gmail.com <mailto:maurobio at gmail.com>> wrote:
> Dear Ben,
> 
> Thanks a lot again for your help and patience. 
> 
> After modifying the call to Basemap according to your suggestion:
> 
> m = Basemap(llcrnrlat=lat_inf, llcrnrlon=lon_inf, urcrnrlat=lat_sup, urcrnrlon=lon_sup)
> 
> the grid is correctly displayed where it belongs (thr west coast of Africa). But now I got a map centered at the Pacific basin, with the grid appearing on the lower left corner of it! In fact, See the attached image. I would like a conventional map centered around the Atlantic basin (ie., with center coordinates at lat_0=0 and long_0=0).
> 
> After using the superb Matplotlib for almost a decade (in the beginning, I even got some help here from the legendary John Hunter!), I nonetheless feel somewhat ashamed of having found a potential bug in the library...
> 
> All the best,
> 
> 2017-10-10 12:20 GMT-03:00 Benjamin Root <ben.v.root at gmail.com <mailto:ben.v.root at gmail.com>>:
> Ah, the problem was that by default, the limits for a Basemap goes from -180 to 180, and -90 to 90. If you do:
> 
> m = Basemap(llcrnrlat=lat_inf, llcrnrlon=lon_inf,
>             urcrnrlat=lat_sup, urcrnrlon=lon_sup)
> 
> Then things line up correctly, and you don't need the origin keyword argument, the transpose, or even the extent argument.
> 
> In fact, there might even be a bug, as I would have expected specifying the extent should have worked regardless of the original bounds. Might need to look into that.
> 
> Cheers!
> Ben Root
> 
> 
> On Mon, Oct 9, 2017 at 7:54 PM, Mauro Cavalcanti <maurobio at gmail.com <mailto:maurobio at gmail.com>> wrote:
> Hi,
> 
> Thanks for you reply and suggestions.
> 
> I changed the imshow call to:
> 
> im = m.imshow(grilla_salida, cmap='summer', extent=(lon_inf, lat_inf, lon_sup, lat_sup), aspect='auto', interpolation='none')
> 
> However, the figure is stil wrong (see attachment).
> 
> Maybe if instead of imshow, should I use meshgrid/pcolormesh?
> 
> Best regards,
> 
> 2017-10-09 11:29 GMT-03:00 Benjamin Root <ben.v.root at gmail.com <mailto:ben.v.root at gmail.com>>:
> First, you shouldn't need to transpose your image... that'll effectively rotate the data by 90 degrees. Second, you didn't specify the extents of your image, so Basemap is putting everything starting at coordinate 0,0 in the default projection.
> 
> If you specify the extent keyword argument to imshow as the (lon1, lat1, lon2, lat2) tuple for the lower-left and upper right corners, you won't even need the origin='lower', and you definitely won't need the transpose.
> 
> Cheers!
> Ben Root
> 
> 
> On Sun, Oct 8, 2017 at 2:33 PM, Mauro Cavalcanti <maurobio at gmail.com <mailto:maurobio at gmail.com>> wrote:
> Dear ALL,
> 
> I have a simple dataset of longitudes/latitudes (see the attached csv file). 
> 
> From such data, I want to generate a grid like this:
> 
> 0 0 0 0 0 0 0 0 0 0 0 0 0 
> 0 0 0 0 0 0 0 0 0 0 0 0 0 
> 0 0 0 0 0 0 0 0 0 0 0 0 0 
> 0 0 0 0 0 0 1 0 0 0 0 0 0 
> 0 0 0 0 0 0 1 1 2 0 0 0 0 
> 0 0 0 0 0 1 1 1 1 0 0 0 0 
> 0 0 0 1 0 1 0 0 0 0 0 0 0 
> 0 0 0 2 0 0 0 0 0 0 0 0 0 
> 0 0 0 0 0 0 0 0 0 1 3 0 0 
> 0 0 0 0 0 0 0 0 0 0 0 0 0 
> 0 0 0 0 0 0 0 0 0 1 0 0 0 
> 0 0 0 0 0 0 0 0 0 1 0 0 0 
> 0 0 0 0 0 0 0 0 0 0 0 0 0 
> 0 0 0 0 0 0 0 0 0 0 1 0 0 
> 0 0 0 0 0 0 0 0 0 0 1 0 0 
> 0 0 0 0 0 0 0 0 0 0 0 0 0 
> 0 0 0 0 0 0 0 0 0 0 0 0 0
> 
> which gives the number of data records in each cell of the grid, using one of the variables in the dataset ("spp") as a categorical (grouping) factor.
> 
> From this grid, I then want to create a heat map, superimposed on a Matplotlib/Basemap.
> 
> I wrote some code which does what I want (see the attachments).
> 
> It (mostly) works, but te problem is that the grid image is not being displayed correctly: as shown in the attached figure, it appears too small, and in the lower left corner of the map, instead of where it should be (the West coast of Africa, along the Gulf of Guinea). 
> 
> Thanks in advance for any assistance you can provide.
> 
> Best regards,
> 
> -- 
> Dr. Mauro J. Cavalcanti
> E-mail: maurobio at gmail.com <mailto:maurobio at gmail.com>
> Web: http://sites.google.com/site/maurobio <http://sites.google.com/site/maurobio>
> "Life is complex. It consists of real and imaginary parts."
> 
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org <mailto:Matplotlib-users at python.org>
> https://mail.python.org/mailman/listinfo/matplotlib-users <https://mail.python.org/mailman/listinfo/matplotlib-users>
> 
> 
> 
> 
> 
> -- 
> Dr. Mauro J. Cavalcanti
> E-mail: maurobio at gmail.com <mailto:maurobio at gmail.com>
> Web: http://sites.google.com/site/maurobio <http://sites.google.com/site/maurobio>
> "Life is complex. It consists of real and imaginary parts."
> 
> 
> 
> 
> -- 
> Dr. Mauro J. Cavalcanti
> E-mail: maurobio at gmail.com <mailto:maurobio at gmail.com>
> Web: http://sites.google.com/site/maurobio <http://sites.google.com/site/maurobio>
> "Life is complex. It consists of real and imaginary parts."
> 
> 
> 
> 
> -- 
> Dr. Mauro J. Cavalcanti
> E-mail: maurobio at gmail.com <mailto:maurobio at gmail.com>
> Web: http://sites.google.com/site/maurobio <http://sites.google.com/site/maurobio>
> "Life is complex. It consists of real and imaginary parts."
> <meshgrid.py><Figure_1.png><testdata.csv>_______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org <mailto:Matplotlib-users at python.org>
> https://mail.python.org/mailman/listinfo/matplotlib-users <https://mail.python.org/mailman/listinfo/matplotlib-users>
--
Jody Klymak    
http://web.uvic.ca/~jklymak/





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From maurobio at gmail.com  Tue Oct 10 19:43:13 2017
From: maurobio at gmail.com (Mauro Cavalcanti)
Date: Tue, 10 Oct 2017 20:43:13 -0300
Subject: [Matplotlib-users] Problem using imshow with Matplotlib/Basemap
In-Reply-To: <D41C77E7-3E28-4A14-AFCC-4995AD0E09D6@uvic.ca>
References: <CAC1JhZZeVg76gwauk_D6RWFNau=WF5c+SexqqgDGX4g-PeXi-Q@mail.gmail.com>
 <CANNq6FnS3Oze=E82axb8bwAhPdSM=5MR_EpJcW-RNpzdXmAEdg@mail.gmail.com>
 <CAC1JhZacB18gaM8tb8LWw0Kn0VTPjpP7gNsn7MwaYkakGEXm1A@mail.gmail.com>
 <CANNq6F=bD48rdS5wgVu8L2f6snAWtQ_6PH4HsrWX=o130exHoQ@mail.gmail.com>
 <CAC1JhZZQAuYHh_DeVW1Kv23ariB8-pHk5d=DK8R4qGEyP-8+gg@mail.gmail.com>
 <CANNq6Fk8Rtcd63AN6Fo-spHM6110YnvGgTGu0M04SgfQkU1eUA@mail.gmail.com>
 <CAC1JhZbu=1Uw-S5eOFG7jz95pDuOW1uvx0ZWuO5EfC6mQoLTAQ@mail.gmail.com>
 <D41C77E7-3E28-4A14-AFCC-4995AD0E09D6@uvic.ca>
Message-ID: <CAC1JhZYZZ3tMh4yWnnijKK9MwVPLShVNH2WVdyCCDtGHu9n3+g@mail.gmail.com>

Dear Jody,

Thank you very much for your suggestion! It looks interesting and I will
play with it. (By the way, I don't consider any suggestion here as "dumb"
and value very much all of them.)

All the best,

2017-10-10 20:10 GMT-03:00 Jody Klymak <jklymak at uvic.ca>:

> I?m not 100% following, but can?t you just `np.roll` your data by N/2 to
> get it between -180 and +180 instead of 0 and 360, and set llcrnrlon=-180
> and urcrnrlon=180?
>
> I didn?t test, so sorry if that is a dumb suggestion...
>
> Cheers,  Jody
>
>
>
>
>
> On 10 Oct 2017, at  16:04 PM, Mauro Cavalcanti <maurobio at gmail.com> wrote:
>
> Dear Ben,
>
> Sure, I am aware of Cartopy and have been following its development.
> However, I am afraid it is not yet mature enough for the kind of
> application I have been working on (a PyQt-based multiplatform desktop
> application for mapping biodiversity data). Documentation is also not as
> good as that already available for Matplotlib/Basemap. But I surely intend
> to take the time to eventually port my application to Cartopy!
>
> That said, I will be very grateful if you can help me with the 'tricky'
> part of my problem. I have also another test version using meshgrid instead
> of imshow. It looks promising because only the gridded data is displayed on
> the map (while imshow covers the whole map). However, as the attached image
> show, I have to fiddle with the grid limits when creating the mesh, and
> even so the grid is not yet plotted in the correct position.
>
> With warmest regards,
>
> 2017-10-10 17:25 GMT-03:00 Benjamin Root <ben.v.root at gmail.com>:
>
>> So, the problem here is that what you are asking for is in conflict with
>> the settings. You are giving bounding box that starts at llcrnrlon=lon_inf
>> and goes to urcrnrlon=lon_sup, and that is the extent of the data you are
>> providing, too. But you want to view it in a different bounds. It is
>> possible to get what you want, but it starts to get tricky at this point.
>>
>> Before venturing further, I should point out to you that Basemap is
>> effectively deprecated. Its support will be terminated in 2020. The CartoPy
>> project, also built on top of matplotlib, is a far superior library, and I
>> think would be much easier for you to use. It doesn't do everything that
>> basemap does yet, but the things that it does do, it does better. I suggest
>> checking it out before we go any further here.
>>
>> Cheers!
>> Ben Root
>>
>>
>> On Tue, Oct 10, 2017 at 2:57 PM, Mauro Cavalcanti <maurobio at gmail.com>
>> wrote:
>>
>>> Dear Ben,
>>>
>>> Thanks a lot again for your help and patience.
>>>
>>> After modifying the call to Basemap according to your suggestion:
>>>
>>> m = Basemap(llcrnrlat=lat_inf, llcrnrlon=lon_inf, urcrnrlat=lat_sup,
>>> urcrnrlon=lon_sup)
>>>
>>> the grid is correctly displayed where it belongs (thr west coast of
>>> Africa). But now I got a map centered at the Pacific basin, with the grid
>>> appearing on the lower left corner of it! In fact, See the attached image.
>>> I would like a conventional map centered around the Atlantic basin (ie.,
>>> with center coordinates at lat_0=0 and long_0=0).
>>>
>>> After using the superb Matplotlib for almost a decade (in the beginning,
>>> I even got some help here from the legendary John Hunter!), I nonetheless
>>> feel somewhat ashamed of having found a potential bug in the library...
>>>
>>> All the best,
>>>
>>> 2017-10-10 12:20 GMT-03:00 Benjamin Root <ben.v.root at gmail.com>:
>>>
>>>> Ah, the problem was that by default, the limits for a Basemap goes from
>>>> -180 to 180, and -90 to 90. If you do:
>>>>
>>>> m = Basemap(llcrnrlat=lat_inf, llcrnrlon=lon_inf,
>>>>             urcrnrlat=lat_sup, urcrnrlon=lon_sup)
>>>>
>>>> Then things line up correctly, and you don't need the origin keyword
>>>> argument, the transpose, or even the extent argument.
>>>>
>>>> In fact, there might even be a bug, as I would have expected specifying
>>>> the extent should have worked regardless of the original bounds. Might need
>>>> to look into that.
>>>>
>>>> Cheers!
>>>> Ben Root
>>>>
>>>>
>>>> On Mon, Oct 9, 2017 at 7:54 PM, Mauro Cavalcanti <maurobio at gmail.com>
>>>> wrote:
>>>>
>>>>> Hi,
>>>>>
>>>>> Thanks for you reply and suggestions.
>>>>>
>>>>> I changed the imshow call to:
>>>>>
>>>>> im = m.imshow(grilla_salida, cmap='summer', extent=(lon_inf, lat_inf,
>>>>> lon_sup, lat_sup), aspect='auto', interpolation='none')
>>>>>
>>>>> However, the figure is stil wrong (see attachment).
>>>>>
>>>>> Maybe if instead of imshow, should I use meshgrid/pcolormesh?
>>>>>
>>>>> Best regards,
>>>>>
>>>>> 2017-10-09 11:29 GMT-03:00 Benjamin Root <ben.v.root at gmail.com>:
>>>>>
>>>>>> First, you shouldn't need to transpose your image... that'll
>>>>>> effectively rotate the data by 90 degrees. Second, you didn't specify the
>>>>>> extents of your image, so Basemap is putting everything starting at
>>>>>> coordinate 0,0 in the default projection.
>>>>>>
>>>>>> If you specify the extent keyword argument to imshow as the (lon1,
>>>>>> lat1, lon2, lat2) tuple for the lower-left and upper right corners, you
>>>>>> won't even need the origin='lower', and you definitely won't need the
>>>>>> transpose.
>>>>>>
>>>>>> Cheers!
>>>>>> Ben Root
>>>>>>
>>>>>>
>>>>>> On Sun, Oct 8, 2017 at 2:33 PM, Mauro Cavalcanti <maurobio at gmail.com>
>>>>>>  wrote:
>>>>>>
>>>>>>> Dear ALL,
>>>>>>>
>>>>>>> I have a simple dataset of longitudes/latitudes (see the attached
>>>>>>> csv file).
>>>>>>>
>>>>>>> From such data, I want to generate a grid like this:
>>>>>>>
>>>>>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>>>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>>>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>>>>> 0 0 0 0 0 0 1 0 0 0 0 0 0
>>>>>>> 0 0 0 0 0 0 1 1 2 0 0 0 0
>>>>>>> 0 0 0 0 0 1 1 1 1 0 0 0 0
>>>>>>> 0 0 0 1 0 1 0 0 0 0 0 0 0
>>>>>>> 0 0 0 2 0 0 0 0 0 0 0 0 0
>>>>>>> 0 0 0 0 0 0 0 0 0 1 3 0 0
>>>>>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>>>>> 0 0 0 0 0 0 0 0 0 1 0 0 0
>>>>>>> 0 0 0 0 0 0 0 0 0 1 0 0 0
>>>>>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>>>>> 0 0 0 0 0 0 0 0 0 0 1 0 0
>>>>>>> 0 0 0 0 0 0 0 0 0 0 1 0 0
>>>>>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>>>>> 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>>>>>
>>>>>>> which gives the number of data records in each cell of the grid,
>>>>>>> using one of the variables in the dataset ("spp") as a categorical
>>>>>>> (grouping) factor.
>>>>>>>
>>>>>>> From this grid, I then want to create a heat map, superimposed on a
>>>>>>> Matplotlib/Basemap.
>>>>>>>
>>>>>>> I wrote some code which does what I want (see the attachments).
>>>>>>>
>>>>>>> It (mostly) works, but te problem is that the grid image is not
>>>>>>> being displayed correctly: as shown in the attached figure, it appears too
>>>>>>> small, and in the lower left corner of the map, instead of where it should
>>>>>>> be (the West coast of Africa, along the Gulf of Guinea).
>>>>>>>
>>>>>>> Thanks in advance for any assistance you can provide.
>>>>>>>
>>>>>>> Best regards,
>>>>>>>
>>>>>>> --
>>>>>>> Dr. Mauro J. Cavalcanti
>>>>>>> E-mail: maurobio at gmail.com
>>>>>>> Web: http://sites.google.com/site/maurobio
>>>>>>> "Life is complex. It consists of real and imaginary parts."
>>>>>>>
>>>>>>> _______________________________________________
>>>>>>> Matplotlib-users mailing list
>>>>>>> Matplotlib-users at python.org
>>>>>>> https://mail.python.org/mailman/listinfo/matplotlib-users
>>>>>>>
>>>>>>>
>>>>>>
>>>>>
>>>>>
>>>>> --
>>>>> Dr. Mauro J. Cavalcanti
>>>>> E-mail: maurobio at gmail.com
>>>>> Web: http://sites.google.com/site/maurobio
>>>>> "Life is complex. It consists of real and imaginary parts."
>>>>>
>>>>
>>>>
>>>
>>>
>>> --
>>> Dr. Mauro J. Cavalcanti
>>> E-mail: maurobio at gmail.com
>>> Web: http://sites.google.com/site/maurobio
>>> "Life is complex. It consists of real and imaginary parts."
>>>
>>
>>
>
>
> --
> Dr. Mauro J. Cavalcanti
> E-mail: maurobio at gmail.com
> Web: http://sites.google.com/site/maurobio
> "Life is complex. It consists of real and imaginary parts."
> <meshgrid.py><Figure_1.png><testdata.csv>_________________
> ______________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org
> https://mail.python.org/mailman/listinfo/matplotlib-users
>
>
> --
> Jody Klymak
> http://web.uvic.ca/~jklymak/
>
>
>
>
>
>


-- 
Dr. Mauro J. Cavalcanti
E-mail: maurobio at gmail.com
Web: http://sites.google.com/site/maurobio
"Life is complex. It consists of real and imaginary parts."
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From Stuart_Mentzer at objexx.com  Tue Oct 10 13:08:29 2017
From: Stuart_Mentzer at objexx.com (Stuart Mentzer)
Date: Tue, 10 Oct 2017 10:08:29 -0700 (MST)
Subject: [Matplotlib-users] Unzoom problem in 2.1.0
In-Reply-To: <CANNq6FkxQJufOe98-najbDY=3v43uv62HqBe2TWJBfhx83o4Mg@mail.gmail.com>
References: <1507606322709-0.post@n5.nabble.com>
 <CANNq6FkxQJufOe98-najbDY=3v43uv62HqBe2TWJBfhx83o4Mg@mail.gmail.com>
Message-ID: <1507655309893-0.post@n5.nabble.com>

Ben,

There is a signal connection set up to call the toolbar's back() method on
right clicks. Using right-click to unzoom is a common plotting UI (e.g., in
Qwt) so we set that up. That is not the problem here.

Both right-click and clicking the back button on the toolbar display the
same buggy behavior I describe.

Thanks,
Stuart



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From vzhao23 at gmail.com  Thu Oct 12 11:59:59 2017
From: vzhao23 at gmail.com (Victor Zhao)
Date: Thu, 12 Oct 2017 11:59:59 -0400
Subject: [Matplotlib-users] MacOS matplotlib broken after `pip install
 matplotlib --upgrade` to version 2.1.0
Message-ID: <CALwhr=SLS9L4qA_sS3+rWvHkd9=+bqH2DS6qWD1QbtizAJkj6Q@mail.gmail.com>

My OS X python installation is through homebrew. I then add python software
using pip. I recently upgraded matplotlib from version 1.4.3 to 2.1.0 via
`pip install matplotlib --upgrade`, and now matplotlib gives me this error
when I try to make a basic plot:

    In [1]: plt.plot(np.arange(10))

---------------------------------------------------------------------------
    TypeError                                 Traceback (most recent call
last)
    <ipython-input-1-a81699eb47e8> in <module>()
    ----> 1 plt.plot(np.arange(10))

    /usr/local/lib/python2.7/site-packages/matplotlib/pyplot.pyc in
plot(*args, **kwargs)
       3227 @_autogen_docstring(Axes.plot)
       3228 def plot(*args, **kwargs):
    -> 3229     ax = gca()
       3230     # Deprecated: allow callers to override the hold state
       3231     # by passing hold=True|False

    /usr/local/lib/python2.7/site-packages/matplotlib/pyplot.pyc in
gca(**kwargs)
        957     matplotlib.figure.Figure.gca : The figure's gca method.
        958     """
    --> 959     return gcf().gca(**kwargs)
        960
        961 # More ways of creating axes:

    /usr/local/lib/python2.7/site-packages/matplotlib/pyplot.pyc in gcf()
        586         return figManager.canvas.figure
        587     else:
    --> 588         return figure()
        589
        590

    /usr/local/lib/python2.7/site-packages/matplotlib/pyplot.pyc in
figure(num, figsize, dpi, facecolor, edgecolor, frameon, FigureClass,
clear, **kwargs)
        554         # FigureManager base class.
        555         if matplotlib.is_interactive():
    --> 556             draw_if_interactive()
        557
        558         if _INSTALL_FIG_OBSERVER:

    /usr/local/lib/python2.7/site-packages/matplotlib/backend_bases.pyc in
draw_if_interactive(cls)
        183             manager = Gcf.get_active()
        184             if manager:
    --> 185                 cls.trigger_manager_draw(manager)
        186
        187     @classmethod

    TypeError: unbound method trigger_manager_draw() must be called with
_BackendMac instance as first argument (got FigureManagerMac instance
instead)

I am not sure how to diagnose this. I do have a working Python 3 matplotlib
however.

Is this actually a bug within version 2.1.0 of matplotlib? Thanks
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From nuncio.m at gmail.com  Fri Oct 13 06:14:04 2017
From: nuncio.m at gmail.com (nuncio)
Date: Fri, 13 Oct 2017 03:14:04 -0700 (MST)
Subject: [Matplotlib-users] contourf with data frames
Message-ID: <1507889644936-0.post@n5.nabble.com>

HI,
I am trying to plot a time height diagram. here is the code used

dates=mpl.dates.date2num(data.index.to_pydatetime())
xh=list(dates)
dfX=pd.DataFrame(xh)
date2= dfX.groupby(np.arange(len(dfX))//400).mean()
dates=num2date(date2)
dfT=pd.DataFrame(data.iloc[:,5].reshape(400,427).T)
#dfT.insert(loc=0,column='',value=date)
hgtl=data.iloc[0:400:,1]
#plot(dates,dfT.iloc[:,0])
contourf(dates,hgtl,dfT.iloc[0:400:,0].T)

this results in the following error
TypeError                                 Traceback (most recent call last)
<ipython-input-133-d1e220b1ce6a> in <module>()
      8 hgtl=data.iloc[0:400:,1]
      9 #plot(dates,dfT.iloc[:,0])
---> 10 contourf(dates,hgtl,dfT.iloc[0:400:,0].T)

/home/nuncio/anaconda2/lib/python2.7/site-packages/matplotlib/pyplot.pyc in
contourf(*args, **kwargs)
   2781         ax.hold(hold)
   2782     try:
-> 2783         ret = ax.contourf(*args, **kwargs)
   2784     finally:
   2785         ax.hold(washold)

/home/nuncio/anaconda2/lib/python2.7/site-packages/matplotlib/__init__.pyc
in inner(ax, *args, **kwargs)
   1810                     warnings.warn(msg % (label_namer,
func.__name__),
   1811                                   RuntimeWarning, stacklevel=2)
-> 1812             return func(ax, *args, **kwargs)
   1813         pre_doc = inner.__doc__
   1814         if pre_doc is None:

/home/nuncio/anaconda2/lib/python2.7/site-packages/matplotlib/axes/_axes.pyc
in contourf(self, *args, **kwargs)
   5650             self.cla()
   5651         kwargs['filled'] = True
-> 5652         return mcontour.QuadContourSet(self, *args, **kwargs)
   5653     contourf.__doc__ = mcontour.QuadContourSet.contour_doc
   5654 

/home/nuncio/anaconda2/lib/python2.7/site-packages/matplotlib/contour.pyc in
__init__(self, ax, *args, **kwargs)
   1422         are described in QuadContourSet.contour_doc.
   1423         """
-> 1424         ContourSet.__init__(self, ax, *args, **kwargs)
   1425 
   1426     def _process_args(self, *args, **kwargs):

/home/nuncio/anaconda2/lib/python2.7/site-packages/matplotlib/contour.pyc in
__init__(self, ax, *args, **kwargs)
    861         self._transform = kwargs.get('transform', None)
    862 
--> 863         self._process_args(*args, **kwargs)
    864         self._process_levels()
    865 

/home/nuncio/anaconda2/lib/python2.7/site-packages/matplotlib/contour.pyc in
_process_args(self, *args, **kwargs)
   1443                 self._corner_mask =
mpl.rcParams['contour.corner_mask']
   1444 
-> 1445             x, y, z = self._contour_args(args, kwargs)
   1446 
   1447             _mask = ma.getmask(z)

/home/nuncio/anaconda2/lib/python2.7/site-packages/matplotlib/contour.pyc in
_contour_args(self, args, kwargs)
   1526             args = args[1:]
   1527         elif Nargs <= 4:
-> 1528             x, y, z = self._check_xyz(args[:3], kwargs)
   1529             args = args[3:]
   1530         else:

/home/nuncio/anaconda2/lib/python2.7/site-packages/matplotlib/contour.pyc in
_check_xyz(self, args, kwargs)
   1555         y = self.ax.convert_yunits(y)
   1556 
-> 1557         x = np.asarray(x, dtype=np.float64)
   1558         y = np.asarray(y, dtype=np.float64)
   1559         z = ma.asarray(args[2], dtype=np.float64)

/home/nuncio/anaconda2/lib/python2.7/site-packages/numpy/core/numeric.pyc in
asarray(a, dtype, order)
    472 
    473     """
--> 474     return array(a, dtype, copy=False, order=order)
    475 
    476 def asanyarray(a, dtype=None, order=None):

*TypeError: float() argument must be a string or a number*

I identify this as the problem with using "dates" in contourf , however I do
not find any problem in running

plot(dates,dfT.iloc[:,0])

How to get around this

THanks and regards
nuncio



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From nuncio.m at gmail.com  Fri Oct 13 07:05:53 2017
From: nuncio.m at gmail.com (nuncio)
Date: Fri, 13 Oct 2017 04:05:53 -0700 (MST)
Subject: [Matplotlib-users] contourf with data frames
Message-ID: <1507892753913-0.post@n5.nabble.com>

I am trying to make a time height contour plot , here is the script used

dates=mpl.dates.date2num(data.index.to_pydatetime())
xh=list(dates)
dfX=pd.DataFrame(xh)
date2= dfX.groupby(np.arange(len(dfX))//400).mean()
dates=num2date(date2)
dfT=pd.DataFrame(data.iloc[:,5].reshape(400,427).T)
#dfT.insert(loc=0,column='',value=date)
hgtl=data.iloc[0:400:,1]
#plot(dates,dfT.iloc[:,0])
contourf(dates,hgtl,dfT.iloc[0:400:,0].T)

this results in the following error


TypeError                                 Traceback (most recent call last)
<ipython-input-140-d1e220b1ce6a> in <module>()
      8 hgtl=data.iloc[0:400:,1]
      9 #plot(dates,dfT.iloc[:,0])
---> 10 contourf(dates,hgtl,dfT.iloc[0:400:,0].T)

/home/nuncio/anaconda2/lib/python2.7/site-packages/matplotlib/pyplot.pyc in
contourf(*args, **kwargs)
   2781         ax.hold(hold)
   2782     try:
-> 2783         ret = ax.contourf(*args, **kwargs)
   2784     finally:
   2785         ax.hold(washold)

/home/nuncio/anaconda2/lib/python2.7/site-packages/matplotlib/__init__.pyc
in inner(ax, *args, **kwargs)
   1810                     warnings.warn(msg % (label_namer,
func.__name__),
   1811                                   RuntimeWarning, stacklevel=2)
-> 1812             return func(ax, *args, **kwargs)
   1813         pre_doc = inner.__doc__
   1814         if pre_doc is None:

/home/nuncio/anaconda2/lib/python2.7/site-packages/matplotlib/axes/_axes.pyc
in contourf(self, *args, **kwargs)
   5650             self.cla()
   5651         kwargs['filled'] = True
-> 5652         return mcontour.QuadContourSet(self, *args, **kwargs)
   5653     contourf.__doc__ = mcontour.QuadContourSet.contour_doc
   5654 

/home/nuncio/anaconda2/lib/python2.7/site-packages/matplotlib/contour.pyc in
__init__(self, ax, *args, **kwargs)
   1422         are described in QuadContourSet.contour_doc.
   1423         """
-> 1424         ContourSet.__init__(self, ax, *args, **kwargs)
   1425 
   1426     def _process_args(self, *args, **kwargs):

/home/nuncio/anaconda2/lib/python2.7/site-packages/matplotlib/contour.pyc in
__init__(self, ax, *args, **kwargs)
    861         self._transform = kwargs.get('transform', None)
    862 
--> 863         self._process_args(*args, **kwargs)
    864         self._process_levels()
    865 

/home/nuncio/anaconda2/lib/python2.7/site-packages/matplotlib/contour.pyc in
_process_args(self, *args, **kwargs)
   1443                 self._corner_mask =
mpl.rcParams['contour.corner_mask']
   1444 
-> 1445             x, y, z = self._contour_args(args, kwargs)
   1446 
   1447             _mask = ma.getmask(z)

/home/nuncio/anaconda2/lib/python2.7/site-packages/matplotlib/contour.pyc in
_contour_args(self, args, kwargs)
   1526             args = args[1:]
   1527         elif Nargs <= 4:
-> 1528             x, y, z = self._check_xyz(args[:3], kwargs)
   1529             args = args[3:]
   1530         else:

/home/nuncio/anaconda2/lib/python2.7/site-packages/matplotlib/contour.pyc in
_check_xyz(self, args, kwargs)
   1555         y = self.ax.convert_yunits(y)
   1556 
-> 1557         x = np.asarray(x, dtype=np.float64)
   1558         y = np.asarray(y, dtype=np.float64)
   1559         z = ma.asarray(args[2], dtype=np.float64)

/home/nuncio/anaconda2/lib/python2.7/site-packages/numpy/core/numeric.pyc in
asarray(a, dtype, order)
    472 
    473     """
--> 474     return array(a, dtype, copy=False, order=order)
    475 
    476 def asanyarray(a, dtype=None, order=None):
*
TypeError: float() argument must be a string or a number*

I undesrstand the problem is with the dates

however 

plot(dates,dfT.iloc[:,0])

runs perfectly

Any suggestions to get around it

THanks and regards
nuncio



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From jklymak at uvic.ca  Fri Oct 13 10:47:13 2017
From: jklymak at uvic.ca (Jody Klymak)
Date: Fri, 13 Oct 2017 07:47:13 -0700
Subject: [Matplotlib-users] MacOS matplotlib broken after `pip install
 matplotlib --upgrade` to version 2.1.0
In-Reply-To: <CALwhr=SLS9L4qA_sS3+rWvHkd9=+bqH2DS6qWD1QbtizAJkj6Q@mail.gmail.com>
References: <CALwhr=SLS9L4qA_sS3+rWvHkd9=+bqH2DS6qWD1QbtizAJkj6Q@mail.gmail.com>
Message-ID: <44163313-E2B6-4D28-AED7-FB5C2A782FC4@uvic.ca>

Hi Victor,

This is apparently a bug that will be patched soon.  I?d roll back to 2.0.2 in the meantime if you aren?t comfortable working off github?.

https://github.com/matplotlib/matplotlib/issues/9345 <https://github.com/matplotlib/matplotlib/issues/9345>

Cheers,   Jody

> On Oct 12, 2017, at  8:59 AM, Victor Zhao <vzhao23 at gmail.com> wrote:
> 
> My OS X python installation is through homebrew. I then add python software using pip. I recently upgraded matplotlib from version 1.4.3 to 2.1.0 via `pip install matplotlib --upgrade`, and now matplotlib gives me this error when I try to make a basic plot:
> 
>     In [1]: plt.plot(np.arange(10))
>     ---------------------------------------------------------------------------
>     TypeError                                 Traceback (most recent call last)
>     <ipython-input-1-a81699eb47e8> in <module>()
>     ----> 1 plt.plot(np.arange(10))
>     
>     /usr/local/lib/python2.7/site-packages/matplotlib/pyplot.pyc in plot(*args, **kwargs)
>        3227 @_autogen_docstring(Axes.plot)
>        3228 def plot(*args, **kwargs):
>     -> 3229     ax = gca()
>        3230     # Deprecated: allow callers to override the hold state
>        3231     # by passing hold=True|False
>     
>     /usr/local/lib/python2.7/site-packages/matplotlib/pyplot.pyc in gca(**kwargs)
>         957     matplotlib.figure.Figure.gca : The figure's gca method.
>         958     """
>     --> 959     return gcf().gca(**kwargs)
>         960
>         961 # More ways of creating axes:
>     
>     /usr/local/lib/python2.7/site-packages/matplotlib/pyplot.pyc in gcf()
>         586         return figManager.canvas.figure
>         587     else:
>     --> 588         return figure()
>         589
>         590
>     
>     /usr/local/lib/python2.7/site-packages/matplotlib/pyplot.pyc in figure(num, figsize, dpi, facecolor, edgecolor, frameon, FigureClass, clear, **kwargs)
>         554         # FigureManager base class.
>         555         if matplotlib.is_interactive():
>     --> 556             draw_if_interactive()
>         557
>         558         if _INSTALL_FIG_OBSERVER:
>     
>     /usr/local/lib/python2.7/site-packages/matplotlib/backend_bases.pyc in draw_if_interactive(cls)
>         183             manager = Gcf.get_active()
>         184             if manager:
>     --> 185                 cls.trigger_manager_draw(manager)
>         186
>         187     @classmethod
>     
>     TypeError: unbound method trigger_manager_draw() must be called with _BackendMac instance as first argument (got FigureManagerMac instance instead)
> 
> I am not sure how to diagnose this. I do have a working Python 3 matplotlib however.
> 
> Is this actually a bug within version 2.1.0 of matplotlib? Thanks
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org
> https://mail.python.org/mailman/listinfo/matplotlib-users

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From tcaswell at gmail.com  Fri Oct 13 17:21:25 2017
From: tcaswell at gmail.com (Thomas Caswell)
Date: Fri, 13 Oct 2017 21:21:25 +0000
Subject: [Matplotlib-users] contourf with data frames
In-Reply-To: <1507892753913-0.post@n5.nabble.com>
References: <1507892753913-0.post@n5.nabble.com>
Message-ID: <CAA48SF9hWrCu_UYqQFdogAGHekSPzKVNXgawE0FpTZmSDbgQ-A@mail.gmail.com>

It looks like you are converting the dates back to datetime objects (`dates
= num2date(date2)`).   This smells like a bug in contour that it does not
properly handle units on the input (but `ax.plot` does)

The easiest way around this (with out patching Matplotlib ;) ) is to plot
with the dates converted to numbers and then manually set the x axis ticker
and locator to be the date related ones.

Tom



On Fri, Oct 13, 2017 at 9:00 AM nuncio <nuncio.m at gmail.com> wrote:

> I am trying to make a time height contour plot , here is the script used
>
> dates=mpl.dates.date2num(data.index.to_pydatetime())
> xh=list(dates)
> dfX=pd.DataFrame(xh)
> date2= dfX.groupby(np.arange(len(dfX))//400).mean()
> dates=num2date(date2)
> dfT=pd.DataFrame(data.iloc[:,5].reshape(400,427).T)
> #dfT.insert(loc=0,column='',value=date)
> hgtl=data.iloc[0:400:,1]
> #plot(dates,dfT.iloc[:,0])
> contourf(dates,hgtl,dfT.iloc[0:400:,0].T)
>
> this results in the following error
>
>
> TypeError                                 Traceback (most recent call last)
> <ipython-input-140-d1e220b1ce6a> in <module>()
>       8 hgtl=data.iloc[0:400:,1]
>       9 #plot(dates,dfT.iloc[:,0])
> ---> 10 contourf(dates,hgtl,dfT.iloc[0:400:,0].T)
>
> /home/nuncio/anaconda2/lib/python2.7/site-packages/matplotlib/pyplot.pyc in
> contourf(*args, **kwargs)
>    2781         ax.hold(hold)
>    2782     try:
> -> 2783         ret = ax.contourf(*args, **kwargs)
>    2784     finally:
>    2785         ax.hold(washold)
>
> /home/nuncio/anaconda2/lib/python2.7/site-packages/matplotlib/__init__.pyc
> in inner(ax, *args, **kwargs)
>    1810                     warnings.warn(msg % (label_namer,
> func.__name__),
>    1811                                   RuntimeWarning, stacklevel=2)
> -> 1812             return func(ax, *args, **kwargs)
>    1813         pre_doc = inner.__doc__
>    1814         if pre_doc is None:
>
>
> /home/nuncio/anaconda2/lib/python2.7/site-packages/matplotlib/axes/_axes.pyc
> in contourf(self, *args, **kwargs)
>    5650             self.cla()
>    5651         kwargs['filled'] = True
> -> 5652         return mcontour.QuadContourSet(self, *args, **kwargs)
>    5653     contourf.__doc__ = mcontour.QuadContourSet.contour_doc
>    5654
>
> /home/nuncio/anaconda2/lib/python2.7/site-packages/matplotlib/contour.pyc
> in
> __init__(self, ax, *args, **kwargs)
>    1422         are described in QuadContourSet.contour_doc.
>    1423         """
> -> 1424         ContourSet.__init__(self, ax, *args, **kwargs)
>    1425
>    1426     def _process_args(self, *args, **kwargs):
>
> /home/nuncio/anaconda2/lib/python2.7/site-packages/matplotlib/contour.pyc
> in
> __init__(self, ax, *args, **kwargs)
>     861         self._transform = kwargs.get('transform', None)
>     862
> --> 863         self._process_args(*args, **kwargs)
>     864         self._process_levels()
>     865
>
> /home/nuncio/anaconda2/lib/python2.7/site-packages/matplotlib/contour.pyc
> in
> _process_args(self, *args, **kwargs)
>    1443                 self._corner_mask =
> mpl.rcParams['contour.corner_mask']
>    1444
> -> 1445             x, y, z = self._contour_args(args, kwargs)
>    1446
>    1447             _mask = ma.getmask(z)
>
> /home/nuncio/anaconda2/lib/python2.7/site-packages/matplotlib/contour.pyc
> in
> _contour_args(self, args, kwargs)
>    1526             args = args[1:]
>    1527         elif Nargs <= 4:
> -> 1528             x, y, z = self._check_xyz(args[:3], kwargs)
>    1529             args = args[3:]
>    1530         else:
>
> /home/nuncio/anaconda2/lib/python2.7/site-packages/matplotlib/contour.pyc
> in
> _check_xyz(self, args, kwargs)
>    1555         y = self.ax.convert_yunits(y)
>    1556
> -> 1557         x = np.asarray(x, dtype=np.float64)
>    1558         y = np.asarray(y, dtype=np.float64)
>    1559         z = ma.asarray(args[2], dtype=np.float64)
>
> /home/nuncio/anaconda2/lib/python2.7/site-packages/numpy/core/numeric.pyc
> in
> asarray(a, dtype, order)
>     472
>     473     """
> --> 474     return array(a, dtype, copy=False, order=order)
>     475
>     476 def asanyarray(a, dtype=None, order=None):
> *
> TypeError: float() argument must be a string or a number*
>
> I undesrstand the problem is with the dates
>
> however
>
> plot(dates,dfT.iloc[:,0])
>
> runs perfectly
>
> Any suggestions to get around it
>
> THanks and regards
> nuncio
>
>
>
> --
> Sent from:
> http://matplotlib.1069221.n5.nabble.com/matplotlib-users-f3.html
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org
> https://mail.python.org/mailman/listinfo/matplotlib-users
>
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From winash12 at gmail.com  Wed Oct 18 16:59:11 2017
From: winash12 at gmail.com (ashwin .D)
Date: Thu, 19 Oct 2017 02:29:11 +0530
Subject: [Matplotlib-users] Obtaining the contour data in terms of latitude
 and longitude
Message-ID: <CAH0LXy4O++PS3morh4B1L-83TvDqRG64YMkDrDe3LKQewFUSiw@mail.gmail.com>

Hello,
         I have data on a 2d lat lon grid(equal grid spacing) in terms of
laitude and longitude and heights as enclosed in the image. I need the
coordinates(in terms of latitude and longitude) of the contours( as shown
in red in the image). From this answer -
https://stackoverflow.com/questions/18304722/python-find-contour-lines-from-matplotlib-pyplot-contour/
will this give me what I am looking for or is there a scaling required to
latitude and longitude that I need to include?

Best regards,
Ashwin.

import numpy as np
def get_contour_verts(cn):
    contours = []
    # for each contour line
    for cc in cn.collections:
        paths = []
        # for each separate section of the contour line
        for pp in cc.get_paths():
            xy = []
            # for each segment of that section
            for vv in pp.iter_segments():
                xy.append(vv[0])
            paths.append(np.vstack(xy))
        contours.append(paths)

    return contours
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From ben.v.root at gmail.com  Wed Oct 18 17:08:16 2017
From: ben.v.root at gmail.com (Benjamin Root)
Date: Wed, 18 Oct 2017 17:08:16 -0400
Subject: [Matplotlib-users] Obtaining the contour data in terms of
 latitude and longitude
In-Reply-To: <CAH0LXy4O++PS3morh4B1L-83TvDqRG64YMkDrDe3LKQewFUSiw@mail.gmail.com>
References: <CAH0LXy4O++PS3morh4B1L-83TvDqRG64YMkDrDe3LKQewFUSiw@mail.gmail.com>
Message-ID: <CANNq6FnAWDDv1vahui+WZY=JYwjEcQpiHDQUPxDKhbk-fEb6tQ@mail.gmail.com>

If you passed the coordinate arrays in with the call to contourf(), then
you will get what you expect.

I should note that you should watch out for path simplification (you'll
want to turn it off). I also recommend using contourf() over contour()
because it handles the edges of the domain better and NaNs.

Cheers!
Ben Root


On Wed, Oct 18, 2017 at 4:59 PM, ashwin .D <winash12 at gmail.com> wrote:

> Hello,
>          I have data on a 2d lat lon grid(equal grid spacing) in terms of
> laitude and longitude and heights as enclosed in the image. I need the
> coordinates(in terms of latitude and longitude) of the contours( as shown
> in red in the image). From this answer - https://stackoverflow.com/
> questions/18304722/python-find-contour-lines-from-
> matplotlib-pyplot-contour/ will this give me what I am looking for or is
> there a scaling required to latitude and longitude that I need to include?
>
> Best regards,
> Ashwin.
>
> import numpy as np
> def get_contour_verts(cn):
>     contours = []
>     # for each contour line
>     for cc in cn.collections:
>         paths = []
>         # for each separate section of the contour line
>         for pp in cc.get_paths():
>             xy = []
>             # for each segment of that section
>             for vv in pp.iter_segments():
>                 xy.append(vv[0])
>             paths.append(np.vstack(xy))
>         contours.append(paths)
>
>     return contours
>
>
>
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org
> https://mail.python.org/mailman/listinfo/matplotlib-users
>
>
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From rmay31 at gmail.com  Wed Oct 18 17:38:42 2017
From: rmay31 at gmail.com (Ryan May)
Date: Wed, 18 Oct 2017 15:38:42 -0600
Subject: [Matplotlib-users] Obtaining the contour data in terms of
 latitude and longitude
In-Reply-To: <CAH0LXy4O++PS3morh4B1L-83TvDqRG64YMkDrDe3LKQewFUSiw@mail.gmail.com>
References: <CAH0LXy4O++PS3morh4B1L-83TvDqRG64YMkDrDe3LKQewFUSiw@mail.gmail.com>
Message-ID: <CAKH0P+X7pjC-pgb63qo9XfR2=5NU8D94nGsgY4ymirhNbqQvyQ@mail.gmail.com>

For this problem, though, I don't think contouring is what you want. I
think you'd be better off looking for some spatial consistency in the
gradient field directly rather than going through contours.

Ryan

On Wed, Oct 18, 2017 at 2:59 PM, ashwin .D <winash12 at gmail.com> wrote:

> Hello,
>          I have data on a 2d lat lon grid(equal grid spacing) in terms of
> laitude and longitude and heights as enclosed in the image. I need the
> coordinates(in terms of latitude and longitude) of the contours( as shown
> in red in the image). From this answer - https://stackoverflow.com/
> questions/18304722/python-find-contour-lines-from-
> matplotlib-pyplot-contour/ will this give me what I am looking for or is
> there a scaling required to latitude and longitude that I need to include?
>
> Best regards,
> Ashwin.
>
> import numpy as np
> def get_contour_verts(cn):
>     contours = []
>     # for each contour line
>     for cc in cn.collections:
>         paths = []
>         # for each separate section of the contour line
>         for pp in cc.get_paths():
>             xy = []
>             # for each segment of that section
>             for vv in pp.iter_segments():
>                 xy.append(vv[0])
>             paths.append(np.vstack(xy))
>         contours.append(paths)
>
>     return contours
>
>
>
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org
> https://mail.python.org/mailman/listinfo/matplotlib-users
>
>


-- 
Ryan May
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From winash12 at gmail.com  Wed Oct 18 17:32:28 2017
From: winash12 at gmail.com (ashwin .D)
Date: Thu, 19 Oct 2017 03:02:28 +0530
Subject: [Matplotlib-users] Obtaining the contour data in terms of
 latitude and longitude
In-Reply-To: <CANNq6FnAWDDv1vahui+WZY=JYwjEcQpiHDQUPxDKhbk-fEb6tQ@mail.gmail.com>
References: <CAH0LXy4O++PS3morh4B1L-83TvDqRG64YMkDrDe3LKQewFUSiw@mail.gmail.com>
 <CANNq6FnAWDDv1vahui+WZY=JYwjEcQpiHDQUPxDKhbk-fEb6tQ@mail.gmail.com>
Message-ID: <CAH0LXy7fMeFfMru5Yq63+rG2gxWF=gFZjvtzeaQnp=A3VN5GnQ@mail.gmail.com>

Ben,
        Thanks ! Can you explain a bit more on what you mean by path
simplification ? Here is the code as I see it right now.

Also I had a second question - cs.levels returns the contour levels
associated with the contours. Is there a dictionary that maps a contour
level to the contour coordinates or is cs.collections in the same order as
cs.levels?

import numpy as np
from netCDF4 import Dataset
import matplotlib.pyplot as plt

clevs = np.arange(5300.,5900.,20.)
cs = plt.contourf(lons, lats, np.squeeze(hgt),clevs)
for cc in cs.collections:
    paths = []
    for pp in cc.get_paths():
        xy = []
        for vv in pp.iter_segments():
            xy.append(vv[0])
            paths.append(np.vstack(xy))
        contours.append(paths)


On Thu, Oct 19, 2017 at 2:38 AM, Benjamin Root <ben.v.root at gmail.com> wrote:

> If you passed the coordinate arrays in with the call to contourf(), then
> you will get what you expect.
>
> I should note that you should watch out for path simplification (you'll
> want to turn it off). I also recommend using contourf() over contour()
> because it handles the edges of the domain better and NaNs.
>
> Cheers!
> Ben Root
>
>
> On Wed, Oct 18, 2017 at 4:59 PM, ashwin .D <winash12 at gmail.com> wrote:
>
>> Hello,
>>          I have data on a 2d lat lon grid(equal grid spacing) in terms of
>> laitude and longitude and heights as enclosed in the image. I need the
>> coordinates(in terms of latitude and longitude) of the contours( as shown
>> in red in the image). From this answer - https://stackoverflow.com/ques
>> tions/18304722/python-find-contour-lines-from-matplotlib-pyplot-contour/
>> will this give me what I am looking for or is there a scaling required to
>> latitude and longitude that I need to include?
>>
>> Best regards,
>> Ashwin.
>>
>> import numpy as np
>> def get_contour_verts(cn):
>>     contours = []
>>     # for each contour line
>>     for cc in cn.collections:
>>         paths = []
>>         # for each separate section of the contour line
>>         for pp in cc.get_paths():
>>             xy = []
>>             # for each segment of that section
>>             for vv in pp.iter_segments():
>>                 xy.append(vv[0])
>>             paths.append(np.vstack(xy))
>>         contours.append(paths)
>>
>>     return contours
>>
>>
>>
>> _______________________________________________
>> Matplotlib-users mailing list
>> Matplotlib-users at python.org
>> https://mail.python.org/mailman/listinfo/matplotlib-users
>>
>>
>
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From winash12 at gmail.com  Thu Oct 19 04:48:24 2017
From: winash12 at gmail.com (ashwin .D)
Date: Thu, 19 Oct 2017 14:18:24 +0530
Subject: [Matplotlib-users] Obtaining the contour data in terms of
 latitude and longitude
In-Reply-To: <CANNq6FnAWDDv1vahui+WZY=JYwjEcQpiHDQUPxDKhbk-fEb6tQ@mail.gmail.com>
References: <CAH0LXy4O++PS3morh4B1L-83TvDqRG64YMkDrDe3LKQewFUSiw@mail.gmail.com>
 <CANNq6FnAWDDv1vahui+WZY=JYwjEcQpiHDQUPxDKhbk-fEb6tQ@mail.gmail.com>
Message-ID: <CAH0LXy6KkztP75o8yTpAdd_zyOcJnpdccZirxHKegHxNzMVoUg@mail.gmail.com>

Hi Ben,
             I believe have a better idea of my doubt  and I am just
thinking loudly here - apologies if all of this is irrelevant. Usually when
you have a raster grid  you need to define it with respect to a origin. So
since in my case I have a global grid the lower left hand corner is taken
to be the origin. This is usually -90 S, 180 W. But you could also have a
subset of the global grid in which case the lower left corner's  latitude
and longitude can be considered as the origin. This is important especially
if you want to know what is grid  orientation of the contour with respect
to the origin. I presume I do not need to worry about this with
matplotlib's contourf  - am I correct on that or put it differently does it
even matter ?

Looking in the code collections.py under lib/matplotlib I see some
references to rasterizing a grid and affine transformations. I am not sure
whether that is related to any of what I just asked.

Best regards,
Ashwin.

On Thu, Oct 19, 2017 at 2:38 AM, Benjamin Root <ben.v.root at gmail.com> wrote:

> If you passed the coordinate arrays in with the call to contourf(), then
> you will get what you expect.
>
> I should note that you should watch out for path simplification (you'll
> want to turn it off). I also recommend using contourf() over contour()
> because it handles the edges of the domain better and NaNs.
>
> Cheers!
> Ben Root
>
>
> On Wed, Oct 18, 2017 at 4:59 PM, ashwin .D <winash12 at gmail.com> wrote:
>
>> Hello,
>>          I have data on a 2d lat lon grid(equal grid spacing) in terms of
>> laitude and longitude and heights as enclosed in the image. I need the
>> coordinates(in terms of latitude and longitude) of the contours( as shown
>> in red in the image). From this answer - https://stackoverflow.com/ques
>> tions/18304722/python-find-contour-lines-from-matplotlib-pyplot-contour/
>> will this give me what I am looking for or is there a scaling required to
>> latitude and longitude that I need to include?
>>
>> Best regards,
>> Ashwin.
>>
>> import numpy as np
>> def get_contour_verts(cn):
>>     contours = []
>>     # for each contour line
>>     for cc in cn.collections:
>>         paths = []
>>         # for each separate section of the contour line
>>         for pp in cc.get_paths():
>>             xy = []
>>             # for each segment of that section
>>             for vv in pp.iter_segments():
>>                 xy.append(vv[0])
>>             paths.append(np.vstack(xy))
>>         contours.append(paths)
>>
>>     return contours
>>
>>
>>
>> _______________________________________________
>> Matplotlib-users mailing list
>> Matplotlib-users at python.org
>> https://mail.python.org/mailman/listinfo/matplotlib-users
>>
>>
>
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From jslavin at cfa.harvard.edu  Thu Oct 19 08:32:36 2017
From: jslavin at cfa.harvard.edu (Slavin, Jonathan)
Date: Thu, 19 Oct 2017 08:32:36 -0400
Subject: [Matplotlib-users] Obtaining the contour data in terms of
 latitude and longitude
Message-ID: <CACcRS=c6ifZv2kgb328MGE=Bk7=Q8mXZ61wCh0u6xKtkXcA2KA@mail.gmail.com>

I'm wondering if we could make this easier by modifying contour/contourf so
that the ContourSet object created had a paths object attribute.  It
doesn't seem like it'd be that hard.  Maybe some would object that it would
slow down the creation of contour plots?

Jon

On Wed, Oct 18, 2017 at 5:08 PM, <matplotlib-users-request at python.org>
wrote:
>
> Date: Thu, 19 Oct 2017 02:29:11 +0530
> From: "ashwin .D" <winash12 at gmail.com>
> To: matplotlib-users at python.org
> Subject: [Matplotlib-users] Obtaining the contour data in terms of
>         latitude and longitude
> Message-ID:
>         <CAH0LXy4O++PS3morh4B1L-83TvDqRG64YMkDrDe3LKQewFUSiw@
> mail.gmail.com>
> Content-Type: text/plain; charset="utf-8"
>
> Hello,
>          I have data on a 2d lat lon grid(equal grid spacing) in terms of
> laitude and longitude and heights as enclosed in the image. I need the
> coordinates(in terms of latitude and longitude) of the contours( as shown
> in red in the image). From this answer -
> https://stackoverflow.com/questions/18304722/python-
> find-contour-lines-from-matplotlib-pyplot-contour/
> will this give me what I am looking for or is there a scaling required to
> latitude and longitude that I need to include?
>
> Best regards,
> Ashwin.
>
> import numpy as np
> def get_contour_verts(cn):
>     contours = []
>     # for each contour line
>     for cc in cn.collections:
>         paths = []
>         # for each separate section of the contour line
>         for pp in cc.get_paths():
>             xy = []
>             # for each segment of that section
>             for vv in pp.iter_segments():
>                 xy.append(vv[0])
>             paths.append(np.vstack(xy))
>         contours.append(paths)
>
>     return contours
> -------------- next part --------------
> An HTML attachment was scrubbed...
> URL: <http://mail.python.org/pipermail/matplotlib-users/
> attachments/20171019/c5cb0664/attachment-0001.html>
> -------------- next part --------------
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> attachments/20171019/c5cb0664/attachment-0001.gif>
>
> ------------------------------
>
> Message: 2
> Date: Wed, 18 Oct 2017 17:08:16 -0400
> From: Benjamin Root <ben.v.root at gmail.com>
> To: "ashwin .D" <winash12 at gmail.com>
> Cc: Matplotlib-users <matplotlib-users at python.org>
> Subject: Re: [Matplotlib-users] Obtaining the contour data in terms of
>         latitude and longitude
> Message-ID:
>         <CANNq6FnAWDDv1vahui+WZY=JYwjEcQpiHDQUPxDKhbk-fEb6tQ at mail.
> gmail.com>
> Content-Type: text/plain; charset="utf-8"
>
> If you passed the coordinate arrays in with the call to contourf(), then
> you will get what you expect.
>
> I should note that you should watch out for path simplification (you'll
> want to turn it off). I also recommend using contourf() over contour()
> because it handles the edges of the domain better and NaNs.
>
> Cheers!
> Ben Root
>
>
> On Wed, Oct 18, 2017 at 4:59 PM, ashwin .D <winash12 at gmail.com> wrote:
>
> > Hello,
> >          I have data on a 2d lat lon grid(equal grid spacing) in terms of
> > laitude and longitude and heights as enclosed in the image. I need the
> > coordinates(in terms of latitude and longitude) of the contours( as shown
> > in red in the image). From this answer - https://stackoverflow.com/
> > questions/18304722/python-find-contour-lines-from-
> > matplotlib-pyplot-contour/ will this give me what I am looking for or is
> > there a scaling required to latitude and longitude that I need to
> include?
> >
> > Best regards,
> > Ashwin.
> >
> > import numpy as np
> > def get_contour_verts(cn):
> >     contours = []
> >     # for each contour line
> >     for cc in cn.collections:
> >         paths = []
> >         # for each separate section of the contour line
> >         for pp in cc.get_paths():
> >             xy = []
> >             # for each segment of that section
> >             for vv in pp.iter_segments():
> >                 xy.append(vv[0])
> >             paths.append(np.vstack(xy))
> >         contours.append(paths)
> >
> >     return contours
> >
> >
> >
> > _______________________________________________
> > Matplotlib-users mailing list
> > Matplotlib-users at python.org
> > https://mail.python.org/mailman/listinfo/matplotlib-users
> >
> >
> -------------- next part --------------
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>
> ------------------------------
>
> Subject: Digest Footer
>
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org
> https://mail.python.org/mailman/listinfo/matplotlib-users
>
>
> ------------------------------
>
> End of Matplotlib-users Digest, Vol 27, Issue 24
> ************************************************
>



-- 
________________________________________________________
Jonathan D. Slavin                 Harvard-Smithsonian CfA
jslavin at cfa.harvard.edu       60 Garden Street, MS 83
phone: (617) 496-7981       Cambridge, MA 02138-1516
cell: (781) 363-0035             USA
________________________________________________________
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From ben.v.root at gmail.com  Thu Oct 19 10:32:35 2017
From: ben.v.root at gmail.com (Benjamin Root)
Date: Thu, 19 Oct 2017 10:32:35 -0400
Subject: [Matplotlib-users] Problem of memory leak using
 FigureCanvasTkAgg
In-Reply-To: <1508423218080.28805@etu.univ-lyon1.fr>
References: <1494500114526.97210@etu.univ-lyon1.fr>
 <CAA48SF9o=5nhn8c-FUYgKcQm041Whm=ZZ2N+JXFVY1osCTn=Wg@mail.gmail.com>
 <CANNq6FkNcreCubg_0pdNn05vUSS4VAjDD=L_=9k90c9v93YUaQ@mail.gmail.com>
 <1508423218080.28805@etu.univ-lyon1.fr>
Message-ID: <CANNq6FkpsddCUy_L4=58Wr4Uw8z5kcjnR5F=nHLP43jAsDixLA@mail.gmail.com>

Did the rate of memory leak change?

On Thu, Oct 19, 2017 at 10:26 AM, GARCIA PIERRE SIMON p0904379 <
pierre.garcia at etu.univ-lyon1.fr> wrote:

> Thank you for your answer but I updated MPL to 2.1.0 and the memory leak
> is still there. Also, I skipped entirely pyplot and the result is strictly
> the same.
>
>
> Pierre
> ------------------------------
> *De :* Benjamin Root <ben.v.root at gmail.com>
> *Envoy? :* lundi 7 ao?t 2017 19:27
> *? :* Thomas Caswell
> *Cc :* GARCIA PIERRE SIMON p0904379; matplotlib-users at python.org
> *Objet :* Re: [Matplotlib-users] Problem of memory leak using
> FigureCanvasTkAgg
>
> The stackoverflow posting notes that mpl v1.1.1 was used, which is 5 years
> old. I am pretty sure we have made a lot of memory leak fixes since then. I
> noted as such on SO.
>
> Ben Root
>
> On Sat, Aug 5, 2017 at 7:48 PM, Thomas Caswell <tcaswell at gmail.com> wrote:
>
>> It is likely there things are not being garbage collected due to either a
>> missed reference surviving or the circular references between the various
>> objects is delaying gc from collecting them (I do not recall the details
>> exactly off the top of my head, but running gc manually with more
>> generations may help).
>>
>> However, if you are embedding in a GUI you are probably better off just
>> doing the embedding directly (and skipping `pyplot` entirely, see
>> https://matplotlib.org/examples/user_interfaces/embedding_in_tk.html ).
>> Re-using the Figures / Axes / Artists to update the data (rather than
>> starting from scratch every time) may also prevent the memory leaks _and_
>> give you a performance boost.
>>
>> Please subscribe to the list so that you can post without moderation.
>>
>> Tom
>>
>> On Sat, Aug 5, 2017 at 12:36 PM GARCIA PIERRE SIMON p0904379 <
>> pierre.garcia at etu.univ-lyon1.fr> wrote:
>>
>>> Hello,
>>>
>>> I'm PhD student in Bioinformatic and I'm working on a project using
>>> matplotlib and Tkinter but I've got a big problem of memory leak using
>>> these modules. I'll be very grateful if you could help me to solve this
>>> problem. Here is the link explaining what happends, with a reduced script:
>>> http://stackoverflow.com/questions/43097378/matplotlib-vs-
>>> tkinter-memory-leak-using-figurecanvastkagg
>>>
>>> Thank you in advance,
>>>
>>>
>>> Sincerly,
>>>
>>>
>>>
>>>
>>> Pierre Garcia, PhD Student
>>> ---------
>>> Bases Mol?culaires et Structurales des Syst?mes infectieux-UMR5086
>>> Institut de Biologie et Chimie des Prot?ines
>>> 7 passage du Vercors
>>> <https://maps.google.com/?q=7+passage+du+Vercors&entry=gmail&source=g>
>>> 69 367 Lyon cedex 07
>>> ---------
>>> Laboratoire de Biom?trie et Biologie Evolutive-UMR5558
>>> UCBL Lyon 1 -B?t. Gr?gory Mendel
>>> 43 bd du 11 novembre 1918
>>> <https://maps.google.com/?q=43+bd+du+11+novembre+1918&entry=gmail&source=g>
>>> 69622 VILLEURBANNE cedex
>>>
>>>
>>> _______________________________________________
>>> Matplotlib-users mailing list
>>> Matplotlib-users at python.org
>>> https://mail.python.org/mailman/listinfo/matplotlib-users
>>>
>>
>> _______________________________________________
>> Matplotlib-users mailing list
>> Matplotlib-users at python.org
>> https://mail.python.org/mailman/listinfo/matplotlib-users
>>
>>
>
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From tcaswell at gmail.com  Thu Oct 19 11:30:26 2017
From: tcaswell at gmail.com (Thomas Caswell)
Date: Thu, 19 Oct 2017 15:30:26 +0000
Subject: [Matplotlib-users] Problem of memory leak using
 FigureCanvasTkAgg
In-Reply-To: <1508425001765.79170@etu.univ-lyon1.fr>
References: <1494500114526.97210@etu.univ-lyon1.fr>
 <CAA48SF9o=5nhn8c-FUYgKcQm041Whm=ZZ2N+JXFVY1osCTn=Wg@mail.gmail.com>
 <CANNq6FkNcreCubg_0pdNn05vUSS4VAjDD=L_=9k90c9v93YUaQ@mail.gmail.com>
 <1508423218080.28805@etu.univ-lyon1.fr>
 <CANNq6FkpsddCUy_L4=58Wr4Uw8z5kcjnR5F=nHLP43jAsDixLA@mail.gmail.com>
 <1508425001765.79170@etu.univ-lyon1.fr>
Message-ID: <CAA48SF_U_irimSS=ZsnVmjDRat0DkpPCcEqgyn5HXG-HOpygvQ@mail.gmail.com>

Have you tried manually adding calls to `gc.collect(2)`?  That may help to
clean up any circular reference cycles hanging around.

Can you replicate this with a simpler script?  The code on SO is hard to
follow.

Tom

On Thu, Oct 19, 2017 at 10:56 AM GARCIA PIERRE SIMON p0904379 <
pierre.garcia at etu.univ-lyon1.fr> wrote:

> Here  is the diagnostic for the old version :
>
> Memory usage: 39832 (kb) Memory usage: 45376 (kb)Memory usage: 46168 (kb) Memory usage: 46960 (kb)Memory usage: 47224 (kb) Memory usage: 47752 (kb)Memory usage: 47752 (kb) Memory usage: 48544 (kb)Memory usage: 48544 (kb) Memory usage: 48808 (kb)Memory usage: 48808 (kb) Memory usage: 49072 (kb)Memory usage: 49072 (kb) Memory usage: 49336 (kb)Memory usage: 49336 (kb) Memory usage: 49864 (kb)Memory usage: 49864 (kb) Memory usage: 50128 (kb)Memory usage: 50128 (kb) Memory usage: 50392 (kb)Memory usage: 50392 (kb) Memory usage: 50656 (kb)
>
> and here is the diagnostic for 2.1.0:
>
> Memory usage: 51808 (kb) Memory usage: 55504 (kb)
> Memory usage: 57088 (kb) Memory usage: 57352 (kb)
> Memory usage: 57880 (kb) Memory usage: 58144 (kb)
> Memory usage: 58672 (kb) Memory usage: 58936 (kb)
> Memory usage: 59464 (kb) Memory usage: 59464 (kb)
> Memory usage: 59992 (kb) Memory usage: 60256 (kb)
> Memory usage: 60520 (kb) Memory usage: 61048 (kb)
> Memory usage: 61312 (kb) Memory usage: 61576 (kb)
> Memory usage: 62104 (kb) Memory usage: 62368 (kb)
> Memory usage: 62632 (kb) Memory usage: 63160 (kb)
> Memory usage: 63424 (kb) Memory usage: 63688 (kb)
> Memory usage: 64216 (kb) Memory usage: 64480 (kb)
>
> It seems that the initial memory used is higher for 2.1.0.
>
> For the delta against n and n+1, the value varies at the begin
> and converge to 264 for 1.1.1 and 792 for 2.1.0.
>
>
> So globally, the rate of leak increases in 2.1.0....
>
>
> Pierre
> ------------------------------
> *De :* Benjamin Root <ben.v.root at gmail.com>
> *Envoy? :* jeudi 19 octobre 2017 16:32
> *? :* GARCIA PIERRE SIMON p0904379
> *Cc :* Thomas Caswell; matplotlib-users at python.org
>
> *Objet :* Re: [Matplotlib-users] Problem of memory leak using
> FigureCanvasTkAgg
> Did the rate of memory leak change?
>
> On Thu, Oct 19, 2017 at 10:26 AM, GARCIA PIERRE SIMON p0904379 <
> pierre.garcia at etu.univ-lyon1.fr> wrote:
>
>> Thank you for your answer but I updated MPL to 2.1.0 and the memory leak
>> is still there. Also, I skipped entirely pyplot and the result is strictly
>> the same.
>>
>>
>> Pierre
>> ------------------------------
>> *De :* Benjamin Root <ben.v.root at gmail.com>
>> *Envoy? :* lundi 7 ao?t 2017 19:27
>> *? :* Thomas Caswell
>> *Cc :* GARCIA PIERRE SIMON p0904379; matplotlib-users at python.org
>> *Objet :* Re: [Matplotlib-users] Problem of memory leak using
>> FigureCanvasTkAgg
>>
>> The stackoverflow posting notes that mpl v1.1.1 was used, which is 5
>> years old. I am pretty sure we have made a lot of memory leak fixes since
>> then. I noted as such on SO.
>>
>> Ben Root
>>
>> On Sat, Aug 5, 2017 at 7:48 PM, Thomas Caswell <tcaswell at gmail.com>
>> wrote:
>>
>>> It is likely there things are not being garbage collected due to either
>>> a missed reference surviving or the circular references between the various
>>> objects is delaying gc from collecting them (I do not recall the details
>>> exactly off the top of my head, but running gc manually with more
>>> generations may help).
>>>
>>> However, if you are embedding in a GUI you are probably better off just
>>> doing the embedding directly (and skipping `pyplot` entirely, see
>>> https://matplotlib.org/examples/user_interfaces/embedding_in_tk.html ).
>>> Re-using the Figures / Axes / Artists to update the data (rather than
>>> starting from scratch every time) may also prevent the memory leaks _and_
>>> give you a performance boost.
>>>
>>> Please subscribe to the list so that you can post without moderation.
>>>
>>> Tom
>>>
>>> On Sat, Aug 5, 2017 at 12:36 PM GARCIA PIERRE SIMON p0904379 <
>>> pierre.garcia at etu.univ-lyon1.fr> wrote:
>>>
>>>> Hello,
>>>>
>>>> I'm PhD student in Bioinformatic and I'm working on a project using
>>>> matplotlib and Tkinter but I've got a big problem of memory leak using
>>>> these modules. I'll be very grateful if you could help me to solve this
>>>> problem. Here is the link explaining what happends, with a reduced script:
>>>>
>>>> http://stackoverflow.com/questions/43097378/matplotlib-vs-tkinter-memory-leak-using-figurecanvastkagg
>>>>
>>>> Thank you in advance,
>>>>
>>>>
>>>> Sincerly,
>>>>
>>>>
>>>>
>>>>
>>>> Pierre Garcia, PhD Student
>>>> ---------
>>>> Bases Mol?culaires et Structurales des Syst?mes infectieux-UMR5086
>>>> Institut de Biologie et Chimie des Prot?ines
>>>> 7 passage du Vercors
>>>> <https://maps.google.com/?q=7+passage+du+Vercors&entry=gmail&source=g>
>>>> 69 367 Lyon cedex 07
>>>> ---------
>>>> Laboratoire de Biom?trie et Biologie Evolutive-UMR5558
>>>> UCBL Lyon 1 -B?t. Gr?gory Mendel
>>>> 43 bd du 11 novembre 1918
>>>> <https://maps.google.com/?q=43+bd+du+11+novembre+1918&entry=gmail&source=g>
>>>> 69622 VILLEURBANNE cedex
>>>>
>>>>
>>>> _______________________________________________
>>>> Matplotlib-users mailing list
>>>> Matplotlib-users at python.org
>>>> https://mail.python.org/mailman/listinfo/matplotlib-users
>>>>
>>>
>>> _______________________________________________
>>> Matplotlib-users mailing list
>>> Matplotlib-users at python.org
>>> https://mail.python.org/mailman/listinfo/matplotlib-users
>>>
>>>
>>
>
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From ianthomas23 at gmail.com  Thu Oct 19 11:57:33 2017
From: ianthomas23 at gmail.com (Ian Thomas)
Date: Thu, 19 Oct 2017 16:57:33 +0100
Subject: [Matplotlib-users] Obtaining the contour data in terms of
 latitude and longitude
In-Reply-To: <CACcRS=c6ifZv2kgb328MGE=Bk7=Q8mXZ61wCh0u6xKtkXcA2KA@mail.gmail.com>
References: <CACcRS=c6ifZv2kgb328MGE=Bk7=Q8mXZ61wCh0u6xKtkXcA2KA@mail.gmail.com>
Message-ID: <CAH53=NYi_x4QAMKWeZyb-QRXPAtjVrBKKshs=BD+nsF2McX4bg@mail.gmail.com>

See https://github.com/matplotlib/matplotlib/issues/367 for discussion and
examples of good and bad ways of doing this.
Ian


On 19 October 2017 at 13:32, Slavin, Jonathan <jslavin at cfa.harvard.edu>
wrote:

> I'm wondering if we could make this easier by modifying contour/contourf
> so that the ContourSet object created had a paths object attribute.  It
> doesn't seem like it'd be that hard.  Maybe some would object that it would
> slow down the creation of contour plots?
>
> Jon
>
> On Wed, Oct 18, 2017 at 5:08 PM, <matplotlib-users-request at python.org>
> wrote:
>>
>> Date: Thu, 19 Oct 2017 02:29:11 +0530
>> From: "ashwin .D" <winash12 at gmail.com>
>> To: matplotlib-users at python.org
>> Subject: [Matplotlib-users] Obtaining the contour data in terms of
>>         latitude and longitude
>> Message-ID:
>>         <CAH0LXy4O++PS3morh4B1L-83TvDqRG64YMkDrDe3LKQewFUSiw at mail.
>> gmail.com>
>> Content-Type: text/plain; charset="utf-8"
>>
>>
>> Hello,
>>          I have data on a 2d lat lon grid(equal grid spacing) in terms of
>> laitude and longitude and heights as enclosed in the image. I need the
>> coordinates(in terms of latitude and longitude) of the contours( as shown
>> in red in the image). From this answer -
>> https://stackoverflow.com/questions/18304722/python-find-
>> contour-lines-from-matplotlib-pyplot-contour/
>> will this give me what I am looking for or is there a scaling required to
>> latitude and longitude that I need to include?
>>
>> Best regards,
>> Ashwin.
>>
>> import numpy as np
>> def get_contour_verts(cn):
>>     contours = []
>>     # for each contour line
>>     for cc in cn.collections:
>>         paths = []
>>         # for each separate section of the contour line
>>         for pp in cc.get_paths():
>>             xy = []
>>             # for each segment of that section
>>             for vv in pp.iter_segments():
>>                 xy.append(vv[0])
>>             paths.append(np.vstack(xy))
>>         contours.append(paths)
>>
>>     return contours
>> -------------- next part --------------
>> An HTML attachment was scrubbed...
>> URL: <http://mail.python.org/pipermail/matplotlib-users/attachmen
>> ts/20171019/c5cb0664/attachment-0001.html>
>> -------------- next part --------------
>> A non-text attachment was scrubbed...
>> Name: sfc_trough1.gif
>> Type: image/gif
>> Size: 13312 bytes
>> Desc: not available
>> URL: <http://mail.python.org/pipermail/matplotlib-users/attachmen
>> ts/20171019/c5cb0664/attachment-0001.gif>
>>
>> ------------------------------
>>
>> Message: 2
>> Date: Wed, 18 Oct 2017 17:08:16 -0400
>> From: Benjamin Root <ben.v.root at gmail.com>
>> To: "ashwin .D" <winash12 at gmail.com>
>> Cc: Matplotlib-users <matplotlib-users at python.org>
>> Subject: Re: [Matplotlib-users] Obtaining the contour data in terms of
>>         latitude and longitude
>> Message-ID:
>>         <CANNq6FnAWDDv1vahui+WZY=JYwjEcQpiHDQUPxDKhbk-fEb6tQ at mail.gm
>> ail.com>
>> Content-Type: text/plain; charset="utf-8"
>>
>>
>> If you passed the coordinate arrays in with the call to contourf(), then
>> you will get what you expect.
>>
>> I should note that you should watch out for path simplification (you'll
>> want to turn it off). I also recommend using contourf() over contour()
>> because it handles the edges of the domain better and NaNs.
>>
>> Cheers!
>> Ben Root
>>
>>
>> On Wed, Oct 18, 2017 at 4:59 PM, ashwin .D <winash12 at gmail.com> wrote:
>>
>> > Hello,
>> >          I have data on a 2d lat lon grid(equal grid spacing) in terms
>> of
>> > laitude and longitude and heights as enclosed in the image. I need the
>> > coordinates(in terms of latitude and longitude) of the contours( as
>> shown
>> > in red in the image). From this answer - https://stackoverflow.com/
>> > questions/18304722/python-find-contour-lines-from-
>> > matplotlib-pyplot-contour/ will this give me what I am looking for or is
>> > there a scaling required to latitude and longitude that I need to
>> include?
>> >
>> > Best regards,
>> > Ashwin.
>> >
>> > import numpy as np
>> > def get_contour_verts(cn):
>> >     contours = []
>> >     # for each contour line
>> >     for cc in cn.collections:
>> >         paths = []
>> >         # for each separate section of the contour line
>> >         for pp in cc.get_paths():
>> >             xy = []
>> >             # for each segment of that section
>> >             for vv in pp.iter_segments():
>> >                 xy.append(vv[0])
>> >             paths.append(np.vstack(xy))
>> >         contours.append(paths)
>> >
>> >     return contours
>> >
>> >
>> >
>> > _______________________________________________
>> > Matplotlib-users mailing list
>> > Matplotlib-users at python.org
>> > https://mail.python.org/mailman/listinfo/matplotlib-users
>> >
>> >
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>> URL: <http://mail.python.org/pipermail/matplotlib-users/attachmen
>> ts/20171018/b9f14b69/attachment.html>
>>
>> ------------------------------
>>
>> Subject: Digest Footer
>>
>> _______________________________________________
>> Matplotlib-users mailing list
>> Matplotlib-users at python.org
>> https://mail.python.org/mailman/listinfo/matplotlib-users
>>
>>
>> ------------------------------
>>
>> End of Matplotlib-users Digest, Vol 27, Issue 24
>> ************************************************
>>
>
>
>
> --
> ________________________________________________________
> Jonathan D. Slavin                 Harvard-Smithsonian CfA
> jslavin at cfa.harvard.edu       60 Garden Street, MS 83
> phone: (617) 496-7981       Cambridge, MA 02138-1516
> cell: (781) 363-0035             USA
> ________________________________________________________
>
>
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org
> https://mail.python.org/mailman/listinfo/matplotlib-users
>
>
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From tcaswell at gmail.com  Fri Oct 20 00:10:59 2017
From: tcaswell at gmail.com (Thomas Caswell)
Date: Fri, 20 Oct 2017 04:10:59 +0000
Subject: [Matplotlib-users] Problem of memory leak using
 FigureCanvasTkAgg
In-Reply-To: <1508429902922.94316@etu.univ-lyon1.fr>
References: <1494500114526.97210@etu.univ-lyon1.fr>
 <CAA48SF9o=5nhn8c-FUYgKcQm041Whm=ZZ2N+JXFVY1osCTn=Wg@mail.gmail.com>
 <CANNq6FkNcreCubg_0pdNn05vUSS4VAjDD=L_=9k90c9v93YUaQ@mail.gmail.com>
 <1508423218080.28805@etu.univ-lyon1.fr>
 <CANNq6FkpsddCUy_L4=58Wr4Uw8z5kcjnR5F=nHLP43jAsDixLA@mail.gmail.com>
 <1508425001765.79170@etu.univ-lyon1.fr>
 <CAA48SF_U_irimSS=ZsnVmjDRat0DkpPCcEqgyn5HXG-HOpygvQ@mail.gmail.com>
 <1508429902922.94316@etu.univ-lyon1.fr>
Message-ID: <CAA48SF9k9jQupFxFbo2WDYSnNPWO83L05cr3fatODZ1dA-8zmg@mail.gmail.com>

I think there was a bug in how you were capturing the clearing the children
of the top level frame.  Tweaking it a bit I can reproduce the leaking, but
not on every cycle:


from matplotlib.backends.backend_tkagg import FigureCanvasTkAgg
import matplotlib.patches as patches
# from Tkinter import Tk, Canvas, Frame, Label, Button
from six.moves import tkinter as Tk
import gc
import resource
from matplotlib.figure import Figure
import sys

list_fig_g = []
list_wid_g = []


def launcher(list_wid, list_fig, frame):
    print('Memory usage: %s (kb)'
          % resource.getrusage(resource.RUSAGE_SELF).ru_maxrss,)
    for f in list_fig:
        f.clf()
        f.clear()
        f._master = None
        del f
    list_fig.clear()

    for i in list_wid:
        print(i, sys.getrefcount(i))
        r = i.winfo_toplevel()
        i.grid_forget()
        i.destroy()
        for k in ('<MouseWheel>', '<Destroy>'):
            r.unbind(k)

        print(i, sys.getrefcount(i))
        del i
    list_wid.clear()

    fig1 = Figure(figsize=(1, 1))
    ax1 = fig1.add_subplot(111, aspect='equal')
    ax1.add_patch(
        patches.Arrow(
            0,            # x
            -0,            # y
            1,            # dx
            0,            # dy
            width=1,
        )
    )
    list_fig.append(fig1)
    # this seems to be the first line that leaks memory
    # but not _always_
    canvas = FigureCanvasTkAgg(fig1, master=None)

    t1, t2, w, h = fig1.bbox.bounds
    widget = canvas.get_tk_widget()

    # widget = Tk.Canvas(
    #     master=frame, width=w, height=h, borderwidth=0,
    #     highlightthickness=0)
    # w2 = Tk.PhotoImage(
    #     master=widget, width=int(w), height=int(h))
    # widget.create_image(w//2, h//2, image=w2)
    widget.grid(row=0, column=1)
    widget.focus_set()
    list_wid.extend(frame.winfo_children())
    # print(w)
    # print('gc: ', gc.collect(2))
    gc.collect(2)
    # canvas.draw()
    print('Memory usage: %s (kb)'
          % resource.getrusage(resource.RUSAGE_SELF).ru_maxrss)
    print()


def bulk(list_wid, list_fig, frame):
    for j in range(100):
        launcher(list_wid, list_fig, frame)


root = Tk.Tk()
Entete5 = Tk.Label(root, width=20, height=1).grid(row=0, column=1)
fond = Tk.Canvas(root, width=800, height=200)
fond.grid(row=1, column=1)
frame_g = Tk.Frame(fond, width=80, height=10)
frame_g.grid(row=0, column=1)
fond.create_window(0, 0, window=frame_g)
Load = Tk.Button(root, text="Load",
                 command=lambda: launcher(list_wid_g, list_fig_g, frame_g))
go = Tk.Button(root, text="bulk",
               command=lambda: bulk(list_wid_g, list_fig_g, frame_g))
Load.grid(row=7, column=4)
go.grid(row=6, column=4)
list_wid = frame_g.winfo_children()
root.mainloop()


(I also added a 'bulk' button to run 100 iterations).  I tried just
creating a FigureCanvasAgg object (the super-class of FigureCanvasTkAgg)
and that seems not to leak.  Doing most of the tk work
that FigureCanvasTkAgg does not seem to leak.

It looks like it is the callbacks that are registered to handle mouse and
key board events, deleteing them from the __init__ seems to prevent the
leak!

Not sure what the right way to clean those up is though, but at least this
is pointing the right direction!

Tom





On Thu, Oct 19, 2017 at 12:18 PM GARCIA PIERRE SIMON p0904379 <
pierre.garcia at etu.univ-lyon1.fr> wrote:

> Adding gc.collect() everywhere in the script doesn't change anything. Here
> is a code that works and that is simpler:
>
>
> #! /usr/bin/python2.7
> #-*- coding: utf-8 -*-
> import matplotlib
> matplotlib.use('Tkagg')
> from matplotlib.backends.backend_tkagg import FigureCanvasTkAgg
> import matplotlib.patches as patches
> from Tkinter import *
> import gc
> import resource
> from matplotlib.figure import Figure
>
> list_fig_g = []
> list_wid_g = []
>
> def Launcher(list_wid,list_fig,frame):
>     print 'Memory usage: %s (kb)' %
> resource.getrusage(resource.RUSAGE_SELF).ru_maxrss,
>     for i in list_wid:
>         i.grid_forget()
>         i.destroy()
>         del i
>     try:
>         list_fig[0].clf()# other round
>         list_fig[0].clear()
>         del list_fig[0]
>     except:
>         print "",# first round
>     fig1 = Figure(figsize=(1,1))
>     ax1 = fig1.add_subplot(111, aspect='equal')
>     ax1.add_patch(
>         patches.Arrow(
>             0,            # x
>             -0,            # y
>             1,            # dx
>             0,            # dy
>             width=1,
>         )
>     )
>     list_fig.append(fig1)
>     canvas = FigureCanvasTkAgg(list_fig[0], master=frame)
>     canvas.get_tk_widget().grid(row=0,column=1)
>     list_wid = frame.winfo_children()
>     print 'Memory usage: %s (kb)' %
> resource.getrusage(resource.RUSAGE_SELF).ru_maxrss
>
> root = Tk()
> Entete5 = Label(root,width=20, height=1).grid(row=0,column=1)
> fond = Canvas(root, width=800, height=200)
> fond.grid(row=1, column=1)
> frame_g = Frame(fond,width=80, height=10)
> frame_g.grid(row=0,column=1)
> fond.create_window(0, 0,  window=frame_g)
> Load = Button (root, text = "Load", command = lambda :
> Launcher(list_wid_g,list_fig_g,frame_g))
> Load.grid(row=7,column=4)
> list_wid = frame_g.winfo_children()
> root.mainloop()
>
>
> and here is the output:
>
>
> Memory usage: 49516 (kb)  Memory usage: 50824 (kb)
> Memory usage: 51880 (kb) Memory usage: 52672 (kb)
> Memory usage: 52936 (kb) Memory usage: 53728 (kb)
> Memory usage: 53992 (kb) Memory usage: 54256 (kb)
> Memory usage: 54520 (kb) Memory usage: 54520 (kb)
> Memory usage: 54520 (kb) Memory usage: 55048 (kb)
> Memory usage: 55312 (kb) Memory usage: 55840 (kb)
> Memory usage: 55840 (kb) Memory usage: 55840 (kb)
> Memory usage: 56104 (kb) Memory usage: 56104 (kb)
> Memory usage: 56104 (kb) Memory usage: 56104 (kb)
> Memory usage: 56368 (kb) Memory usage: 56368 (kb)
> Memory usage: 56368 (kb) Memory usage: 56632 (kb)
> Memory usage: 56896 (kb) Memory usage: 57424 (kb)
> Memory usage: 57688 (kb) Memory usage: 57688 (kb)
> Memory usage: 57688 (kb) Memory usage: 57952 (kb)
> Memory usage: 58216 (kb) Memory usage: 59008 (kb)
> Memory usage: 59008 (kb) Memory usage: 59008 (kb)
> Memory usage: 59008 (kb) Memory usage: 59272 (kb)
> Memory usage: 59536 (kb) Memory usage: 60328 (kb)
>
> Pierre
> ------------------------------
> *De :* Thomas Caswell <tcaswell at gmail.com>
> *Envoy? :* jeudi 19 octobre 2017 17:30
> *? :* GARCIA PIERRE SIMON p0904379; Benjamin Root
> *Cc :* matplotlib-users at python.org
>
> *Objet :* Re: [Matplotlib-users] Problem of memory leak using
> FigureCanvasTkAgg
> Have you tried manually adding calls to `gc.collect(2)`?  That may help to
> clean up any circular reference cycles hanging around.
>
> Can you replicate this with a simpler script?  The code on SO is hard to
> follow.
>
> Tom
>
> On Thu, Oct 19, 2017 at 10:56 AM GARCIA PIERRE SIMON p0904379 <
> pierre.garcia at etu.univ-lyon1.fr> wrote:
>
>> Here  is the diagnostic for the old version :
>>
>> Memory usage: 39832 (kb) Memory usage: 45376 (kb)Memory usage: 46168 (kb) Memory usage: 46960 (kb)Memory usage: 47224 (kb) Memory usage: 47752 (kb)Memory usage: 47752 (kb) Memory usage: 48544 (kb)Memory usage: 48544 (kb) Memory usage: 48808 (kb)Memory usage: 48808 (kb) Memory usage: 49072 (kb)Memory usage: 49072 (kb) Memory usage: 49336 (kb)Memory usage: 49336 (kb) Memory usage: 49864 (kb)Memory usage: 49864 (kb) Memory usage: 50128 (kb)Memory usage: 50128 (kb) Memory usage: 50392 (kb)Memory usage: 50392 (kb) Memory usage: 50656 (kb)
>>
>> and here is the diagnostic for 2.1.0:
>>
>> Memory usage: 51808 (kb) Memory usage: 55504 (kb)
>> Memory usage: 57088 (kb) Memory usage: 57352 (kb)
>> Memory usage: 57880 (kb) Memory usage: 58144 (kb)
>> Memory usage: 58672 (kb) Memory usage: 58936 (kb)
>> Memory usage: 59464 (kb) Memory usage: 59464 (kb)
>> Memory usage: 59992 (kb) Memory usage: 60256 (kb)
>> Memory usage: 60520 (kb) Memory usage: 61048 (kb)
>> Memory usage: 61312 (kb) Memory usage: 61576 (kb)
>> Memory usage: 62104 (kb) Memory usage: 62368 (kb)
>> Memory usage: 62632 (kb) Memory usage: 63160 (kb)
>> Memory usage: 63424 (kb) Memory usage: 63688 (kb)
>> Memory usage: 64216 (kb) Memory usage: 64480 (kb)
>>
>> It seems that the initial memory used is higher for 2.1.0.
>>
>> For the delta against n and n+1, the value varies at the begin
>> and converge to 264 for 1.1.1 and 792 for 2.1.0.
>>
>>
>> So globally, the rate of leak increases in 2.1.0....
>>
>>
>> Pierre
>> ------------------------------
>> *De :* Benjamin Root <ben.v.root at gmail.com>
>> *Envoy? :* jeudi 19 octobre 2017 16:32
>> *? :* GARCIA PIERRE SIMON p0904379
>> *Cc :* Thomas Caswell; matplotlib-users at python.org
>>
>> *Objet :* Re: [Matplotlib-users] Problem of memory leak using
>> FigureCanvasTkAgg
>> Did the rate of memory leak change?
>>
>> On Thu, Oct 19, 2017 at 10:26 AM, GARCIA PIERRE SIMON p0904379 <
>> pierre.garcia at etu.univ-lyon1.fr> wrote:
>>
>>> Thank you for your answer but I updated MPL to 2.1.0 and the memory leak
>>> is still there. Also, I skipped entirely pyplot and the result is strictly
>>> the same.
>>>
>>>
>>> Pierre
>>> ------------------------------
>>> *De :* Benjamin Root <ben.v.root at gmail.com>
>>> *Envoy? :* lundi 7 ao?t 2017 19:27
>>> *? :* Thomas Caswell
>>> *Cc :* GARCIA PIERRE SIMON p0904379; matplotlib-users at python.org
>>> *Objet :* Re: [Matplotlib-users] Problem of memory leak using
>>> FigureCanvasTkAgg
>>>
>>> The stackoverflow posting notes that mpl v1.1.1 was used, which is 5
>>> years old. I am pretty sure we have made a lot of memory leak fixes since
>>> then. I noted as such on SO.
>>>
>>> Ben Root
>>>
>>> On Sat, Aug 5, 2017 at 7:48 PM, Thomas Caswell <tcaswell at gmail.com>
>>> wrote:
>>>
>>>> It is likely there things are not being garbage collected due to either
>>>> a missed reference surviving or the circular references between the various
>>>> objects is delaying gc from collecting them (I do not recall the details
>>>> exactly off the top of my head, but running gc manually with more
>>>> generations may help).
>>>>
>>>> However, if you are embedding in a GUI you are probably better off just
>>>> doing the embedding directly (and skipping `pyplot` entirely, see
>>>> https://matplotlib.org/examples/user_interfaces/embedding_in_tk.html ).
>>>> Re-using the Figures / Axes / Artists to update the data (rather than
>>>> starting from scratch every time) may also prevent the memory leaks _and_
>>>> give you a performance boost.
>>>>
>>>> Please subscribe to the list so that you can post without moderation.
>>>>
>>>> Tom
>>>>
>>>> On Sat, Aug 5, 2017 at 12:36 PM GARCIA PIERRE SIMON p0904379 <
>>>> pierre.garcia at etu.univ-lyon1.fr> wrote:
>>>>
>>>>> Hello,
>>>>>
>>>>> I'm PhD student in Bioinformatic and I'm working on a project using
>>>>> matplotlib and Tkinter but I've got a big problem of memory leak using
>>>>> these modules. I'll be very grateful if you could help me to solve this
>>>>> problem. Here is the link explaining what happends, with a reduced script:
>>>>>
>>>>> http://stackoverflow.com/questions/43097378/matplotlib-vs-tkinter-memory-leak-using-figurecanvastkagg
>>>>>
>>>>> Thank you in advance,
>>>>>
>>>>>
>>>>> Sincerly,
>>>>>
>>>>>
>>>>>
>>>>>
>>>>> Pierre Garcia, PhD Student
>>>>> ---------
>>>>> Bases Mol?culaires et Structurales des Syst?mes infectieux-UMR5086
>>>>> Institut de Biologie et Chimie des Prot?ines
>>>>> 7 passage du Vercors
>>>>> <https://maps.google.com/?q=7+passage+du+Vercors&entry=gmail&source=g>
>>>>> 69 367 Lyon cedex 07
>>>>> ---------
>>>>> Laboratoire de Biom?trie et Biologie Evolutive-UMR5558
>>>>> UCBL Lyon 1 -B?t. Gr?gory Mendel
>>>>> 43 bd du 11 novembre 1918
>>>>> <https://maps.google.com/?q=43+bd+du+11+novembre+1918&entry=gmail&source=g>
>>>>> 69622 VILLEURBANNE cedex
>>>>>
>>>>>
>>>>> _______________________________________________
>>>>> Matplotlib-users mailing list
>>>>> Matplotlib-users at python.org
>>>>> https://mail.python.org/mailman/listinfo/matplotlib-users
>>>>>
>>>>
>>>> _______________________________________________
>>>> Matplotlib-users mailing list
>>>> Matplotlib-users at python.org
>>>> https://mail.python.org/mailman/listinfo/matplotlib-users
>>>>
>>>>
>>>
>>
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From eulergaussriemann at gmail.com  Fri Oct 20 17:38:52 2017
From: eulergaussriemann at gmail.com (David Goldsmith)
Date: Fri, 20 Oct 2017 14:38:52 -0700
Subject: [Matplotlib-users] Why is canvas.print_figure to a png reduce the
 real size of my image
Message-ID: <CA+_2Pn1rYc=y8imytzjDfVJ2Z3gro-k92JqMhKoBHHr9YkHdPQ@mail.gmail.com>

Hi!  I spent a good part of yesterday trying to figure this out on my own,
without success, so I'm posting.  I have a 3601x2401 pixel, 300 DPI figure
I export to a png using print_figure; here's sample code:

PLB.imshow(rslt) # rslt is a 3601x2401 int-valued array
curfig = PLB.Figure
fig = PLB.gcf()
# next three lines are in my code; don't seem relevant, but I include
them in case they are

ax = fig.gca()
ax.xaxis.set_ticks([])
ax.yaxis.set_ticks([])

fig.canvas.print_figure(fn + '.png',
                        dpi=300,

                        bbox_inches='tight',

                        pad_inches=-1.0 / 72)

The resulting png is neither the right number of pixels by pixels, nor
the right number of inches by incheses; what's more the resulting png
differs depending on whether I use imshow or matshow (which I also
tried).

What's going on, and how do I "fix" it?  (If I really am getting the
full resolution figure, despite what the png is telling me are the
pixel counts, why are the physical dimensions off, and why is the png
telling me that the pixel counts are different?  If it has something
to do with compression, is there some  undocumented way to say "no
compression" and will that have the desired effect?)

Thanks!

DLG
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From jklymak at uvic.ca  Fri Oct 20 17:59:59 2017
From: jklymak at uvic.ca (Jody Klymak)
Date: Fri, 20 Oct 2017 14:59:59 -0700
Subject: [Matplotlib-users] Why is canvas.print_figure to a png reduce
 the real size of my image
In-Reply-To: <CA+_2Pn1rYc=y8imytzjDfVJ2Z3gro-k92JqMhKoBHHr9YkHdPQ@mail.gmail.com>
References: <CA+_2Pn1rYc=y8imytzjDfVJ2Z3gro-k92JqMhKoBHHr9YkHdPQ@mail.gmail.com>
Message-ID: <45743492-C845-4493-A8EE-585FB33040C3@uvic.ca>

Hi David,

What are you trying to achieve as the output?  An image with no white space?  

`bbox_inches=tight` changes the size of the figure to get rid of white space around it.  Prob not what you really want.

Drilling down into `canvas` shouldn?t really be necessary. 

What is `PLB`?  etc?

This will get you pretty close if you just want an image w/ no white space.  But I?m not 100% sure that every pixel has been preserved?


```
import matplotlib.pyplot as plt
import numpy as np

im = np.floor(np.random.rand(3601,2401)*100)

fig, ax = plt.subplots(figsize = (2401/300, 3601/300), dpi=300)

# turn off axis and labels
ax.set_axis_off()
# fill the figure
ax.set_position([0., 0., 1., 1.])
ax.imshow(im)
fig.savefig('Test.png', dpi=300)
```




> On 20 Oct 2017, at  14:38 PM, David Goldsmith <eulergaussriemann at gmail.com> wrote:
> 
> Hi!  I spent a good part of yesterday trying to figure this out on my own, without success, so I'm posting.  I have a 3601x2401 pixel, 300 DPI figure I export to a png using print_figure; here's sample code:
> 
> PLB.imshow(rslt) # rslt is a 3601x2401 int-valued array
> curfig = PLB.Figure
> fig = PLB.gcf() 
> # next three lines are in my code; don't seem relevant, but I include them in case they are
> ax = fig.gca()
> ax.xaxis.set_ticks([]) 
> ax.yaxis.set_ticks([])
> 
> fig.canvas.print_figure(fn + '.png',
>                         dpi=300,
>                         bbox_inches='tight', 
>                         pad_inches=-1.0 / 72)
> The resulting png is neither the right number of pixels by pixels, nor the right number of inches by incheses; what's more the resulting png differs depending on whether I use imshow or matshow (which I also tried).
> What's going on, and how do I "fix" it?  (If I really am getting the full resolution figure, despite what the png is telling me are the pixel counts, why are the physical dimensions off, and why is the png telling me that the pixel counts are different?  If it has something to do with compression, is there some  undocumented way to say "no compression" and will that have the desired effect?)
> Thanks!
> DLG 
> 
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org
> https://mail.python.org/mailman/listinfo/matplotlib-users

--
Jody Klymak    
http://web.uvic.ca/~jklymak/





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From warren.weckesser at gmail.com  Fri Oct 20 18:28:29 2017
From: warren.weckesser at gmail.com (Warren Weckesser)
Date: Fri, 20 Oct 2017 18:28:29 -0400
Subject: [Matplotlib-users] Why is canvas.print_figure to a png reduce
 the real size of my image
In-Reply-To: <CA+_2Pn1rYc=y8imytzjDfVJ2Z3gro-k92JqMhKoBHHr9YkHdPQ@mail.gmail.com>
References: <CA+_2Pn1rYc=y8imytzjDfVJ2Z3gro-k92JqMhKoBHHr9YkHdPQ@mail.gmail.com>
Message-ID: <CAGzF1ucFurk4CV=fDo_5K4CSFztLmK_cUo_dyYXcvc0skGzS=g@mail.gmail.com>

On Fri, Oct 20, 2017 at 5:38 PM, David Goldsmith <
eulergaussriemann at gmail.com> wrote:

> Hi!  I spent a good part of yesterday trying to figure this out on my own,
> without success, so I'm posting.  I have a 3601x2401 pixel, 300 DPI figure
> I export to a png using print_figure; here's sample code:
>
>

David,

If you are trying to write the numpy array directly to a PNG file, an
alternative to matplotlib is a package that I created called numpngw:
https://pypi.python.org/pypi/numpngw (github:
https://github.com/WarrenWeckesser/numpngw).  Both those links show several
examples of its use.   numpngw uses just Python and NumPy, so it is easy to
install.

Warren


PLB.imshow(rslt) # rslt is a 3601x2401 int-valued array
> curfig = PLB.Figure
> fig = PLB.gcf()
> # next three lines are in my code; don't seem relevant, but I include them in case they are
>
> ax = fig.gca()
> ax.xaxis.set_ticks([])
> ax.yaxis.set_ticks([])
>
> fig.canvas.print_figure(fn + '.png',
>                         dpi=300,
>
>                         bbox_inches='tight',
>
>                         pad_inches=-1.0 / 72)
>
> The resulting png is neither the right number of pixels by pixels, nor the right number of inches by incheses; what's more the resulting png differs depending on whether I use imshow or matshow (which I also tried).
>
> What's going on, and how do I "fix" it?  (If I really am getting the full resolution figure, despite what the png is telling me are the pixel counts, why are the physical dimensions off, and why is the png telling me that the pixel counts are different?  If it has something to do with compression, is there some  undocumented way to say "no compression" and will that have the desired effect?)
>
> Thanks!
>
> DLG
>
>
>
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org
> https://mail.python.org/mailman/listinfo/matplotlib-users
>
>
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From eulergaussriemann at gmail.com  Fri Oct 20 18:33:49 2017
From: eulergaussriemann at gmail.com (David Goldsmith)
Date: Fri, 20 Oct 2017 22:33:49 +0000
Subject: [Matplotlib-users] Why is canvas.print_figure to a png reduce
 the real size of my image
In-Reply-To: <45743492-C845-4493-A8EE-585FB33040C3@uvic.ca>
References: <CA+_2Pn1rYc=y8imytzjDfVJ2Z3gro-k92JqMhKoBHHr9YkHdPQ@mail.gmail.com>
 <45743492-C845-4493-A8EE-585FB33040C3@uvic.ca>
Message-ID: <CA+_2Pn3Be5Tx_5sNEk82aaMcTvyAff=_pVPXV7k0Sh8vkPKeOA@mail.gmail.com>

On Fri, Oct 20, 2017 at 3:00 PM Jody Klymak <jklymak at uvic.ca> wrote:

> Hi David,
>
> What are you trying to achieve as the output?  An image with no white
> space?
>

Yes, and no edge, but one that replaces that white space with my figure.
 in other words, if what is going on is that my figure w/ the white space
is what is 3600x2400, how do I get the full 3600x2400 to be my figure? (I
don?t care if it puts white space around that because I figured out an
?alternative? way to get rid of that).  But mostly, I?d just like to
understand what?s going on, because what?s happening now makes absolutely
no sense to me whatsoever.


> `bbox_inches=tight` changes the size of the figure to get rid of white
> space around it.  Prob not what you really want.
>

OK, so that may be the problem: what?s the ?proper? way to ?get rid of? the
white space, module what I said above.

>

> Drilling down into `canvas` shouldn?t really be necessary.
>

I?m all ears.

>
> What is `PLB`?  etc?
>

import pylab as PLB

>
> This will get you pretty close if you just want an image w/ no white
> space.  But I?m not 100% sure that every pixel has been preserved?
>
>
> ```
> import matplotlib.pyplot as plt
> import numpy as np
>
> im = np.floor(np.random.rand(3601,2401)*100)
>
> fig, ax = plt.subplots(figsize = (2401/300, 3601/300), dpi=300)
>
> # turn off axis and labels
> ax.set_axis_off()
> # fill the figure
> ax.set_position([0., 0., 1., 1.])
> ax.imshow(im)
> fig.savefig('Test.png', dpi=300)
>

Ok, I?ll try that, thanks.

DLG

> ```
>
>
>
>
> On 20 Oct 2017, at  14:38 PM, David Goldsmith <eulergaussriemann at gmail.com>
> wrote:
>
> Hi!  I spent a good part of yesterday trying to figure this out on my own,
> without success, so I'm posting.  I have a 3601x2401 pixel, 300 DPI figure
> I export to a png using print_figure; here's sample code:
>
> PLB.imshow(rslt) # rslt is a 3601x2401 int-valued array
> curfig = PLB.Figure
> fig = PLB.gcf()
> # next three lines are in my code; don't seem relevant, but I include them in case they are
>
> ax = fig.gca()
> ax.xaxis.set_ticks([])
> ax.yaxis.set_ticks([])
>
> fig.canvas.print_figure(fn + '.png',
>                         dpi=300,
>
>                         bbox_inches='tight',
>
>                         pad_inches=-1.0 / 72)
>
> The resulting png is neither the right number of pixels by pixels, nor the right number of inches by incheses; what's more the resulting png differs depending on whether I use imshow or matshow (which I also tried).
>
> What's going on, and how do I "fix" it?  (If I really am getting the full resolution figure, despite what the png is telling me are the pixel counts, why are the physical dimensions off, and why is the png telling me that the pixel counts are different?  If it has something to do with compression, is there some  undocumented way to say "no compression" and will that have the desired effect?)
>
> Thanks!
>
> DLG
>
>
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org
> https://mail.python.org/mailman/listinfo/matplotlib-users
>
>
> --
> Jody Klymak
> http://web.uvic.ca/~jklymak/
>
>
>
>
>
>
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From eulergaussriemann at gmail.com  Fri Oct 20 18:36:30 2017
From: eulergaussriemann at gmail.com (David Goldsmith)
Date: Fri, 20 Oct 2017 22:36:30 +0000
Subject: [Matplotlib-users] Why is canvas.print_figure to a png reduce
 the real size of my image
In-Reply-To: <CAGzF1ucFurk4CV=fDo_5K4CSFztLmK_cUo_dyYXcvc0skGzS=g@mail.gmail.com>
References: <CA+_2Pn1rYc=y8imytzjDfVJ2Z3gro-k92JqMhKoBHHr9YkHdPQ@mail.gmail.com>
 <CAGzF1ucFurk4CV=fDo_5K4CSFztLmK_cUo_dyYXcvc0skGzS=g@mail.gmail.com>
Message-ID: <CA+_2Pn3Ak67qX8W0PDOuyA6VK=f7Xt3npehheskoSVpkL2q-vw@mail.gmail.com>

Thanks.  Does it have pylab?s range of built in color maps?  Does it even
use colormaps, or do I have to write my own (or ?copy? them from MPL).

DLG

On Fri, Oct 20, 2017 at 3:28 PM Warren Weckesser <warren.weckesser at gmail.com>
wrote:

> On Fri, Oct 20, 2017 at 5:38 PM, David Goldsmith <
> eulergaussriemann at gmail.com> wrote:
>
>> Hi!  I spent a good part of yesterday trying to figure this out on my
>> own, without success, so I'm posting.  I have a 3601x2401 pixel, 300 DPI
>> figure I export to a png using print_figure; here's sample code:
>>
>>
>
> David,
>
> If you are trying to write the numpy array directly to a PNG file, an
> alternative to matplotlib is a package that I created called numpngw:
> https://pypi.python.org/pypi/numpngw (github:
> https://github.com/WarrenWeckesser/numpngw).  Both those links show
> several examples of its use.   numpngw uses just Python and NumPy, so it is
> easy to install.
>
> Warren
>
>
> PLB.imshow(rslt) # rslt is a 3601x2401 int-valued array
>> curfig = PLB.Figure
>> fig = PLB.gcf()
>> # next three lines are in my code; don't seem relevant, but I include them in case they are
>>
>> ax = fig.gca()
>> ax.xaxis.set_ticks([])
>> ax.yaxis.set_ticks([])
>>
>> fig.canvas.print_figure(fn + '.png',
>>                         dpi=300,
>>
>>                         bbox_inches='tight',
>>
>>                         pad_inches=-1.0 / 72)
>>
>> The resulting png is neither the right number of pixels by pixels, nor the right number of inches by incheses; what's more the resulting png differs depending on whether I use imshow or matshow (which I also tried).
>>
>> What's going on, and how do I "fix" it?  (If I really am getting the full resolution figure, despite what the png is telling me are the pixel counts, why are the physical dimensions off, and why is the png telling me that the pixel counts are different?  If it has something to do with compression, is there some  undocumented way to say "no compression" and will that have the desired effect?)
>>
>> Thanks!
>>
>> DLG
>>
>>
>>
>> _______________________________________________
>> Matplotlib-users mailing list
>> Matplotlib-users at python.org
>> https://mail.python.org/mailman/listinfo/matplotlib-users
>>
>>
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From jklymak at uvic.ca  Fri Oct 20 18:44:54 2017
From: jklymak at uvic.ca (Jody Klymak)
Date: Fri, 20 Oct 2017 15:44:54 -0700
Subject: [Matplotlib-users] Why is canvas.print_figure to a png reduce
 the real size of my image
In-Reply-To: <CA+_2Pn3Be5Tx_5sNEk82aaMcTvyAff=_pVPXV7k0Sh8vkPKeOA@mail.gmail.com>
References: <CA+_2Pn1rYc=y8imytzjDfVJ2Z3gro-k92JqMhKoBHHr9YkHdPQ@mail.gmail.com>
 <45743492-C845-4493-A8EE-585FB33040C3@uvic.ca>
 <CA+_2Pn3Be5Tx_5sNEk82aaMcTvyAff=_pVPXV7k0Sh8vkPKeOA@mail.gmail.com>
Message-ID: <686398FC-DCB6-4367-AA99-E841CC4846B8@uvic.ca>



> On 20 Oct 2017, at  15:33 PM, David Goldsmith <eulergaussriemann at gmail.com> wrote:
> 
>  But mostly, I?d just like to understand what?s going on, because what?s happening now makes absolutely no sense to me whatsoever.

Ha!  

I think you are approaching this from the point of view of an image processing library or something like Photoshop where you want your image to be 6x8 and 300 dpi, and you expect 1800x2400 pixels.  

Thats not *really* what matplotlib is for. It really is for presentation of scientific data.  So, I have a 500x600 matrix of data, I may use `imshow` to plot it up, but most people don?t care about the dpi of those pixels, so they will just set the dpi to something that shows all the data clearly.  Much more important is the ability to have axis labels, annotations, etc.  

Its not to say that you can?t be careful and get what you want.  I think what I sent you is pretty close.  But the library is not trying to conserve pixels.  

Anyway, what I sent should be clear enough.  I think what you were missing was setting the figure size to the size you wanted before trying to draw on it.

Cheers,   Jody




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From eulergaussriemann at gmail.com  Fri Oct 20 19:02:19 2017
From: eulergaussriemann at gmail.com (David Goldsmith)
Date: Fri, 20 Oct 2017 23:02:19 +0000
Subject: [Matplotlib-users] Why is canvas.print_figure to a png reduce
 the real size of my image
In-Reply-To: <686398FC-DCB6-4367-AA99-E841CC4846B8@uvic.ca>
References: <CA+_2Pn1rYc=y8imytzjDfVJ2Z3gro-k92JqMhKoBHHr9YkHdPQ@mail.gmail.com>
 <45743492-C845-4493-A8EE-585FB33040C3@uvic.ca>
 <CA+_2Pn3Be5Tx_5sNEk82aaMcTvyAff=_pVPXV7k0Sh8vkPKeOA@mail.gmail.com>
 <686398FC-DCB6-4367-AA99-E841CC4846B8@uvic.ca>
Message-ID: <CA+_2Pn1M_8vryWWzCfhEfX152rDBQ=2TCNr=FgZ7ebMFS881gQ@mail.gmail.com>

OK, thanks for your help.

What I?m doing is somewhere in between: I?m making my own ?fractals,? ?from
scratch?; so if im going to the trouble of computing a 3600x2400 grid so
that I can have a 300 DPI image end up being 12x8, that?s what I want to
end up with (i don?t want to compute any more than I have to, but by the
same token, I want to see everything I?m computing).  Anyway, i?ll see if
that does what I?m after and reply again if not.  (Though I?d still like to
know what?s going on under the hood, so that I can exploit that if I
choose.)

Thanks again!

DLG

On Fri, Oct 20, 2017 at 3:44 PM Jody Klymak <jklymak at uvic.ca> wrote:

>
>
> On 20 Oct 2017, at  15:33 PM, David Goldsmith <eulergaussriemann at gmail.com>
> wrote:
>
>  But mostly, I?d just like to understand what?s going on, because what?s
> happening now makes absolutely no sense to me whatsoever.
>
>>
> Ha!
>
> I think you are approaching this from the point of view of an image
> processing library or something like Photoshop where you want your image to
> be 6x8 and 300 dpi, and you expect 1800x2400 pixels.
>
> Thats not *really* what matplotlib is for. It really is for presentation
> of scientific data.  So, I have a 500x600 matrix of data, I may use
> `imshow` to plot it up, but most people don?t care about the dpi of those
> pixels, so they will just set the dpi to something that shows all the data
> clearly.  Much more important is the ability to have axis labels,
> annotations, etc.
>
> Its not to say that you can?t be careful and get what you want.  I think
> what I sent you is pretty close.  But the library is not trying to conserve
> pixels.
>
> Anyway, what I sent should be clear enough.  I think what you were missing
> was setting the figure size to the size you wanted before trying to draw on
> it.
>
> Cheers,   Jody
>
>
>
>
>
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From jklymak at uvic.ca  Fri Oct 20 20:32:57 2017
From: jklymak at uvic.ca (Jody Klymak)
Date: Fri, 20 Oct 2017 17:32:57 -0700
Subject: [Matplotlib-users] Why is canvas.print_figure to a png reduce
 the real size of my image
In-Reply-To: <CA+_2Pn1M_8vryWWzCfhEfX152rDBQ=2TCNr=FgZ7ebMFS881gQ@mail.gmail.com>
References: <CA+_2Pn1rYc=y8imytzjDfVJ2Z3gro-k92JqMhKoBHHr9YkHdPQ@mail.gmail.com>
 <45743492-C845-4493-A8EE-585FB33040C3@uvic.ca>
 <CA+_2Pn3Be5Tx_5sNEk82aaMcTvyAff=_pVPXV7k0Sh8vkPKeOA@mail.gmail.com>
 <686398FC-DCB6-4367-AA99-E841CC4846B8@uvic.ca>
 <CA+_2Pn1M_8vryWWzCfhEfX152rDBQ=2TCNr=FgZ7ebMFS881gQ@mail.gmail.com>
Message-ID: <ACEE7E73-867E-40F2-A147-F20D15662707@uvic.ca>

Fair enough.  Just to be safe, just do dpi=600!  It?ll be big, but...

Cheers,   Jody

> On Oct 20, 2017, at  16:02 PM, David Goldsmith <eulergaussriemann at gmail.com> wrote:
> 
> OK, thanks for your help.
> 
> What I?m doing is somewhere in between: I?m making my own ?fractals,? ?from scratch?; so if im going to the trouble of computing a 3600x2400 grid so that I can have a 300 DPI image end up being 12x8, that?s what I want to end up with (i don?t want to compute any more than I have to, but by the same token, I want to see everything I?m computing).  Anyway, i?ll see if that does what I?m after and reply again if not.  (Though I?d still like to know what?s going on under the hood, so that I can exploit that if I choose.)
> 
> Thanks again!
> 
> DLG
> 
> On Fri, Oct 20, 2017 at 3:44 PM Jody Klymak <jklymak at uvic.ca <mailto:jklymak at uvic.ca>> wrote:
> 
> 
>> On 20 Oct 2017, at  15:33 PM, David Goldsmith <eulergaussriemann at gmail.com <mailto:eulergaussriemann at gmail.com>> wrote:
>> 
>>  But mostly, I?d just like to understand what?s going on, because what?s happening now makes absolutely no sense to me whatsoever.
> 
> Ha!  
> 
> I think you are approaching this from the point of view of an image processing library or something like Photoshop where you want your image to be 6x8 and 300 dpi, and you expect 1800x2400 pixels.  
> 
> Thats not *really* what matplotlib is for. It really is for presentation of scientific data.  So, I have a 500x600 matrix of data, I may use `imshow` to plot it up, but most people don?t care about the dpi of those pixels, so they will just set the dpi to something that shows all the data clearly.  Much more important is the ability to have axis labels, annotations, etc.  
> 
> Its not to say that you can?t be careful and get what you want.  I think what I sent you is pretty close.  But the library is not trying to conserve pixels.  
> 
> Anyway, what I sent should be clear enough.  I think what you were missing was setting the figure size to the size you wanted before trying to draw on it.
> 
> Cheers,   Jody
> 
> 
> 
> 

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From warren.weckesser at gmail.com  Sat Oct 21 08:27:09 2017
From: warren.weckesser at gmail.com (Warren Weckesser)
Date: Sat, 21 Oct 2017 08:27:09 -0400
Subject: [Matplotlib-users] Why is canvas.print_figure to a png reduce
 the real size of my image
In-Reply-To: <CA+_2Pn3Ak67qX8W0PDOuyA6VK=f7Xt3npehheskoSVpkL2q-vw@mail.gmail.com>
References: <CA+_2Pn1rYc=y8imytzjDfVJ2Z3gro-k92JqMhKoBHHr9YkHdPQ@mail.gmail.com>
 <CAGzF1ucFurk4CV=fDo_5K4CSFztLmK_cUo_dyYXcvc0skGzS=g@mail.gmail.com>
 <CA+_2Pn3Ak67qX8W0PDOuyA6VK=f7Xt3npehheskoSVpkL2q-vw@mail.gmail.com>
Message-ID: <CAGzF1uee_=pLPbz5bMQjdZHSr6riiFQvVbBoOMD969ZLL3aAhg@mail.gmail.com>

On Fri, Oct 20, 2017 at 6:36 PM, David Goldsmith <
eulergaussriemann at gmail.com> wrote:

> Thanks.  Does it have pylab?s range of built in color maps?  Does it even
> use colormaps, or do I have to write my own (or ?copy? them from MPL).
>

No, numpngw doesn't use colormaps.  You'd have to map and scale the image
to an array of integers yourself.  You could use matplotlib for the
mapping.  For example, if foo is a 2-D array of floats with values between
0 and 1, you could do something like

    mapped_foo = mpl.cm.viridis(foo)  # Use matplotlib's 'viridis'
colormap.
    maxint = 2**16
    img = np.floor(maxint*mapped_foo)
    img[img == maxint] = maxint - 1
    numpngw.write_png("foo.png", img.astype(np.uint16))

That would create a 16 bit RGB PNG file.

Warren




> DLG
>
> On Fri, Oct 20, 2017 at 3:28 PM Warren Weckesser <
> warren.weckesser at gmail.com> wrote:
>
>> On Fri, Oct 20, 2017 at 5:38 PM, David Goldsmith <
>> eulergaussriemann at gmail.com> wrote:
>>
>>> Hi!  I spent a good part of yesterday trying to figure this out on my
>>> own, without success, so I'm posting.  I have a 3601x2401 pixel, 300 DPI
>>> figure I export to a png using print_figure; here's sample code:
>>>
>>>
>>
>> David,
>>
>> If you are trying to write the numpy array directly to a PNG file, an
>> alternative to matplotlib is a package that I created called numpngw:
>> https://pypi.python.org/pypi/numpngw (github: https://github.com/
>> WarrenWeckesser/numpngw).  Both those links show several examples of its
>> use.   numpngw uses just Python and NumPy, so it is easy to install.
>>
>> Warren
>>
>>
>> PLB.imshow(rslt) # rslt is a 3601x2401 int-valued array
>>> curfig = PLB.Figure
>>> fig = PLB.gcf()
>>> # next three lines are in my code; don't seem relevant, but I include them in case they are
>>>
>>> ax = fig.gca()
>>> ax.xaxis.set_ticks([])
>>> ax.yaxis.set_ticks([])
>>>
>>> fig.canvas.print_figure(fn + '.png',
>>>                         dpi=300,
>>>
>>>                         bbox_inches='tight',
>>>
>>>                         pad_inches=-1.0 / 72)
>>>
>>> The resulting png is neither the right number of pixels by pixels, nor the right number of inches by incheses; what's more the resulting png differs depending on whether I use imshow or matshow (which I also tried).
>>>
>>> What's going on, and how do I "fix" it?  (If I really am getting the full resolution figure, despite what the png is telling me are the pixel counts, why are the physical dimensions off, and why is the png telling me that the pixel counts are different?  If it has something to do with compression, is there some  undocumented way to say "no compression" and will that have the desired effect?)
>>>
>>> Thanks!
>>>
>>> DLG
>>>
>>>
>>>
>>> _______________________________________________
>>> Matplotlib-users mailing list
>>> Matplotlib-users at python.org
>>> https://mail.python.org/mailman/listinfo/matplotlib-users
>>>
>>>
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From ardeal.liang at philips.com  Fri Oct 27 03:50:45 2017
From: ardeal.liang at philips.com (Ardeal LIANG)
Date: Fri, 27 Oct 2017 07:50:45 +0000
Subject: [Matplotlib-users] Scientific notation digits at left-bottom corner
 of figure
Message-ID: <AM5P121MB00340983CBCDCAA55F7F700FF75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>

Hi,

If I put the mouse on the figure, x and y axis are displayed on the left-bottom corner of the figure.
but the x axis is displayed as scientific notation, I want it to be displayed as decimal notation.
how to make x axis or y axis values displayed as decimal notation on the left-bottom corner?


[cid:image003.jpg at 01D34F3B.58162F30]

Best Regards,
Ardeal Liang


________________________________
The information contained in this email may be confidential and/or legally protected under applicable law. The message is intended solely for the addressee(s). If you are not the intended recipient, you are hereby notified that any use, forwarding, dissemination, or reproduction of this email is strictly prohibited and may be unlawful. If you are not the intended recipient, please contact the sender by return e-mail and destroy all copies of the original email.
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From efiring at hawaii.edu  Fri Oct 27 03:59:43 2017
From: efiring at hawaii.edu (Eric Firing)
Date: Thu, 26 Oct 2017 21:59:43 -1000
Subject: [Matplotlib-users] Scientific notation digits at left-bottom
 corner of figure
In-Reply-To: <AM5P121MB00340983CBCDCAA55F7F700FF75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
References: <AM5P121MB00340983CBCDCAA55F7F700FF75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
Message-ID: <60935ee1-4b69-76a5-4c71-d2901cb00b2e@hawaii.edu>

On 2017/10/26 9:50 PM, Ardeal LIANG wrote:
> Hi,
> 
> If I put the mouse on the figure, x and y axis are displayed on the 
> left-bottom corner of the figure.
> but the x axis is displayed as scientific notation, I want it to be 
> displayed as decimal notation.
> 
> how to make x axis or y axis values displayed as decimal notation on the 
> left-bottom corner?
> 
> *Best Regards,*
> 
> *Ardeal Liang*

Something like this, assuming your Axes is ax:

from matplotlib.ticker import ScalarFormatter

fmt_no_sci = ScalarFormatter(useOffset=False)
fmt_no_sci.set_scientific(False)

ax.fmt_xaxis = fmt_no_sci



Eric

From ardeal.liang at philips.com  Fri Oct 27 04:45:31 2017
From: ardeal.liang at philips.com (Ardeal LIANG)
Date: Fri, 27 Oct 2017 08:45:31 +0000
Subject: [Matplotlib-users] Scientific notation digits at left-bottom
 corner of figure
In-Reply-To: <60935ee1-4b69-76a5-4c71-d2901cb00b2e@hawaii.edu>
References: <AM5P121MB00340983CBCDCAA55F7F700FF75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <60935ee1-4b69-76a5-4c71-d2901cb00b2e@hawaii.edu>
Message-ID: <AM5P121MB0034F03A31E4303B00B1D3AEF75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>

Hi Eric,



I add you code to my code:
fmt_no_sci = ScalarFormatter(useOffset=False)
fmt_no_sci.set_scientific(False)
# ax.fmt_xaxis = fmt_no_sci

fig = plt.figure()
ax = fig.add_subplot(111)
ax.plot(testdata)
ax.fmt_xaxis = fmt_no_sci
plt.show(0)



The x axis value displayed at the left-bottom corner of the figure is still scientific notation, but not decimal notation.

Please check the x axis value displayed at the left-bottom corner of the figure in the red circle.

[cid:image001.jpg at 01D34F42.FD042720]







Best Regards,

Ardeal Liang



-----Original Message-----
From: Matplotlib-users [mailto:matplotlib-users-bounces+ardeal.liang=philips.com at python.org] On Behalf Of Eric Firing
Sent: 2017?10?27? 16:00
To: matplotlib-users at python.org
Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure



On 2017/10/26 9:50 PM, Ardeal LIANG wrote:

> Hi,

>

> If I put the mouse on the figure, x and y axis are displayed on the

> left-bottom corner of the figure.

> but the x axis is displayed as scientific notation, I want it to be

> displayed as decimal notation.

>

> how to make x axis or y axis values displayed as decimal notation on

> the left-bottom corner?

>

> *Best Regards,*

>

> *Ardeal Liang*



Something like this, assuming your Axes is ax:



from matplotlib.ticker import ScalarFormatter



fmt_no_sci = ScalarFormatter(useOffset=False)

fmt_no_sci.set_scientific(False)



ax.fmt_xaxis = fmt_no_sci







Eric

_______________________________________________

Matplotlib-users mailing list

Matplotlib-users at python.org<mailto:Matplotlib-users at python.org>

https://mail.python.org/mailman/listinfo/matplotlib-users

________________________________
The information contained in this email may be confidential and/or legally protected under applicable law. The message is intended solely for the addressee(s). If you are not the intended recipient, you are hereby notified that any use, forwarding, dissemination, or reproduction of this email is strictly prohibited and may be unlawful. If you are not the intended recipient, please contact the sender by return e-mail and destroy all copies of the original email.
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From berceanu at runbox.com  Fri Oct 27 07:17:54 2017
From: berceanu at runbox.com (Andrei Berceanu)
Date: Fri, 27 Oct 2017 13:17:54 +0200 (CEST)
Subject: [Matplotlib-users] loop through x and y axes
Message-ID: <E1e82eA-00052Q-70@rmmprod05.runbox>

Hi all,

Please consider this code:

fig, (ax1, ax2) = plt.subplots(ncols=2, figsize=[14,4])

for ax in (ax1, ax2):
    ax.set_xlabel(labels[0])
    ax.set_ylabel(labels[1])

This could be simplified if I could just loop though the x and y axes of each subplot, something like:

for index, ax in enumerate((ax1, ax2)):
    ax.set_label[index](labels[index])

Is there some way of achieving this in Matplotlib? From what I found, it seems one can only refer to the axes using x and y, and so I don't know how to loop over both of them.

Thank you,
Andrei

From joferkington at gmail.com  Fri Oct 27 11:05:52 2017
From: joferkington at gmail.com (Joe Kington)
Date: Fri, 27 Oct 2017 10:05:52 -0500
Subject: [Matplotlib-users] loop through x and y axes
In-Reply-To: <E1e82eA-00052Q-70@rmmprod05.runbox>
References: <E1e82eA-00052Q-70@rmmprod05.runbox>
Message-ID: <CACe1pJeqHqWg54tQCgcYKM8ApsjRSXkNZCTL8KyO-UFXwHKLuw@mail.gmail.com>

It sounds like you want to use the ``set_label_text`` method of the
``Axis`` instances directly?  Perhaps something like:


import matplotlib.pyplot as plt

labels = ['xlabel', 'ylabel']

fig, axes = plt.subplots(ncols=2)
for ax in axes:
    for axis, label in zip([ax.xaxis, ax.yaxis], labels):
        axis.set(label_text=label)

fig.tight_layout()
plt.show()



However, that's not really much cleaner than doing:


import matplotlib.pyplot as plt

fig, axes = plt.subplots(ncols=2)
for ax in axes:
    ax.set(xlabel='xlabel', ylabel='ylabel')

fig.tight_layout()
plt.show()


However, I might be misunderstanding your question entirely...  Hope that
helps a bit, anyway.
-Joe

On Fri, Oct 27, 2017 at 6:17 AM, Andrei Berceanu <berceanu at runbox.com>
wrote:

> Hi all,
>
> Please consider this code:
>
> fig, (ax1, ax2) = plt.subplots(ncols=2, figsize=[14,4])
>
> for ax in (ax1, ax2):
>     ax.set_xlabel(labels[0])
>     ax.set_ylabel(labels[1])
>
> This could be simplified if I could just loop though the x and y axes of
> each subplot, something like:
>
> for index, ax in enumerate((ax1, ax2)):
>     ax.set_label[index](labels[index])
>
> Is there some way of achieving this in Matplotlib? From what I found, it
> seems one can only refer to the axes using x and y, and so I don't know how
> to loop over both of them.
>
> Thank you,
> Andrei
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org
> https://mail.python.org/mailman/listinfo/matplotlib-users
>
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From ardeal.liang at philips.com  Fri Oct 27 04:07:40 2017
From: ardeal.liang at philips.com (Ardeal LIANG)
Date: Fri, 27 Oct 2017 08:07:40 +0000
Subject: [Matplotlib-users] Scientific notation digits at left-bottom
 corner of figure
In-Reply-To: <60935ee1-4b69-76a5-4c71-d2901cb00b2e@hawaii.edu>
References: <AM5P121MB00340983CBCDCAA55F7F700FF75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <60935ee1-4b69-76a5-4c71-d2901cb00b2e@hawaii.edu>
Message-ID: <AM5P121MB0034BD7560CCE138784C96B7F75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>

Hi Eric,



I added your code to my code:




from matplotlib.ticker import ScalarFormatter

fmt_no_sci = ScalarFormatter(useOffset=False)
fmt_no_sci.set_scientific(False)

# ax.fmt_xaxis = fmt_no_sci

prow = 4
pcol = 1
pidx = 0
fig = plt.figure()
pidx += 1
axx = fig.add_subplot(prow, pcol, pidx, ylabel='rawdata')
axx.plot(np.abs(testdata))

axx.fmt_xaxis = fmt_no_sci





The x axis value displayed at the left-bottom corner of the figure is still scientific notation, but not decimal notation.







Best Regards,

Ardeal Liang



-----Original Message-----
From: Matplotlib-users [mailto:matplotlib-users-bounces+ardeal.liang=philips.com at python.org] On Behalf Of Eric Firing
Sent: 2017?10?27? 16:00
To: matplotlib-users at python.org
Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure



On 2017/10/26 9:50 PM, Ardeal LIANG wrote:

> Hi,

>

> If I put the mouse on the figure, x and y axis are displayed on the

> left-bottom corner of the figure.

> but the x axis is displayed as scientific notation, I want it to be

> displayed as decimal notation.

>

> how to make x axis or y axis values displayed as decimal notation on

> the left-bottom corner?

>

> *Best Regards,*

>

> *Ardeal Liang*



Something like this, assuming your Axes is ax:



from matplotlib.ticker import ScalarFormatter



fmt_no_sci = ScalarFormatter(useOffset=False)

fmt_no_sci.set_scientific(False)



ax.fmt_xaxis = fmt_no_sci







Eric

_______________________________________________

Matplotlib-users mailing list

Matplotlib-users at python.org<mailto:Matplotlib-users at python.org>

https://mail.python.org/mailman/listinfo/matplotlib-users

________________________________
The information contained in this email may be confidential and/or legally protected under applicable law. The message is intended solely for the addressee(s). If you are not the intended recipient, you are hereby notified that any use, forwarding, dissemination, or reproduction of this email is strictly prohibited and may be unlawful. If you are not the intended recipient, please contact the sender by return e-mail and destroy all copies of the original email.
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From jklymak at uvic.ca  Fri Oct 27 11:39:38 2017
From: jklymak at uvic.ca (Jody Klymak)
Date: Fri, 27 Oct 2017 08:39:38 -0700
Subject: [Matplotlib-users] loop through x and y axes
In-Reply-To: <E1e82eA-00052Q-70@rmmprod05.runbox>
References: <E1e82eA-00052Q-70@rmmprod05.runbox>
Message-ID: <1D56B271-0D0E-43B6-90DB-F51ACB0A6633@uvic.ca>


How about, as recently pointed out by @anntzer on github:

```python
fig, axs = plt.subplots(1, 2, figsize=[14, 4])
plt.setp(axs, ?xlabel?, labels[0], ?ylabel?, labels[1])
```

or, suppose you don?t want to repeat labels except on outer 

```python
fig, axs = plt.subplots(3, 3, figsize=[14, 4])
plt.setp(axs[-1,:], ?xlabel?, labels[0])
plt.setp(axs[0,:], ?ylabel?, labels[1])
```

Cheers,  Jody


> On 27 Oct 2017, at  4:17 AM, Andrei Berceanu <berceanu at runbox.com> wrote:
> 
> Hi all,
> 
> Please consider this code:
> 
> fig, (ax1, ax2) = plt.subplots(ncols=2, figsize=[14,4])
> 
> for ax in (ax1, ax2):
>    ax.set_xlabel(labels[0])
>    ax.set_ylabel(labels[1])
> 
> This could be simplified if I could just loop though the x and y axes of each subplot, something like:
> 
> for index, ax in enumerate((ax1, ax2)):
>    ax.set_label[index](labels[index])
> 
> Is there some way of achieving this in Matplotlib? From what I found, it seems one can only refer to the axes using x and y, and so I don't know how to loop over both of them.
> 
> Thank you,
> Andrei
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org
> https://mail.python.org/mailman/listinfo/matplotlib-users

--
Jody Klymak    
http://web.uvic.ca/~jklymak/






From efiring at hawaii.edu  Fri Oct 27 13:22:42 2017
From: efiring at hawaii.edu (Eric Firing)
Date: Fri, 27 Oct 2017 07:22:42 -1000
Subject: [Matplotlib-users] Scientific notation digits at left-bottom
 corner of figure
In-Reply-To: <AM5P121MB0034BD7560CCE138784C96B7F75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
References: <AM5P121MB00340983CBCDCAA55F7F700FF75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <60935ee1-4b69-76a5-4c71-d2901cb00b2e@hawaii.edu>
 <AM5P121MB0034BD7560CCE138784C96B7F75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
Message-ID: <135de16c-f71b-78aa-40bc-2b7e6279c0d6@hawaii.edu>

I'm getting a different result, but on the basic point we now agree: 
the ScalarFormatter is not the right tool for this.  What you need is a 
function that takes a number and returns the kind of string you want.

Examples:

def format_as_int(num):
     return str(int(num))

def format_float_no_sci(num):
     return "%f" % num

Eric

On 2017/10/26 10:07 PM, Ardeal LIANG wrote:
> Hi Eric,
> 
> I added your code to my code:
> 
> *from *matplotlib.ticker *import *ScalarFormatter
> 
> fmt_no_sci = ScalarFormatter(useOffset=False)
> fmt_no_sci.set_scientific(False)
> 
> /# ax.fmt_xaxis = fmt_no_sci
> 
> /prow = 4
> pcol = 1
> pidx = 0
> fig = plt.figure()
> pidx += 1
> axx = fig.add_subplot(prow, pcol, pidx, ylabel=*'rawdata'*)
> axx.plot(np.abs(testdata))
> 
> axx.fmt_xaxis = fmt_no_sci
> 
> The *x axis value displayed at the left-bottom corner of the figure*is 
> still scientific notation, but not decimal notation.
> 
> Best Regards,
> 
> Ardeal Liang
> 
> -----Original Message-----
> From: Matplotlib-users 
> [mailto:matplotlib-users-bounces+ardeal.liang=philips.com at python.org] On 
> Behalf Of Eric Firing
> Sent: 2017?10?27? 16:00
> To: matplotlib-users at python.org
> Subject: Re: [Matplotlib-users] Scientific notation digits at 
> left-bottom corner of figure
> 
> On 2017/10/26 9:50 PM, Ardeal LIANG wrote:
> 
>  > Hi,
> 
>  >
> 
>  > If I put the mouse on the figure, x and y axis are displayed on the
> 
>  > left-bottom corner of the figure.
> 
>  > but the x axis is displayed as scientific notation, I want it to be
> 
>  > displayed as decimal notation.
> 
>  >
> 
>  > how to make x axis or y axis values displayed as decimal notation on
> 
>  > the left-bottom corner?
> 
>  >
> 
>  > *Best Regards,*
> 
>  >
> 
>  > *Ardeal Liang*
> 
> Something like this, assuming your Axes is ax:
> 
> from matplotlib.ticker import ScalarFormatter
> 
> fmt_no_sci = ScalarFormatter(useOffset=False)
> 
> fmt_no_sci.set_scientific(False)
> 
> ax.fmt_xaxis = fmt_no_sci
> 
> Eric
> 
> _______________________________________________
> 
> Matplotlib-users mailing list
> 
> Matplotlib-users at python.org <mailto:Matplotlib-users at python.org>
> 
> https://mail.python.org/mailman/listinfo/matplotlib-users
> 
> 
> ------------------------------------------------------------------------
> The information contained in this email may be confidential and/or 
> legally protected under applicable law. The message is intended solely 
> for the addressee(s). If you are not the intended recipient, you are 
> hereby notified that any use, forwarding, dissemination, or reproduction 
> of this email is strictly prohibited and may be unlawful. If you are not 
> the intended recipient, please contact the sender by return e-mail and 
> destroy all copies of the original email.


From ardeal.liang at philips.com  Sun Oct 29 20:46:04 2017
From: ardeal.liang at philips.com (Ardeal LIANG)
Date: Mon, 30 Oct 2017 00:46:04 +0000
Subject: [Matplotlib-users] Scientific notation digits at left-bottom
 corner of figure
In-Reply-To: <135de16c-f71b-78aa-40bc-2b7e6279c0d6@hawaii.edu>
References: <AM5P121MB00340983CBCDCAA55F7F700FF75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <60935ee1-4b69-76a5-4c71-d2901cb00b2e@hawaii.edu>
 <AM5P121MB0034BD7560CCE138784C96B7F75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <135de16c-f71b-78aa-40bc-2b7e6279c0d6@hawaii.edu>
Message-ID: <AM5P121MB0034152073291DA0B39CBC0AF7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>

HI Eric,

Did you review the picture attached in my former email?
I suspect you go to the wrong direction about what I want to ask.

Best Regards,
Ardeal Liang

-----Original Message-----
From: Eric Firing [mailto:efiring at hawaii.edu]
Sent: 2017?10?28? 1:23
To: Ardeal LIANG <ardeal.liang at philips.com>; matplotlib-users at python.org
Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure

I'm getting a different result, but on the basic point we now agree:
the ScalarFormatter is not the right tool for this.  What you need is a function that takes a number and returns the kind of string you want.

Examples:

def format_as_int(num):
     return str(int(num))

def format_float_no_sci(num):
     return "%f" % num

Eric

On 2017/10/26 10:07 PM, Ardeal LIANG wrote:
> Hi Eric,
>
> I added your code to my code:
>
> *from *matplotlib.ticker *import *ScalarFormatter
>
> fmt_no_sci = ScalarFormatter(useOffset=False)
> fmt_no_sci.set_scientific(False)
>
> /# ax.fmt_xaxis = fmt_no_sci
>
> /prow = 4
> pcol = 1
> pidx = 0
> fig = plt.figure()
> pidx += 1
> axx = fig.add_subplot(prow, pcol, pidx, ylabel=*'rawdata'*)
> axx.plot(np.abs(testdata))
>
> axx.fmt_xaxis = fmt_no_sci
>
> The *x axis value displayed at the left-bottom corner of the figure*is
> still scientific notation, but not decimal notation.
>
> Best Regards,
>
> Ardeal Liang
>
> -----Original Message-----
> From: Matplotlib-users
> [mailto:matplotlib-users-bounces+ardeal.liang=philips.com at python.org]
> On Behalf Of Eric Firing
> Sent: 2017?10?27? 16:00
> To: matplotlib-users at python.org
> Subject: Re: [Matplotlib-users] Scientific notation digits at
> left-bottom corner of figure
>
> On 2017/10/26 9:50 PM, Ardeal LIANG wrote:
>
>  > Hi,
>
>  >
>
>  > If I put the mouse on the figure, x and y axis are displayed on the
>
>  > left-bottom corner of the figure.
>
>  > but the x axis is displayed as scientific notation, I want it to be
>
>  > displayed as decimal notation.
>
>  >
>
>  > how to make x axis or y axis values displayed as decimal notation
> on
>
>  > the left-bottom corner?
>
>  >
>
>  > *Best Regards,*
>
>  >
>
>  > *Ardeal Liang*
>
> Something like this, assuming your Axes is ax:
>
> from matplotlib.ticker import ScalarFormatter
>
> fmt_no_sci = ScalarFormatter(useOffset=False)
>
> fmt_no_sci.set_scientific(False)
>
> ax.fmt_xaxis = fmt_no_sci
>
> Eric
>
> _______________________________________________
>
> Matplotlib-users mailing list
>
> Matplotlib-users at python.org <mailto:Matplotlib-users at python.org>
>
> https://mail.python.org/mailman/listinfo/matplotlib-users
>
>
> ----------------------------------------------------------------------
> -- The information contained in this email may be confidential and/or
> legally protected under applicable law. The message is intended solely
> for the addressee(s). If you are not the intended recipient, you are
> hereby notified that any use, forwarding, dissemination, or
> reproduction of this email is strictly prohibited and may be unlawful.
> If you are not the intended recipient, please contact the sender by
> return e-mail and destroy all copies of the original email.


________________________________
The information contained in this email may be confidential and/or legally protected under applicable law. The message is intended solely for the addressee(s). If you are not the intended recipient, you are hereby notified that any use, forwarding, dissemination, or reproduction of this email is strictly prohibited and may be unlawful. If you are not the intended recipient, please contact the sender by return e-mail and destroy all copies of the original email.

From jklymak at uvic.ca  Sun Oct 29 22:24:14 2017
From: jklymak at uvic.ca (Jody Klymak)
Date: Sun, 29 Oct 2017 19:24:14 -0700
Subject: [Matplotlib-users] Scientific notation digits at left-bottom
 corner of figure
In-Reply-To: <AM5P121MB0034152073291DA0B39CBC0AF7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
References: <AM5P121MB00340983CBCDCAA55F7F700FF75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <60935ee1-4b69-76a5-4c71-d2901cb00b2e@hawaii.edu>
 <AM5P121MB0034BD7560CCE138784C96B7F75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <135de16c-f71b-78aa-40bc-2b7e6279c0d6@hawaii.edu>
 <AM5P121MB0034152073291DA0B39CBC0AF7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
Message-ID: <9ABC007B-267A-4A2D-8342-62917D281C88@uvic.ca>



> On Oct 29, 2017, at  17:46 PM, Ardeal LIANG <ardeal.liang at philips.com> wrote:
> 
> HI Eric,
> 
> Did you review the picture attached in my former email?
> I suspect you go to the wrong direction about what I want to ask.
> 
> Best Regards,
> Ardeal Liang

Hi Ardeal,

Its not clear to me what you want.  Do you want 3.2 x 10^3 instead of 0.0032?  The former is scientific and the latter decimal.

Thanks,  Jody



> 
> -----Original Message-----
> From: Eric Firing [mailto:efiring at hawaii.edu <mailto:efiring at hawaii.edu>]
> Sent: 2017?10?28? 1:23
> To: Ardeal LIANG <ardeal.liang at philips.com <mailto:ardeal.liang at philips.com>>; matplotlib-users at python.org <mailto:matplotlib-users at python.org>
> Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure
> 
> I'm getting a different result, but on the basic point we now agree:
> the ScalarFormatter is not the right tool for this.  What you need is a function that takes a number and returns the kind of string you want.
> 
> Examples:
> 
> def format_as_int(num):
>     return str(int(num))
> 
> def format_float_no_sci(num):
>     return "%f" % num
> 
> Eric
> 
> On 2017/10/26 10:07 PM, Ardeal LIANG wrote:
>> Hi Eric,
>> 
>> I added your code to my code:
>> 
>> *from *matplotlib.ticker *import *ScalarFormatter
>> 
>> fmt_no_sci = ScalarFormatter(useOffset=False)
>> fmt_no_sci.set_scientific(False)
>> 
>> /# ax.fmt_xaxis = fmt_no_sci
>> 
>> /prow = 4
>> pcol = 1
>> pidx = 0
>> fig = plt.figure()
>> pidx += 1
>> axx = fig.add_subplot(prow, pcol, pidx, ylabel=*'rawdata'*)
>> axx.plot(np.abs(testdata))
>> 
>> axx.fmt_xaxis = fmt_no_sci
>> 
>> The *x axis value displayed at the left-bottom corner of the figure*is
>> still scientific notation, but not decimal notation.
>> 
>> Best Regards,
>> 
>> Ardeal Liang
>> 
>> -----Original Message-----
>> From: Matplotlib-users
>> [mailto:matplotlib-users-bounces+ardeal.liang=philips.com at python.org]
>> On Behalf Of Eric Firing
>> Sent: 2017?10?27? 16:00
>> To: matplotlib-users at python.org
>> Subject: Re: [Matplotlib-users] Scientific notation digits at
>> left-bottom corner of figure
>> 
>> On 2017/10/26 9:50 PM, Ardeal LIANG wrote:
>> 
>>> Hi,
>> 
>>> 
>> 
>>> If I put the mouse on the figure, x and y axis are displayed on the
>> 
>>> left-bottom corner of the figure.
>> 
>>> but the x axis is displayed as scientific notation, I want it to be
>> 
>>> displayed as decimal notation.
>> 
>>> 
>> 
>>> how to make x axis or y axis values displayed as decimal notation
>> on
>> 
>>> the left-bottom corner?
>> 
>>> 
>> 
>>> *Best Regards,*
>> 
>>> 
>> 
>>> *Ardeal Liang*
>> 
>> Something like this, assuming your Axes is ax:
>> 
>> from matplotlib.ticker import ScalarFormatter
>> 
>> fmt_no_sci = ScalarFormatter(useOffset=False)
>> 
>> fmt_no_sci.set_scientific(False)
>> 
>> ax.fmt_xaxis = fmt_no_sci
>> 
>> Eric
>> 
>> _______________________________________________
>> 
>> Matplotlib-users mailing list
>> 
>> Matplotlib-users at python.org <mailto:Matplotlib-users at python.org>
>> 
>> https://mail.python.org/mailman/listinfo/matplotlib-users
>> 
>> 
>> ----------------------------------------------------------------------
>> -- The information contained in this email may be confidential and/or
>> legally protected under applicable law. The message is intended solely
>> for the addressee(s). If you are not the intended recipient, you are
>> hereby notified that any use, forwarding, dissemination, or
>> reproduction of this email is strictly prohibited and may be unlawful.
>> If you are not the intended recipient, please contact the sender by
>> return e-mail and destroy all copies of the original email.
> 
> 
> ________________________________
> The information contained in this email may be confidential and/or legally protected under applicable law. The message is intended solely for the addressee(s). If you are not the intended recipient, you are hereby notified that any use, forwarding, dissemination, or reproduction of this email is strictly prohibited and may be unlawful. If you are not the intended recipient, please contact the sender by return e-mail and destroy all copies of the original email.
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org <mailto:Matplotlib-users at python.org>
> https://mail.python.org/mailman/listinfo/matplotlib-users <https://mail.python.org/mailman/listinfo/matplotlib-users>
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From mckauf at gmail.com  Sun Oct 29 22:31:30 2017
From: mckauf at gmail.com (Mike Kaufman)
Date: Sun, 29 Oct 2017 22:31:30 -0400
Subject: [Matplotlib-users] Scientific notation digits at left-bottom
 corner of figure
In-Reply-To: <9ABC007B-267A-4A2D-8342-62917D281C88@uvic.ca>
References: <AM5P121MB00340983CBCDCAA55F7F700FF75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <60935ee1-4b69-76a5-4c71-d2901cb00b2e@hawaii.edu>
 <AM5P121MB0034BD7560CCE138784C96B7F75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <135de16c-f71b-78aa-40bc-2b7e6279c0d6@hawaii.edu>
 <AM5P121MB0034152073291DA0B39CBC0AF7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <9ABC007B-267A-4A2D-8342-62917D281C88@uvic.ca>
Message-ID: <6d55cfbf-71bd-5dd1-5b41-7ed989362cd6@gmail.com>

I believe he's talking about the coordinates label in the navigation 
toolbar...

M

On 10/29/17 10:24 PM, Jody Klymak wrote:
>
>
>> On Oct 29, 2017, at  17:46 PM, Ardeal LIANG <ardeal.liang at philips.com
>> <mailto:ardeal.liang at philips.com>> wrote:
>>
>> HI Eric,
>>
>> Did you review the picture attached in my former email?
>> I suspect you go to the wrong direction about what I want to ask.
>>
>> Best Regards,
>> Ardeal Liang
>
> Hi Ardeal,
>
> Its not clear to me what you want.  Do you want 3.2 x 10^3 instead of
> 0.0032?  The former is scientific and the latter decimal.
>
> Thanks,  Jody
>
>
>
>>
>> -----Original Message-----
>> From: Eric Firing [mailto:efiring at hawaii.edu]
>> Sent: 2017?10?28? 1:23
>> To: Ardeal LIANG <ardeal.liang at philips.com
>> <mailto:ardeal.liang at philips.com>>; matplotlib-users at python.org
>> <mailto:matplotlib-users at python.org>
>> Subject: Re: [Matplotlib-users] Scientific notation digits at
>> left-bottom corner of figure
>>
>> I'm getting a different result, but on the basic point we now agree:
>> the ScalarFormatter is not the right tool for this.  What you need is
>> a function that takes a number and returns the kind of string you want.
>>
>> Examples:
>>
>> def format_as_int(num):
>>     return str(int(num))
>>
>> def format_float_no_sci(num):
>>     return "%f" % num
>>
>> Eric
>>
>> On 2017/10/26 10:07 PM, Ardeal LIANG wrote:
>>> Hi Eric,
>>>
>>> I added your code to my code:
>>>
>>> *from *matplotlib.ticker *import *ScalarFormatter
>>>
>>> fmt_no_sci = ScalarFormatter(useOffset=False)
>>> fmt_no_sci.set_scientific(False)
>>>
>>> /# ax.fmt_xaxis = fmt_no_sci
>>>
>>> /prow = 4
>>> pcol = 1
>>> pidx = 0
>>> fig = plt.figure()
>>> pidx += 1
>>> axx = fig.add_subplot(prow, pcol, pidx, ylabel=*'rawdata'*)
>>> axx.plot(np.abs(testdata))
>>>
>>> axx.fmt_xaxis = fmt_no_sci
>>>
>>> The *x axis value displayed at the left-bottom corner of the figure*is
>>> still scientific notation, but not decimal notation.
>>>
>>> Best Regards,
>>>
>>> Ardeal Liang
>>>
>>> -----Original Message-----
>>> From: Matplotlib-users
>>> [mailto:matplotlib-users-bounces+ardeal.liang=philips.com at python.org]
>>> On Behalf Of Eric Firing
>>> Sent: 2017?10?27? 16:00
>>> To: matplotlib-users at python.org <mailto:matplotlib-users at python.org>
>>> Subject: Re: [Matplotlib-users] Scientific notation digits at
>>> left-bottom corner of figure
>>>
>>> On 2017/10/26 9:50 PM, Ardeal LIANG wrote:
>>>
>>>> Hi,
>>>
>>>>
>>>
>>>> If I put the mouse on the figure, x and y axis are displayed on the
>>>
>>>> left-bottom corner of the figure.
>>>
>>>> but the x axis is displayed as scientific notation, I want it to be
>>>
>>>> displayed as decimal notation.
>>>
>>>>
>>>
>>>> how to make x axis or y axis values displayed as decimal notation
>>> on
>>>
>>>> the left-bottom corner?
>>>
>>>>
>>>
>>>> *Best Regards,*
>>>
>>>>
>>>
>>>> *Ardeal Liang*
>>>
>>> Something like this, assuming your Axes is ax:
>>>
>>> from matplotlib.ticker import ScalarFormatter
>>>
>>> fmt_no_sci = ScalarFormatter(useOffset=False)
>>>
>>> fmt_no_sci.set_scientific(False)
>>>
>>> ax.fmt_xaxis = fmt_no_sci
>>>
>>> Eric
>>>
>>> _______________________________________________
>>>
>>> Matplotlib-users mailing list
>>>
>>> Matplotlib-users at python.org <mailto:Matplotlib-users at python.org>
>>> <mailto:Matplotlib-users at python.org>
>>>
>>> https://mail.python.org/mailman/listinfo/matplotlib-users
>>>
>>>
>>> ----------------------------------------------------------------------
>>> -- The information contained in this email may be confidential and/or
>>> legally protected under applicable law. The message is intended solely
>>> for the addressee(s). If you are not the intended recipient, you are
>>> hereby notified that any use, forwarding, dissemination, or
>>> reproduction of this email is strictly prohibited and may be unlawful.
>>> If you are not the intended recipient, please contact the sender by
>>> return e-mail and destroy all copies of the original email.
>>
>>
>> ________________________________
>> The information contained in this email may be confidential and/or
>> legally protected under applicable law. The message is intended solely
>> for the addressee(s). If you are not the intended recipient, you are
>> hereby notified that any use, forwarding, dissemination, or
>> reproduction of this email is strictly prohibited and may be unlawful.
>> If you are not the intended recipient, please contact the sender by
>> return e-mail and destroy all copies of the original email.
>> _______________________________________________
>> Matplotlib-users mailing list
>> Matplotlib-users at python.org <mailto:Matplotlib-users at python.org>
>> https://mail.python.org/mailman/listinfo/matplotlib-users
>
>
>
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org
> https://mail.python.org/mailman/listinfo/matplotlib-users
>


From ardeal.liang at philips.com  Sun Oct 29 22:36:42 2017
From: ardeal.liang at philips.com (Ardeal LIANG)
Date: Mon, 30 Oct 2017 02:36:42 +0000
Subject: [Matplotlib-users] Scientific notation digits at left-bottom
 corner of figure
In-Reply-To: <E181C6E8-1F33-44C3-A169-3D6F30BC56CF@uvic.ca>
References: <AM5P121MB00340983CBCDCAA55F7F700FF75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <60935ee1-4b69-76a5-4c71-d2901cb00b2e@hawaii.edu>
 <AM5P121MB0034BD7560CCE138784C96B7F75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <135de16c-f71b-78aa-40bc-2b7e6279c0d6@hawaii.edu>
 <AM5P121MB0034152073291DA0B39CBC0AF7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <9ABC007B-267A-4A2D-8342-62917D281C88@uvic.ca>
 <AM5P121MB00343424E73AB679A96D5EA4F7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <E181C6E8-1F33-44C3-A169-3D6F30BC56CF@uvic.ca>
Message-ID: <AM5P121MB00349E346C7704FE4425D062F7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>

Hi Jody,

I apologize if I said it backwards!
I think you now get what I want.

I tried your code. It didn?t work still.

Best Regards,
Ardeal Liang

From: Jody Klymak [mailto:jklymak at uvic.ca]
Sent: 2017?10?30? 10:34
To: Ardeal LIANG <ardeal.liang at philips.com>
Cc: Firing Eric <efiring at hawaii.edu>; matplotlib-users at python.org
Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure

Did you try:

from matplotlib.ticker import ScalarFormatter

fmt_no_sci = ScalarFormatter(useOffset=False)
fmt_no_sci.set_scientific(True)  # NOTE True instead of False.

I think you just said it backwards: You want scientific ntoation, not decimal notation.

Cheers,   Jody



On Oct 29, 2017, at  19:29 PM, Ardeal LIANG <ardeal.liang at philips.com<mailto:ardeal.liang at philips.com>> wrote:

Hi Jody,

Please check the red ellipse in the picture.

I want the value 1.85092e+06 to be 1850922 in the red ellipse in the following picture.


fmt_no_sci = ScalarFormatter(useOffset=False)
fmt_no_sci.set_scientific(False)
# ax.fmt_xaxis = fmt_no_sci

fig = plt.figure()
ax = fig.add_subplot(111)
ax.plot(testdata)
ax.fmt_xaxis = fmt_no_sci
plt.show(0)

The x axis value displayed at the left-bottom corner of the figure is still scientific notation, but not decimal notation.
Please check the x axis value displayed at the left-bottom corner of the figure in the red circle.
<image001.jpg>




Best Regards,
Ardeal Liang

From: Jody Klymak [mailto:jklymak at uvic.ca]
Sent: 2017?10?30? 10:24
To: Ardeal LIANG <ardeal.liang at philips.com<mailto:ardeal.liang at philips.com>>
Cc: Firing Eric <efiring at hawaii.edu<mailto:efiring at hawaii.edu>>; matplotlib-users at python.org<mailto:matplotlib-users at python.org>
Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure





On Oct 29, 2017, at  17:46 PM, Ardeal LIANG <ardeal.liang at philips.com<mailto:ardeal.liang at philips.com>> wrote:

HI Eric,

Did you review the picture attached in my former email?
I suspect you go to the wrong direction about what I want to ask.

Best Regards,
Ardeal Liang

Hi Ardeal,

Its not clear to me what you want.  Do you want 3.2 x 10^3 instead of 0.0032?  The former is scientific and the latter decimal.

Thanks,  Jody






-----Original Message-----
From: Eric Firing [mailto:efiring at hawaii.edu]
Sent: 2017?10?28? 1:23
To: Ardeal LIANG <ardeal.liang at philips.com<mailto:ardeal.liang at philips.com>>; matplotlib-users at python.org<mailto:matplotlib-users at python.org>
Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure

I'm getting a different result, but on the basic point we now agree:
the ScalarFormatter is not the right tool for this.  What you need is a function that takes a number and returns the kind of string you want.

Examples:

def format_as_int(num):
    return str(int(num))

def format_float_no_sci(num):
    return "%f" % num

Eric

On 2017/10/26 10:07 PM, Ardeal LIANG wrote:


Hi Eric,

I added your code to my code:

*from *matplotlib.ticker *import *ScalarFormatter

fmt_no_sci = ScalarFormatter(useOffset=False)
fmt_no_sci.set_scientific(False)

/# ax.fmt_xaxis = fmt_no_sci

/prow = 4
pcol = 1
pidx = 0
fig = plt.figure()
pidx += 1
axx = fig.add_subplot(prow, pcol, pidx, ylabel=*'rawdata'*)
axx.plot(np.abs(testdata))

axx.fmt_xaxis = fmt_no_sci

The *x axis value displayed at the left-bottom corner of the figure*is
still scientific notation, but not decimal notation.

Best Regards,

Ardeal Liang

-----Original Message-----
From: Matplotlib-users
[mailto:matplotlib-users-bounces+ardeal.liang=philips.com at python.org]
On Behalf Of Eric Firing
Sent: 2017?10?27? 16:00
To: matplotlib-users at python.org<mailto:matplotlib-users at python.org>
Subject: Re: [Matplotlib-users] Scientific notation digits at
left-bottom corner of figure

On 2017/10/26 9:50 PM, Ardeal LIANG wrote:



Hi,







If I put the mouse on the figure, x and y axis are displayed on the



left-bottom corner of the figure.



but the x axis is displayed as scientific notation, I want it to be



displayed as decimal notation.







how to make x axis or y axis values displayed as decimal notation
on



the left-bottom corner?







*Best Regards,*







*Ardeal Liang*

Something like this, assuming your Axes is ax:

from matplotlib.ticker import ScalarFormatter

fmt_no_sci = ScalarFormatter(useOffset=False)

fmt_no_sci.set_scientific(False)

ax.fmt_xaxis = fmt_no_sci

Eric

_______________________________________________

Matplotlib-users mailing list

Matplotlib-users at python.org<mailto:Matplotlib-users at python.org> <mailto:Matplotlib-users at python.org>

https://mail.python.org/mailman/listinfo/matplotlib-users


----------------------------------------------------------------------
-- The information contained in this email may be confidential and/or
legally protected under applicable law. The message is intended solely
for the addressee(s). If you are not the intended recipient, you are
hereby notified that any use, forwarding, dissemination, or
reproduction of this email is strictly prohibited and may be unlawful.
If you are not the intended recipient, please contact the sender by
return e-mail and destroy all copies of the original email.


________________________________
The information contained in this email may be confidential and/or legally protected under applicable law. The message is intended solely for the addressee(s). If you are not the intended recipient, you are hereby notified that any use, forwarding, dissemination, or reproduction of this email is strictly prohibited and may be unlawful. If you are not the intended recipient, please contact the sender by return e-mail and destroy all copies of the original email.
_______________________________________________
Matplotlib-users mailing list
Matplotlib-users at python.org<mailto:Matplotlib-users at python.org>
https://mail.python.org/mailman/listinfo/matplotlib-users


________________________________
The information contained in this email may be confidential and/or legally protected under applicable law. The message is intended solely for the addressee(s). If you are not the intended recipient, you are hereby notified that any use, forwarding, dissemination, or reproduction of this email is strictly prohibited and may be unlawful. If you are not the intended recipient, please contact the sender by return e-mail and destroy all copies of the original email.


________________________________
The information contained in this email may be confidential and/or legally protected under applicable law. The message is intended solely for the addressee(s). If you are not the intended recipient, you are hereby notified that any use, forwarding, dissemination, or reproduction of this email is strictly prohibited and may be unlawful. If you are not the intended recipient, please contact the sender by return e-mail and destroy all copies of the original email.
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From jklymak at uvic.ca  Sun Oct 29 22:33:46 2017
From: jklymak at uvic.ca (Jody Klymak)
Date: Sun, 29 Oct 2017 19:33:46 -0700
Subject: [Matplotlib-users] Scientific notation digits at left-bottom
 corner of figure
In-Reply-To: <AM5P121MB00343424E73AB679A96D5EA4F7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
References: <AM5P121MB00340983CBCDCAA55F7F700FF75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <60935ee1-4b69-76a5-4c71-d2901cb00b2e@hawaii.edu>
 <AM5P121MB0034BD7560CCE138784C96B7F75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <135de16c-f71b-78aa-40bc-2b7e6279c0d6@hawaii.edu>
 <AM5P121MB0034152073291DA0B39CBC0AF7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <9ABC007B-267A-4A2D-8342-62917D281C88@uvic.ca>
 <AM5P121MB00343424E73AB679A96D5EA4F7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
Message-ID: <E181C6E8-1F33-44C3-A169-3D6F30BC56CF@uvic.ca>

Did you try:

from matplotlib.ticker import ScalarFormatter

fmt_no_sci = ScalarFormatter(useOffset=False)
fmt_no_sci.set_scientific(True)  # NOTE True instead of False.

I think you just said it backwards: You want scientific ntoation, not decimal notation.

Cheers,   Jody


> On Oct 29, 2017, at  19:29 PM, Ardeal LIANG <ardeal.liang at philips.com> wrote:
> 
> Hi Jody,
>  
> Please check the red ellipse in the picture.
>  
> I want the value 1.85092e+06 to be 1850922 in the red ellipse in the following picture.
>  
>  
> fmt_no_sci = ScalarFormatter(useOffset=False)
> fmt_no_sci.set_scientific(False)
> # ax.fmt_xaxis = fmt_no_sci
> 
> fig = plt.figure()
> ax = fig.add_subplot(111)
> ax.plot(testdata)
> ax.fmt_xaxis = fmt_no_sci
> plt.show(0)
>  
> The x axis value displayed at the left-bottom corner of the figure is still scientific notation, but not decimal notation.
> Please check the x axis value displayed at the left-bottom corner of the figure in the red circle.
> <image001.jpg>
>  
>  
>  
>  
> Best Regards,
> Ardeal Liang
>  
> From: Jody Klymak [mailto:jklymak at uvic.ca] 
> Sent: 2017?10?30? 10:24
> To: Ardeal LIANG <ardeal.liang at philips.com>
> Cc: Firing Eric <efiring at hawaii.edu>; matplotlib-users at python.org
> Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure
>  
>  
> 
> 
> On Oct 29, 2017, at  17:46 PM, Ardeal LIANG <ardeal.liang at philips.com <mailto:ardeal.liang at philips.com>> wrote:
>  
> HI Eric,
> 
> Did you review the picture attached in my former email?
> I suspect you go to the wrong direction about what I want to ask.
> 
> Best Regards,
> Ardeal Liang
>  
> Hi Ardeal,
>  
> Its not clear to me what you want.  Do you want 3.2 x 10^3 instead of 0.0032?  The former is scientific and the latter decimal.
>  
> Thanks,  Jody
>  
>  
> 
> 
> 
> -----Original Message-----
> From: Eric Firing [mailto:efiring at hawaii.edu <mailto:efiring at hawaii.edu>]
> Sent: 2017?10?28? 1:23
> To: Ardeal LIANG <ardeal.liang at philips.com <mailto:ardeal.liang at philips.com>>; matplotlib-users at python.org <mailto:matplotlib-users at python.org>
> Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure
> 
> I'm getting a different result, but on the basic point we now agree:
> the ScalarFormatter is not the right tool for this.  What you need is a function that takes a number and returns the kind of string you want.
> 
> Examples:
> 
> def format_as_int(num):
>     return str(int(num))
> 
> def format_float_no_sci(num):
>     return "%f" % num
> 
> Eric
> 
> On 2017/10/26 10:07 PM, Ardeal LIANG wrote:
> 
> Hi Eric,
> 
> I added your code to my code:
> 
> *from *matplotlib.ticker *import *ScalarFormatter
> 
> fmt_no_sci = ScalarFormatter(useOffset=False)
> fmt_no_sci.set_scientific(False)
> 
> /# ax.fmt_xaxis = fmt_no_sci
> 
> /prow = 4
> pcol = 1
> pidx = 0
> fig = plt.figure()
> pidx += 1
> axx = fig.add_subplot(prow, pcol, pidx, ylabel=*'rawdata'*)
> axx.plot(np.abs(testdata))
> 
> axx.fmt_xaxis = fmt_no_sci
> 
> The *x axis value displayed at the left-bottom corner of the figure*is
> still scientific notation, but not decimal notation.
> 
> Best Regards,
> 
> Ardeal Liang
> 
> -----Original Message-----
> From: Matplotlib-users
> [mailto:matplotlib-users-bounces+ardeal.liang=philips.com at python.org <mailto:matplotlib-users-bounces+ardeal.liang=philips.com at python.org>]
> On Behalf Of Eric Firing
> Sent: 2017?10?27? 16:00
> To: matplotlib-users at python.org <mailto:matplotlib-users at python.org>
> Subject: Re: [Matplotlib-users] Scientific notation digits at
> left-bottom corner of figure
> 
> On 2017/10/26 9:50 PM, Ardeal LIANG wrote:
> 
> 
> Hi,
> 
> 
>  
> 
> 
> If I put the mouse on the figure, x and y axis are displayed on the
> 
> 
> left-bottom corner of the figure.
> 
> 
> but the x axis is displayed as scientific notation, I want it to be
> 
> 
> displayed as decimal notation.
> 
> 
>  
> 
> 
> how to make x axis or y axis values displayed as decimal notation
> on
> 
> 
> the left-bottom corner?
> 
> 
>  
> 
> 
> *Best Regards,*
> 
> 
>  
> 
> 
> *Ardeal Liang*
> 
> Something like this, assuming your Axes is ax:
> 
> from matplotlib.ticker import ScalarFormatter
> 
> fmt_no_sci = ScalarFormatter(useOffset=False)
> 
> fmt_no_sci.set_scientific(False)
> 
> ax.fmt_xaxis = fmt_no_sci
> 
> Eric
> 
> _______________________________________________
> 
> Matplotlib-users mailing list
> 
> Matplotlib-users at python.org <mailto:Matplotlib-users at python.org> <mailto:Matplotlib-users at python.org <mailto:Matplotlib-users at python.org>>
> 
> https://mail.python.org/mailman/listinfo/matplotlib-users <https://mail.python.org/mailman/listinfo/matplotlib-users>
> 
> 
> ----------------------------------------------------------------------
> -- The information contained in this email may be confidential and/or
> legally protected under applicable law. The message is intended solely
> for the addressee(s). If you are not the intended recipient, you are
> hereby notified that any use, forwarding, dissemination, or
> reproduction of this email is strictly prohibited and may be unlawful.
> If you are not the intended recipient, please contact the sender by
> return e-mail and destroy all copies of the original email.
> 
> 
> ________________________________
> The information contained in this email may be confidential and/or legally protected under applicable law. The message is intended solely for the addressee(s). If you are not the intended recipient, you are hereby notified that any use, forwarding, dissemination, or reproduction of this email is strictly prohibited and may be unlawful. If you are not the intended recipient, please contact the sender by return e-mail and destroy all copies of the original email.
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org <mailto:Matplotlib-users at python.org>
> https://mail.python.org/mailman/listinfo/matplotlib-users <https://mail.python.org/mailman/listinfo/matplotlib-users>
>  
> 
> The information contained in this email may be confidential and/or legally protected under applicable law. The message is intended solely for the addressee(s). If you are not the intended recipient, you are hereby notified that any use, forwarding, dissemination, or reproduction of this email is strictly prohibited and may be unlawful. If you are not the intended recipient, please contact the sender by return e-mail and destroy all copies of the original email.

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From jklymak at uvic.ca  Sun Oct 29 22:44:13 2017
From: jklymak at uvic.ca (Jody Klymak)
Date: Sun, 29 Oct 2017 19:44:13 -0700
Subject: [Matplotlib-users] Scientific notation digits at left-bottom
 corner of figure
In-Reply-To: <AM5P121MB00349E346C7704FE4425D062F7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
References: <AM5P121MB00340983CBCDCAA55F7F700FF75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <60935ee1-4b69-76a5-4c71-d2901cb00b2e@hawaii.edu>
 <AM5P121MB0034BD7560CCE138784C96B7F75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <135de16c-f71b-78aa-40bc-2b7e6279c0d6@hawaii.edu>
 <AM5P121MB0034152073291DA0B39CBC0AF7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <9ABC007B-267A-4A2D-8342-62917D281C88@uvic.ca>
 <AM5P121MB00343424E73AB679A96D5EA4F7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <E181C6E8-1F33-44C3-A169-3D6F30BC56CF@uvic.ca>
 <AM5P121MB00349E346C7704FE4425D062F7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
Message-ID: <8DF3FBE8-BBD1-4051-AEB2-CD1C6786CABB@uvic.ca>

Maybe?


def format_coord(x, y):
    return 'x=%4.4e, y=%4.4e?%(x, y))  # or whatever formatting you want

ax.format_coord = format_coord


Cheers,   Jody



> On Oct 29, 2017, at  19:36 PM, Ardeal LIANG <ardeal.liang at philips.com> wrote:
> 
> Hi Jody,
>  
> I apologize if I said it backwards!
> I think you now get what I want.
>  
> I tried your code. It didn?t work still.
>  
> Best Regards,
> Ardeal Liang
>  
> From: Jody Klymak [mailto:jklymak at uvic.ca] 
> Sent: 2017?10?30? 10:34
> To: Ardeal LIANG <ardeal.liang at philips.com>
> Cc: Firing Eric <efiring at hawaii.edu>; matplotlib-users at python.org
> Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure
>  
> Did you try:
> 
> from matplotlib.ticker import ScalarFormatter
> 
> fmt_no_sci = ScalarFormatter(useOffset=False)
> fmt_no_sci.set_scientific(True)  # NOTE True instead of False.
> 
> I think you just said it backwards: You want scientific ntoation, not decimal notation.
> 
> Cheers,   Jody
>  
> 
> 
> On Oct 29, 2017, at  19:29 PM, Ardeal LIANG <ardeal.liang at philips.com <mailto:ardeal.liang at philips.com>> wrote:
>  
> Hi Jody,
>  
> Please check the red ellipse in the picture.
>  
> I want the value 1.85092e+06 to be 1850922 in the red ellipse in the following picture.
>  
>  
> fmt_no_sci = ScalarFormatter(useOffset=False)
> fmt_no_sci.set_scientific(False)
> # ax.fmt_xaxis = fmt_no_sci
> 
> fig = plt.figure()
> ax = fig.add_subplot(111)
> ax.plot(testdata)
> ax.fmt_xaxis = fmt_no_sci
> plt.show(0)
>  
> The x axis value displayed at the left-bottom corner of the figure is still scientific notation, but not decimal notation.
> Please check the x axis value displayed at the left-bottom corner of the figure in the red circle.
> <image001.jpg>
>  
>  
>  
>  
> Best Regards,
> Ardeal Liang
>  
> From: Jody Klymak [mailto:jklymak at uvic.ca <mailto:jklymak at uvic.ca>] 
> Sent: 2017?10?30? 10:24
> To: Ardeal LIANG <ardeal.liang at philips.com <mailto:ardeal.liang at philips.com>>
> Cc: Firing Eric <efiring at hawaii.edu <mailto:efiring at hawaii.edu>>; matplotlib-users at python.org <mailto:matplotlib-users at python.org>
> Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure
>  
>  
> 
> 
> 
> On Oct 29, 2017, at  17:46 PM, Ardeal LIANG <ardeal.liang at philips.com <mailto:ardeal.liang at philips.com>> wrote:
>  
> HI Eric,
> 
> Did you review the picture attached in my former email?
> I suspect you go to the wrong direction about what I want to ask.
> 
> Best Regards,
> Ardeal Liang
>  
> Hi Ardeal,
>  
> Its not clear to me what you want.  Do you want 3.2 x 10^3 instead of 0.0032?  The former is scientific and the latter decimal.
>  
> Thanks,  Jody
>  
>  
> 
> 
> 
> 
> -----Original Message-----
> From: Eric Firing [mailto:efiring at hawaii.edu <mailto:efiring at hawaii.edu>]
> Sent: 2017?10?28? 1:23
> To: Ardeal LIANG <ardeal.liang at philips.com <mailto:ardeal.liang at philips.com>>; matplotlib-users at python.org <mailto:matplotlib-users at python.org>
> Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure
> 
> I'm getting a different result, but on the basic point we now agree:
> the ScalarFormatter is not the right tool for this.  What you need is a function that takes a number and returns the kind of string you want.
> 
> Examples:
> 
> def format_as_int(num):
>     return str(int(num))
> 
> def format_float_no_sci(num):
>     return "%f" % num
> 
> Eric
> 
> On 2017/10/26 10:07 PM, Ardeal LIANG wrote:
> 
> 
> Hi Eric,
> 
> I added your code to my code:
> 
> *from *matplotlib.ticker *import *ScalarFormatter
> 
> fmt_no_sci = ScalarFormatter(useOffset=False)
> fmt_no_sci.set_scientific(False)
> 
> /# ax.fmt_xaxis = fmt_no_sci
> 
> /prow = 4
> pcol = 1
> pidx = 0
> fig = plt.figure()
> pidx += 1
> axx = fig.add_subplot(prow, pcol, pidx, ylabel=*'rawdata'*)
> axx.plot(np.abs(testdata))
> 
> axx.fmt_xaxis = fmt_no_sci
> 
> The *x axis value displayed at the left-bottom corner of the figure*is
> still scientific notation, but not decimal notation.
> 
> Best Regards,
> 
> Ardeal Liang
> 
> -----Original Message-----
> From: Matplotlib-users
> [mailto:matplotlib-users-bounces+ardeal.liang=philips.com at python.org <mailto:matplotlib-users-bounces+ardeal.liang=philips.com at python.org>]
> On Behalf Of Eric Firing
> Sent: 2017?10?27? 16:00
> To: matplotlib-users at python.org <mailto:matplotlib-users at python.org>
> Subject: Re: [Matplotlib-users] Scientific notation digits at
> left-bottom corner of figure
> 
> On 2017/10/26 9:50 PM, Ardeal LIANG wrote:
> 
> 
> 
> Hi,
> 
> 
> 
>  
> 
> 
> 
> If I put the mouse on the figure, x and y axis are displayed on the
> 
> 
> 
> left-bottom corner of the figure.
> 
> 
> 
> but the x axis is displayed as scientific notation, I want it to be
> 
> 
> 
> displayed as decimal notation.
> 
> 
> 
>  
> 
> 
> 
> how to make x axis or y axis values displayed as decimal notation
> on
> 
> 
> 
> the left-bottom corner?
> 
> 
> 
>  
> 
> 
> 
> *Best Regards,*
> 
> 
> 
>  
> 
> 
> 
> *Ardeal Liang*
> 
> Something like this, assuming your Axes is ax:
> 
> from matplotlib.ticker import ScalarFormatter
> 
> fmt_no_sci = ScalarFormatter(useOffset=False)
> 
> fmt_no_sci.set_scientific(False)
> 
> ax.fmt_xaxis = fmt_no_sci
> 
> Eric
> 
> _______________________________________________
> 
> Matplotlib-users mailing list
> 
> Matplotlib-users at python.org <mailto:Matplotlib-users at python.org> <mailto:Matplotlib-users at python.org <mailto:Matplotlib-users at python.org>>
> 
> https://mail.python.org/mailman/listinfo/matplotlib-users <https://mail.python.org/mailman/listinfo/matplotlib-users>
> 
> 
> ----------------------------------------------------------------------
> -- The information contained in this email may be confidential and/or
> legally protected under applicable law. The message is intended solely
> for the addressee(s). If you are not the intended recipient, you are
> hereby notified that any use, forwarding, dissemination, or
> reproduction of this email is strictly prohibited and may be unlawful.
> If you are not the intended recipient, please contact the sender by
> return e-mail and destroy all copies of the original email.
> 
> 
> ________________________________
> The information contained in this email may be confidential and/or legally protected under applicable law. The message is intended solely for the addressee(s). If you are not the intended recipient, you are hereby notified that any use, forwarding, dissemination, or reproduction of this email is strictly prohibited and may be unlawful. If you are not the intended recipient, please contact the sender by return e-mail and destroy all copies of the original email.
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org <mailto:Matplotlib-users at python.org>
> https://mail.python.org/mailman/listinfo/matplotlib-users <https://mail.python.org/mailman/listinfo/matplotlib-users>
>  
>  
> The information contained in this email may be confidential and/or legally protected under applicable law. The message is intended solely for the addressee(s). If you are not the intended recipient, you are hereby notified that any use, forwarding, dissemination, or reproduction of this email is strictly prohibited and may be unlawful. If you are not the intended recipient, please contact the sender by return e-mail and destroy all copies of the original email.
>  
> 
> The information contained in this email may be confidential and/or legally protected under applicable law. The message is intended solely for the addressee(s). If you are not the intended recipient, you are hereby notified that any use, forwarding, dissemination, or reproduction of this email is strictly prohibited and may be unlawful. If you are not the intended recipient, please contact the sender by return e-mail and destroy all copies of the original email.

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From jklymak at uvic.ca  Sun Oct 29 22:53:18 2017
From: jklymak at uvic.ca (Jody Klymak)
Date: Sun, 29 Oct 2017 19:53:18 -0700
Subject: [Matplotlib-users] Scientific notation digits at left-bottom
 corner of figure
In-Reply-To: <AM5P121MB00343512D40FC5D82C5FB96CF7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
References: <AM5P121MB00340983CBCDCAA55F7F700FF75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <60935ee1-4b69-76a5-4c71-d2901cb00b2e@hawaii.edu>
 <AM5P121MB0034BD7560CCE138784C96B7F75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <135de16c-f71b-78aa-40bc-2b7e6279c0d6@hawaii.edu>
 <AM5P121MB0034152073291DA0B39CBC0AF7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <9ABC007B-267A-4A2D-8342-62917D281C88@uvic.ca>
 <AM5P121MB00343424E73AB679A96D5EA4F7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <E181C6E8-1F33-44C3-A169-3D6F30BC56CF@uvic.ca>
 <AM5P121MB00349E346C7704FE4425D062F7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <8DF3FBE8-BBD1-4051-AEB2-CD1C6786CABB@uvic.ca>
 <AM5P121MB00343512D40FC5D82C5FB96CF7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
Message-ID: <44BCB453-6271-46A9-B7A8-47F7AD311D64@uvic.ca>

ax.format_coord is the function that formats the numbers shown in the red ellipse of your picture.  Given that knowledge you can put anything in that red ellipse you want.

Cheers,   Jody

> On Oct 29, 2017, at  19:46 PM, Ardeal LIANG <ardeal.liang at philips.com> wrote:
> 
> Hi Jody,
>  
> Did you check the picture I attached? You are still in the wrong direction about what I want?????..
> Please check the digit in the red ellipse in my picture!!!!!!!!!!!!!!
>  
> Best Regards,
> Ardeal Liang
>  
> From: Jody Klymak [mailto:jklymak at uvic.ca <mailto:jklymak at uvic.ca>] 
> Sent: 2017?10?30? 10:44
> To: Ardeal LIANG <ardeal.liang at philips.com <mailto:ardeal.liang at philips.com>>
> Cc: Firing Eric <efiring at hawaii.edu <mailto:efiring at hawaii.edu>>; matplotlib-users at python.org <mailto:matplotlib-users at python.org>
> Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure
>  
> Maybe?
>  
>  
> def format_coord(x, y):
>     return 'x=%4.4e, y=%4.4e?%(x, y))  # or whatever formatting you want
>  
> ax.format_coord = format_coord
>  
>  
> Cheers,   Jody
>  
>  
> 
> 
> On Oct 29, 2017, at  19:36 PM, Ardeal LIANG <ardeal.liang at philips.com <mailto:ardeal.liang at philips.com>> wrote:
>  
> Hi Jody,
>  
> I apologize if I said it backwards!
> I think you now get what I want.
>  
> I tried your code. It didn?t work still.
>  
> Best Regards,
> Ardeal Liang
>  
> From: Jody Klymak [mailto:jklymak at uvic.ca <mailto:jklymak at uvic.ca>] 
> Sent: 2017?10?30? 10:34
> To: Ardeal LIANG <ardeal.liang at philips.com <mailto:ardeal.liang at philips.com>>
> Cc: Firing Eric <efiring at hawaii.edu <mailto:efiring at hawaii.edu>>; matplotlib-users at python.org <mailto:matplotlib-users at python.org>
> Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure
>  
> Did you try:
> 
> from matplotlib.ticker import ScalarFormatter
> 
> fmt_no_sci = ScalarFormatter(useOffset=False)
> fmt_no_sci.set_scientific(True)  # NOTE True instead of False.
> 
> I think you just said it backwards: You want scientific ntoation, not decimal notation.
> 
> Cheers,   Jody
>  
> 
> 
> 
> On Oct 29, 2017, at  19:29 PM, Ardeal LIANG <ardeal.liang at philips.com <mailto:ardeal.liang at philips.com>> wrote:
>  
> Hi Jody,
>  
> Please check the red ellipse in the picture.
>  
> I want the value 1.85092e+06 to be 1850922 in the red ellipse in the following picture.
>  
>  
> fmt_no_sci = ScalarFormatter(useOffset=False)
> fmt_no_sci.set_scientific(False)
> # ax.fmt_xaxis = fmt_no_sci
> 
> fig = plt.figure()
> ax = fig.add_subplot(111)
> ax.plot(testdata)
> ax.fmt_xaxis = fmt_no_sci
> plt.show(0)
>  
> The x axis value displayed at the left-bottom corner of the figure is still scientific notation, but not decimal notation.
> Please check the x axis value displayed at the left-bottom corner of the figure in the red circle.
> <image001.jpg>
>  
>  
>  
>  
> Best Regards,
> Ardeal Liang
>  
> From: Jody Klymak [mailto:jklymak at uvic.ca <mailto:jklymak at uvic.ca>] 
> Sent: 2017?10?30? 10:24
> To: Ardeal LIANG <ardeal.liang at philips.com <mailto:ardeal.liang at philips.com>>
> Cc: Firing Eric <efiring at hawaii.edu <mailto:efiring at hawaii.edu>>; matplotlib-users at python.org <mailto:matplotlib-users at python.org>
> Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure
>  
>  
> 
> 
> 
> 
> On Oct 29, 2017, at  17:46 PM, Ardeal LIANG <ardeal.liang at philips.com <mailto:ardeal.liang at philips.com>> wrote:
>  
> HI Eric,
> 
> Did you review the picture attached in my former email?
> I suspect you go to the wrong direction about what I want to ask.
> 
> Best Regards,
> Ardeal Liang
>  
> Hi Ardeal,
>  
> Its not clear to me what you want.  Do you want 3.2 x 10^3 instead of 0.0032?  The former is scientific and the latter decimal.
>  
> Thanks,  Jody
>  
>  
> 
> 
> 
> 
> 
> -----Original Message-----
> From: Eric Firing [mailto:efiring at hawaii.edu <mailto:efiring at hawaii.edu>]
> Sent: 2017?10?28? 1:23
> To: Ardeal LIANG <ardeal.liang at philips.com <mailto:ardeal.liang at philips.com>>; matplotlib-users at python.org <mailto:matplotlib-users at python.org>
> Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure
> 
> I'm getting a different result, but on the basic point we now agree:
> the ScalarFormatter is not the right tool for this.  What you need is a function that takes a number and returns the kind of string you want.
> 
> Examples:
> 
> def format_as_int(num):
>     return str(int(num))
> 
> def format_float_no_sci(num):
>     return "%f" % num
> 
> Eric
> 
> On 2017/10/26 10:07 PM, Ardeal LIANG wrote:
> 
> 
> 
> Hi Eric,
> 
> I added your code to my code:
> 
> *from *matplotlib.ticker *import *ScalarFormatter
> 
> fmt_no_sci = ScalarFormatter(useOffset=False)
> fmt_no_sci.set_scientific(False)
> 
> /# ax.fmt_xaxis = fmt_no_sci
> 
> /prow = 4
> pcol = 1
> pidx = 0
> fig = plt.figure()
> pidx += 1
> axx = fig.add_subplot(prow, pcol, pidx, ylabel=*'rawdata'*)
> axx.plot(np.abs(testdata))
> 
> axx.fmt_xaxis = fmt_no_sci
> 
> The *x axis value displayed at the left-bottom corner of the figure*is
> still scientific notation, but not decimal notation.
> 
> Best Regards,
> 
> Ardeal Liang
> 
> -----Original Message-----
> From: Matplotlib-users
> [mailto:matplotlib-users-bounces+ardeal.liang=philips.com at python.org <mailto:matplotlib-users-bounces+ardeal.liang=philips.com at python.org>]
> On Behalf Of Eric Firing
> Sent: 2017?10?27? 16:00
> To: matplotlib-users at python.org <mailto:matplotlib-users at python.org>
> Subject: Re: [Matplotlib-users] Scientific notation digits at
> left-bottom corner of figure
> 
> On 2017/10/26 9:50 PM, Ardeal LIANG wrote:
> 
> 
> 
> 
> Hi,
> 
> 
> 
> 
>  
> 
> 
> 
> 
> If I put the mouse on the figure, x and y axis are displayed on the
> 
> 
> 
> 
> left-bottom corner of the figure.
> 
> 
> 
> 
> but the x axis is displayed as scientific notation, I want it to be
> 
> 
> 
> 
> displayed as decimal notation.
> 
> 
> 
> 
>  
> 
> 
> 
> 
> how to make x axis or y axis values displayed as decimal notation
> on
> 
> 
> 
> 
> the left-bottom corner?
> 
> 
> 
> 
>  
> 
> 
> 
> 
> *Best Regards,*
> 
> 
> 
> 
>  
> 
> 
> 
> 
> *Ardeal Liang*
> 
> Something like this, assuming your Axes is ax:
> 
> from matplotlib.ticker import ScalarFormatter
> 
> fmt_no_sci = ScalarFormatter(useOffset=False)
> 
> fmt_no_sci.set_scientific(False)
> 
> ax.fmt_xaxis = fmt_no_sci
> 
> Eric
> 
> _______________________________________________
> 
> Matplotlib-users mailing list
> 
> Matplotlib-users at python.org <mailto:Matplotlib-users at python.org> <mailto:Matplotlib-users at python.org <mailto:Matplotlib-users at python.org>>
> 
> https://mail.python.org/mailman/listinfo/matplotlib-users <https://mail.python.org/mailman/listinfo/matplotlib-users>
> 
> 
> ----------------------------------------------------------------------
> -- The information contained in this email may be confidential and/or
> legally protected under applicable law. The message is intended solely
> for the addressee(s). If you are not the intended recipient, you are
> hereby notified that any use, forwarding, dissemination, or
> reproduction of this email is strictly prohibited and may be unlawful.
> If you are not the intended recipient, please contact the sender by
> return e-mail and destroy all copies of the original email.
> 
> 
> ________________________________
> The information contained in this email may be confidential and/or legally protected under applicable law. The message is intended solely for the addressee(s). If you are not the intended recipient, you are hereby notified that any use, forwarding, dissemination, or reproduction of this email is strictly prohibited and may be unlawful. If you are not the intended recipient, please contact the sender by return e-mail and destroy all copies of the original email.
> _______________________________________________
> Matplotlib-users mailing list
> Matplotlib-users at python.org <mailto:Matplotlib-users at python.org>
> https://mail.python.org/mailman/listinfo/matplotlib-users <https://mail.python.org/mailman/listinfo/matplotlib-users>
>  
>  
> The information contained in this email may be confidential and/or legally protected under applicable law. The message is intended solely for the addressee(s). If you are not the intended recipient, you are hereby notified that any use, forwarding, dissemination, or reproduction of this email is strictly prohibited and may be unlawful. If you are not the intended recipient, please contact the sender by return e-mail and destroy all copies of the original email.
>  
>  
> The information contained in this email may be confidential and/or legally protected under applicable law. The message is intended solely for the addressee(s). If you are not the intended recipient, you are hereby notified that any use, forwarding, dissemination, or reproduction of this email is strictly prohibited and may be unlawful. If you are not the intended recipient, please contact the sender by return e-mail and destroy all copies of the original email.
>  
> 
> The information contained in this email may be confidential and/or legally protected under applicable law. The message is intended solely for the addressee(s). If you are not the intended recipient, you are hereby notified that any use, forwarding, dissemination, or reproduction of this email is strictly prohibited and may be unlawful. If you are not the intended recipient, please contact the sender by return e-mail and destroy all copies of the original email.

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From ardeal.liang at philips.com  Sun Oct 29 22:42:06 2017
From: ardeal.liang at philips.com (Ardeal LIANG)
Date: Mon, 30 Oct 2017 02:42:06 +0000
Subject: [Matplotlib-users] Scientific notation digits at left-bottom
 corner of figure
In-Reply-To: <AM5P121MB00349E346C7704FE4425D062F7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
References: <AM5P121MB00340983CBCDCAA55F7F700FF75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <60935ee1-4b69-76a5-4c71-d2901cb00b2e@hawaii.edu>
 <AM5P121MB0034BD7560CCE138784C96B7F75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <135de16c-f71b-78aa-40bc-2b7e6279c0d6@hawaii.edu>
 <AM5P121MB0034152073291DA0B39CBC0AF7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <9ABC007B-267A-4A2D-8342-62917D281C88@uvic.ca>
 <AM5P121MB00343424E73AB679A96D5EA4F7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <E181C6E8-1F33-44C3-A169-3D6F30BC56CF@uvic.ca>
 <AM5P121MB00349E346C7704FE4425D062F7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
Message-ID: <AM5P121MB00344C63CFE723DE8DD2CA5EF7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>

Hi All,

You can try to run the following code.
If I put the mouse on the figure, the digit of x axis displayed on the left-bottom corner is scientific notation which is not what I want. I want it to be decimal notation.


testdata = np.arange(0, 10000000, 1)

# fmt_no_sci = ScalarFormatter(useOffset=False)
# fmt_no_sci.set_scientific(False)
# ax.fmt_xaxis = fmt_no_sci


from matplotlib.ticker import ScalarFormatter

fmt_no_sci = ScalarFormatter(useOffset=False)
fmt_no_sci.set_scientific(True)  # NOTE True instead of False.



fig = plt.figure()
ax = fig.add_subplot(111)
ax.plot(testdata)
ax.fmt_xaxis = fmt_no_sci
plt.show(0)



Best Regards,
Ardeal Liang

From: Ardeal LIANG [mailto:ardeal.liang at lighting.com]
Sent: 2017?10?30? 10:37
To: Jody Klymak <jklymak at uvic.ca>; Ardeal LIANG <ardeal.liang at philips.com>
Cc: Firing Eric <efiring at hawaii.edu>; matplotlib-users at python.org
Subject: RE: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure

Hi Jody,

I apologize if I said it backwards!
I think you now get what I want.

I tried your code. It didn?t work still.

Best Regards,
Ardeal Liang

From: Jody Klymak [mailto:jklymak at uvic.ca]
Sent: 2017?10?30? 10:34
To: Ardeal LIANG <ardeal.liang at philips.com<mailto:ardeal.liang at philips.com>>
Cc: Firing Eric <efiring at hawaii.edu<mailto:efiring at hawaii.edu>>; matplotlib-users at python.org<mailto:matplotlib-users at python.org>
Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure

Did you try:

from matplotlib.ticker import ScalarFormatter

fmt_no_sci = ScalarFormatter(useOffset=False)
fmt_no_sci.set_scientific(True)  # NOTE True instead of False.

I think you just said it backwards: You want scientific ntoation, not decimal notation.

Cheers,   Jody


On Oct 29, 2017, at  19:29 PM, Ardeal LIANG <ardeal.liang at philips.com<mailto:ardeal.liang at philips.com>> wrote:

Hi Jody,

Please check the red ellipse in the picture.

I want the value 1.85092e+06 to be 1850922 in the red ellipse in the following picture.


fmt_no_sci = ScalarFormatter(useOffset=False)
fmt_no_sci.set_scientific(False)
# ax.fmt_xaxis = fmt_no_sci

fig = plt.figure()
ax = fig.add_subplot(111)
ax.plot(testdata)
ax.fmt_xaxis = fmt_no_sci
plt.show(0)

The x axis value displayed at the left-bottom corner of the figure is still scientific notation, but not decimal notation.
Please check the x axis value displayed at the left-bottom corner of the figure in the red circle.
<image001.jpg>




Best Regards,
Ardeal Liang

From: Jody Klymak [mailto:jklymak at uvic.ca]
Sent: 2017?10?30? 10:24
To: Ardeal LIANG <ardeal.liang at philips.com<mailto:ardeal.liang at philips.com>>
Cc: Firing Eric <efiring at hawaii.edu<mailto:efiring at hawaii.edu>>; matplotlib-users at python.org<mailto:matplotlib-users at python.org>
Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure




On Oct 29, 2017, at  17:46 PM, Ardeal LIANG <ardeal.liang at philips.com<mailto:ardeal.liang at philips.com>> wrote:

HI Eric,

Did you review the picture attached in my former email?
I suspect you go to the wrong direction about what I want to ask.

Best Regards,
Ardeal Liang

Hi Ardeal,

Its not clear to me what you want.  Do you want 3.2 x 10^3 instead of 0.0032?  The former is scientific and the latter decimal.

Thanks,  Jody





-----Original Message-----
From: Eric Firing [mailto:efiring at hawaii.edu]
Sent: 2017?10?28? 1:23
To: Ardeal LIANG <ardeal.liang at philips.com<mailto:ardeal.liang at philips.com>>; matplotlib-users at python.org<mailto:matplotlib-users at python.org>
Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure

I'm getting a different result, but on the basic point we now agree:
the ScalarFormatter is not the right tool for this.  What you need is a function that takes a number and returns the kind of string you want.

Examples:

def format_as_int(num):
    return str(int(num))

def format_float_no_sci(num):
    return "%f" % num

Eric

On 2017/10/26 10:07 PM, Ardeal LIANG wrote:

Hi Eric,

I added your code to my code:

*from *matplotlib.ticker *import *ScalarFormatter

fmt_no_sci = ScalarFormatter(useOffset=False)
fmt_no_sci.set_scientific(False)

/# ax.fmt_xaxis = fmt_no_sci

/prow = 4
pcol = 1
pidx = 0
fig = plt.figure()
pidx += 1
axx = fig.add_subplot(prow, pcol, pidx, ylabel=*'rawdata'*)
axx.plot(np.abs(testdata))

axx.fmt_xaxis = fmt_no_sci

The *x axis value displayed at the left-bottom corner of the figure*is
still scientific notation, but not decimal notation.

Best Regards,

Ardeal Liang

-----Original Message-----
From: Matplotlib-users
[mailto:matplotlib-users-bounces+ardeal.liang=philips.com at python.org]
On Behalf Of Eric Firing
Sent: 2017?10?27? 16:00
To: matplotlib-users at python.org<mailto:matplotlib-users at python.org>
Subject: Re: [Matplotlib-users] Scientific notation digits at
left-bottom corner of figure

On 2017/10/26 9:50 PM, Ardeal LIANG wrote:


Hi,





If I put the mouse on the figure, x and y axis are displayed on the


left-bottom corner of the figure.


but the x axis is displayed as scientific notation, I want it to be


displayed as decimal notation.





how to make x axis or y axis values displayed as decimal notation
on


the left-bottom corner?





*Best Regards,*





*Ardeal Liang*

Something like this, assuming your Axes is ax:

from matplotlib.ticker import ScalarFormatter

fmt_no_sci = ScalarFormatter(useOffset=False)

fmt_no_sci.set_scientific(False)

ax.fmt_xaxis = fmt_no_sci

Eric

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From ardeal.liang at philips.com  Sun Oct 29 22:29:27 2017
From: ardeal.liang at philips.com (Ardeal LIANG)
Date: Mon, 30 Oct 2017 02:29:27 +0000
Subject: [Matplotlib-users] Scientific notation digits at left-bottom
 corner of figure
In-Reply-To: <9ABC007B-267A-4A2D-8342-62917D281C88@uvic.ca>
References: <AM5P121MB00340983CBCDCAA55F7F700FF75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <60935ee1-4b69-76a5-4c71-d2901cb00b2e@hawaii.edu>
 <AM5P121MB0034BD7560CCE138784C96B7F75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <135de16c-f71b-78aa-40bc-2b7e6279c0d6@hawaii.edu>
 <AM5P121MB0034152073291DA0B39CBC0AF7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <9ABC007B-267A-4A2D-8342-62917D281C88@uvic.ca>
Message-ID: <AM5P121MB00343424E73AB679A96D5EA4F7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>

Hi Jody,

Please check the red ellipse in the picture.

I want the value 1.85092e+06 to be 1850922 in the red ellipse in the following picture.


fmt_no_sci = ScalarFormatter(useOffset=False)
fmt_no_sci.set_scientific(False)
# ax.fmt_xaxis = fmt_no_sci

fig = plt.figure()
ax = fig.add_subplot(111)
ax.plot(testdata)
ax.fmt_xaxis = fmt_no_sci
plt.show(0)



The x axis value displayed at the left-bottom corner of the figure is still scientific notation, but not decimal notation.

Please check the x axis value displayed at the left-bottom corner of the figure in the red circle.

[cid:image001.jpg at 01D35169.F2F33F40]






Best Regards,
Ardeal Liang

From: Jody Klymak [mailto:jklymak at uvic.ca]
Sent: 2017?10?30? 10:24
To: Ardeal LIANG <ardeal.liang at philips.com>
Cc: Firing Eric <efiring at hawaii.edu>; matplotlib-users at python.org
Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure




On Oct 29, 2017, at  17:46 PM, Ardeal LIANG <ardeal.liang at philips.com<mailto:ardeal.liang at philips.com>> wrote:

HI Eric,

Did you review the picture attached in my former email?
I suspect you go to the wrong direction about what I want to ask.

Best Regards,
Ardeal Liang

Hi Ardeal,

Its not clear to me what you want.  Do you want 3.2 x 10^3 instead of 0.0032?  The former is scientific and the latter decimal.

Thanks,  Jody





-----Original Message-----
From: Eric Firing [mailto:efiring at hawaii.edu]
Sent: 2017?10?28? 1:23
To: Ardeal LIANG <ardeal.liang at philips.com<mailto:ardeal.liang at philips.com>>; matplotlib-users at python.org<mailto:matplotlib-users at python.org>
Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure

I'm getting a different result, but on the basic point we now agree:
the ScalarFormatter is not the right tool for this.  What you need is a function that takes a number and returns the kind of string you want.

Examples:

def format_as_int(num):
    return str(int(num))

def format_float_no_sci(num):
    return "%f" % num

Eric

On 2017/10/26 10:07 PM, Ardeal LIANG wrote:

Hi Eric,

I added your code to my code:

*from *matplotlib.ticker *import *ScalarFormatter

fmt_no_sci = ScalarFormatter(useOffset=False)
fmt_no_sci.set_scientific(False)

/# ax.fmt_xaxis = fmt_no_sci

/prow = 4
pcol = 1
pidx = 0
fig = plt.figure()
pidx += 1
axx = fig.add_subplot(prow, pcol, pidx, ylabel=*'rawdata'*)
axx.plot(np.abs(testdata))

axx.fmt_xaxis = fmt_no_sci

The *x axis value displayed at the left-bottom corner of the figure*is
still scientific notation, but not decimal notation.

Best Regards,

Ardeal Liang

-----Original Message-----
From: Matplotlib-users
[mailto:matplotlib-users-bounces+ardeal.liang=philips.com at python.org]
On Behalf Of Eric Firing
Sent: 2017?10?27? 16:00
To: matplotlib-users at python.org<mailto:matplotlib-users at python.org>
Subject: Re: [Matplotlib-users] Scientific notation digits at
left-bottom corner of figure

On 2017/10/26 9:50 PM, Ardeal LIANG wrote:


Hi,





If I put the mouse on the figure, x and y axis are displayed on the


left-bottom corner of the figure.


but the x axis is displayed as scientific notation, I want it to be


displayed as decimal notation.





how to make x axis or y axis values displayed as decimal notation
on


the left-bottom corner?





*Best Regards,*





*Ardeal Liang*

Something like this, assuming your Axes is ax:

from matplotlib.ticker import ScalarFormatter

fmt_no_sci = ScalarFormatter(useOffset=False)

fmt_no_sci.set_scientific(False)

ax.fmt_xaxis = fmt_no_sci

Eric

_______________________________________________

Matplotlib-users mailing list

Matplotlib-users at python.org<mailto:Matplotlib-users at python.org> <mailto:Matplotlib-users at python.org>

https://mail.python.org/mailman/listinfo/matplotlib-users


----------------------------------------------------------------------
-- The information contained in this email may be confidential and/or
legally protected under applicable law. The message is intended solely
for the addressee(s). If you are not the intended recipient, you are
hereby notified that any use, forwarding, dissemination, or
reproduction of this email is strictly prohibited and may be unlawful.
If you are not the intended recipient, please contact the sender by
return e-mail and destroy all copies of the original email.


________________________________
The information contained in this email may be confidential and/or legally protected under applicable law. The message is intended solely for the addressee(s). If you are not the intended recipient, you are hereby notified that any use, forwarding, dissemination, or reproduction of this email is strictly prohibited and may be unlawful. If you are not the intended recipient, please contact the sender by return e-mail and destroy all copies of the original email.
_______________________________________________
Matplotlib-users mailing list
Matplotlib-users at python.org<mailto:Matplotlib-users at python.org>
https://mail.python.org/mailman/listinfo/matplotlib-users


________________________________
The information contained in this email may be confidential and/or legally protected under applicable law. The message is intended solely for the addressee(s). If you are not the intended recipient, you are hereby notified that any use, forwarding, dissemination, or reproduction of this email is strictly prohibited and may be unlawful. If you are not the intended recipient, please contact the sender by return e-mail and destroy all copies of the original email.
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From ardeal.liang at philips.com  Sun Oct 29 22:46:23 2017
From: ardeal.liang at philips.com (Ardeal LIANG)
Date: Mon, 30 Oct 2017 02:46:23 +0000
Subject: [Matplotlib-users] Scientific notation digits at left-bottom
 corner of figure
In-Reply-To: <8DF3FBE8-BBD1-4051-AEB2-CD1C6786CABB@uvic.ca>
References: <AM5P121MB00340983CBCDCAA55F7F700FF75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <60935ee1-4b69-76a5-4c71-d2901cb00b2e@hawaii.edu>
 <AM5P121MB0034BD7560CCE138784C96B7F75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <135de16c-f71b-78aa-40bc-2b7e6279c0d6@hawaii.edu>
 <AM5P121MB0034152073291DA0B39CBC0AF7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <9ABC007B-267A-4A2D-8342-62917D281C88@uvic.ca>
 <AM5P121MB00343424E73AB679A96D5EA4F7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <E181C6E8-1F33-44C3-A169-3D6F30BC56CF@uvic.ca>
 <AM5P121MB00349E346C7704FE4425D062F7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <8DF3FBE8-BBD1-4051-AEB2-CD1C6786CABB@uvic.ca>
Message-ID: <AM5P121MB00343512D40FC5D82C5FB96CF7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>

Hi Jody,

Did you check the picture I attached? You are still in the wrong direction about what I want?????..
Please check the digit in the red ellipse in my picture!!!!!!!!!!!!!!

Best Regards,
Ardeal Liang

From: Jody Klymak [mailto:jklymak at uvic.ca]
Sent: 2017?10?30? 10:44
To: Ardeal LIANG <ardeal.liang at philips.com>
Cc: Firing Eric <efiring at hawaii.edu>; matplotlib-users at python.org
Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure

Maybe?


def format_coord(x, y):
    return 'x=%4.4e, y=%4.4e?%(x, y))  # or whatever formatting you want

ax.format_coord = format_coord


Cheers,   Jody




On Oct 29, 2017, at  19:36 PM, Ardeal LIANG <ardeal.liang at philips.com<mailto:ardeal.liang at philips.com>> wrote:

Hi Jody,

I apologize if I said it backwards!
I think you now get what I want.

I tried your code. It didn?t work still.

Best Regards,
Ardeal Liang

From: Jody Klymak [mailto:jklymak at uvic.ca]
Sent: 2017?10?30? 10:34
To: Ardeal LIANG <ardeal.liang at philips.com<mailto:ardeal.liang at philips.com>>
Cc: Firing Eric <efiring at hawaii.edu<mailto:efiring at hawaii.edu>>; matplotlib-users at python.org<mailto:matplotlib-users at python.org>
Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure

Did you try:

from matplotlib.ticker import ScalarFormatter

fmt_no_sci = ScalarFormatter(useOffset=False)
fmt_no_sci.set_scientific(True)  # NOTE True instead of False.

I think you just said it backwards: You want scientific ntoation, not decimal notation.

Cheers,   Jody




On Oct 29, 2017, at  19:29 PM, Ardeal LIANG <ardeal.liang at philips.com<mailto:ardeal.liang at philips.com>> wrote:

Hi Jody,

Please check the red ellipse in the picture.

I want the value 1.85092e+06 to be 1850922 in the red ellipse in the following picture.


fmt_no_sci = ScalarFormatter(useOffset=False)
fmt_no_sci.set_scientific(False)
# ax.fmt_xaxis = fmt_no_sci

fig = plt.figure()
ax = fig.add_subplot(111)
ax.plot(testdata)
ax.fmt_xaxis = fmt_no_sci
plt.show(0)

The x axis value displayed at the left-bottom corner of the figure is still scientific notation, but not decimal notation.
Please check the x axis value displayed at the left-bottom corner of the figure in the red circle.
<image001.jpg>




Best Regards,
Ardeal Liang

From: Jody Klymak [mailto:jklymak at uvic.ca]
Sent: 2017?10?30? 10:24
To: Ardeal LIANG <ardeal.liang at philips.com<mailto:ardeal.liang at philips.com>>
Cc: Firing Eric <efiring at hawaii.edu<mailto:efiring at hawaii.edu>>; matplotlib-users at python.org<mailto:matplotlib-users at python.org>
Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure






On Oct 29, 2017, at  17:46 PM, Ardeal LIANG <ardeal.liang at philips.com<mailto:ardeal.liang at philips.com>> wrote:

HI Eric,

Did you review the picture attached in my former email?
I suspect you go to the wrong direction about what I want to ask.

Best Regards,
Ardeal Liang

Hi Ardeal,

Its not clear to me what you want.  Do you want 3.2 x 10^3 instead of 0.0032?  The former is scientific and the latter decimal.

Thanks,  Jody







-----Original Message-----
From: Eric Firing [mailto:efiring at hawaii.edu]
Sent: 2017?10?28? 1:23
To: Ardeal LIANG <ardeal.liang at philips.com<mailto:ardeal.liang at philips.com>>; matplotlib-users at python.org<mailto:matplotlib-users at python.org>
Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure

I'm getting a different result, but on the basic point we now agree:
the ScalarFormatter is not the right tool for this.  What you need is a function that takes a number and returns the kind of string you want.

Examples:

def format_as_int(num):
    return str(int(num))

def format_float_no_sci(num):
    return "%f" % num

Eric

On 2017/10/26 10:07 PM, Ardeal LIANG wrote:



Hi Eric,

I added your code to my code:

*from *matplotlib.ticker *import *ScalarFormatter

fmt_no_sci = ScalarFormatter(useOffset=False)
fmt_no_sci.set_scientific(False)

/# ax.fmt_xaxis = fmt_no_sci

/prow = 4
pcol = 1
pidx = 0
fig = plt.figure()
pidx += 1
axx = fig.add_subplot(prow, pcol, pidx, ylabel=*'rawdata'*)
axx.plot(np.abs(testdata))

axx.fmt_xaxis = fmt_no_sci

The *x axis value displayed at the left-bottom corner of the figure*is
still scientific notation, but not decimal notation.

Best Regards,

Ardeal Liang

-----Original Message-----
From: Matplotlib-users
[mailto:matplotlib-users-bounces+ardeal.liang=philips.com at python.org]
On Behalf Of Eric Firing
Sent: 2017?10?27? 16:00
To: matplotlib-users at python.org<mailto:matplotlib-users at python.org>
Subject: Re: [Matplotlib-users] Scientific notation digits at
left-bottom corner of figure

On 2017/10/26 9:50 PM, Ardeal LIANG wrote:




Hi,









If I put the mouse on the figure, x and y axis are displayed on the




left-bottom corner of the figure.




but the x axis is displayed as scientific notation, I want it to be




displayed as decimal notation.









how to make x axis or y axis values displayed as decimal notation
on




the left-bottom corner?









*Best Regards,*









*Ardeal Liang*

Something like this, assuming your Axes is ax:

from matplotlib.ticker import ScalarFormatter

fmt_no_sci = ScalarFormatter(useOffset=False)

fmt_no_sci.set_scientific(False)

ax.fmt_xaxis = fmt_no_sci

Eric

_______________________________________________

Matplotlib-users mailing list

Matplotlib-users at python.org<mailto:Matplotlib-users at python.org> <mailto:Matplotlib-users at python.org>

https://mail.python.org/mailman/listinfo/matplotlib-users


----------------------------------------------------------------------
-- The information contained in this email may be confidential and/or
legally protected under applicable law. The message is intended solely
for the addressee(s). If you are not the intended recipient, you are
hereby notified that any use, forwarding, dissemination, or
reproduction of this email is strictly prohibited and may be unlawful.
If you are not the intended recipient, please contact the sender by
return e-mail and destroy all copies of the original email.


________________________________
The information contained in this email may be confidential and/or legally protected under applicable law. The message is intended solely for the addressee(s). If you are not the intended recipient, you are hereby notified that any use, forwarding, dissemination, or reproduction of this email is strictly prohibited and may be unlawful. If you are not the intended recipient, please contact the sender by return e-mail and destroy all copies of the original email.
_______________________________________________
Matplotlib-users mailing list
Matplotlib-users at python.org<mailto:Matplotlib-users at python.org>
https://mail.python.org/mailman/listinfo/matplotlib-users


________________________________
The information contained in this email may be confidential and/or legally protected under applicable law. The message is intended solely for the addressee(s). If you are not the intended recipient, you are hereby notified that any use, forwarding, dissemination, or reproduction of this email is strictly prohibited and may be unlawful. If you are not the intended recipient, please contact the sender by return e-mail and destroy all copies of the original email.


________________________________
The information contained in this email may be confidential and/or legally protected under applicable law. The message is intended solely for the addressee(s). If you are not the intended recipient, you are hereby notified that any use, forwarding, dissemination, or reproduction of this email is strictly prohibited and may be unlawful. If you are not the intended recipient, please contact the sender by return e-mail and destroy all copies of the original email.


________________________________
The information contained in this email may be confidential and/or legally protected under applicable law. The message is intended solely for the addressee(s). If you are not the intended recipient, you are hereby notified that any use, forwarding, dissemination, or reproduction of this email is strictly prohibited and may be unlawful. If you are not the intended recipient, please contact the sender by return e-mail and destroy all copies of the original email.
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From ardeal.liang at philips.com  Sun Oct 29 22:55:38 2017
From: ardeal.liang at philips.com (Ardeal LIANG)
Date: Mon, 30 Oct 2017 02:55:38 +0000
Subject: [Matplotlib-users] Scientific notation digits at left-bottom
 corner of figure
In-Reply-To: <44BCB453-6271-46A9-B7A8-47F7AD311D64@uvic.ca>
References: <AM5P121MB00340983CBCDCAA55F7F700FF75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <60935ee1-4b69-76a5-4c71-d2901cb00b2e@hawaii.edu>
 <AM5P121MB0034BD7560CCE138784C96B7F75A0@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <135de16c-f71b-78aa-40bc-2b7e6279c0d6@hawaii.edu>
 <AM5P121MB0034152073291DA0B39CBC0AF7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <9ABC007B-267A-4A2D-8342-62917D281C88@uvic.ca>
 <AM5P121MB00343424E73AB679A96D5EA4F7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <E181C6E8-1F33-44C3-A169-3D6F30BC56CF@uvic.ca>
 <AM5P121MB00349E346C7704FE4425D062F7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <8DF3FBE8-BBD1-4051-AEB2-CD1C6786CABB@uvic.ca>
 <AM5P121MB00343512D40FC5D82C5FB96CF7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>
 <44BCB453-6271-46A9-B7A8-47F7AD311D64@uvic.ca>
Message-ID: <AM5P121MB00344A850554338185DF766BF7590@AM5P121MB0034.EURP121.PROD.OUTLOOK.COM>

This is the error message if I add your code to my test code:
Traceback (most recent call last):
  File "C:\Program Files\JetBrains\PyCharm Community Edition 2017.1.5\helpers\pydev\pydevd.py", line 1591, in <module>
    globals = debugger.run(setup['file'], None, None, is_module)
  File "C:\Program Files\JetBrains\PyCharm Community Edition 2017.1.5\helpers\pydev\pydevd.py", line 1018, in run
    pydev_imports.execfile(file, globals, locals)  # execute the script
  File "C:/work/code_python/mwsignal/misc_experiment.py", line 49
SyntaxError: Non-ASCII character '\xe2' in file C:/work/code_python/mwsignal/misc_experiment.py on line 49, but no encoding declared; see http://python.org/dev/peps/pep-0263/ for details



Best Regards,
Ardeal Liang

From: Jody Klymak [mailto:jklymak at uvic.ca]
Sent: 2017?10?30? 10:53
To: Ardeal LIANG <ardeal.liang at philips.com>
Cc: matplotlib-users at python.org
Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure

ax.format_coord is the function that formats the numbers shown in the red ellipse of your picture.  Given that knowledge you can put anything in that red ellipse you want.

Cheers,   Jody


On Oct 29, 2017, at  19:46 PM, Ardeal LIANG <ardeal.liang at philips.com<mailto:ardeal.liang at philips.com>> wrote:

Hi Jody,

Did you check the picture I attached? You are still in the wrong direction about what I want?????..
Please check the digit in the red ellipse in my picture!!!!!!!!!!!!!!

Best Regards,
Ardeal Liang

From: Jody Klymak [mailto:jklymak at uvic.ca]
Sent: 2017?10?30? 10:44
To: Ardeal LIANG <ardeal.liang at philips.com<mailto:ardeal.liang at philips.com>>
Cc: Firing Eric <efiring at hawaii.edu<mailto:efiring at hawaii.edu>>; matplotlib-users at python.org<mailto:matplotlib-users at python.org>
Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure

Maybe?


def format_coord(x, y):
    return 'x=%4.4e, y=%4.4e?%(x, y))  # or whatever formatting you want

ax.format_coord = format_coord


Cheers,   Jody





On Oct 29, 2017, at  19:36 PM, Ardeal LIANG <ardeal.liang at philips.com<mailto:ardeal.liang at philips.com>> wrote:

Hi Jody,

I apologize if I said it backwards!
I think you now get what I want.

I tried your code. It didn?t work still.

Best Regards,
Ardeal Liang

From: Jody Klymak [mailto:jklymak at uvic.ca]
Sent: 2017?10?30? 10:34
To: Ardeal LIANG <ardeal.liang at philips.com<mailto:ardeal.liang at philips.com>>
Cc: Firing Eric <efiring at hawaii.edu<mailto:efiring at hawaii.edu>>; matplotlib-users at python.org<mailto:matplotlib-users at python.org>
Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure

Did you try:

from matplotlib.ticker import ScalarFormatter

fmt_no_sci = ScalarFormatter(useOffset=False)
fmt_no_sci.set_scientific(True)  # NOTE True instead of False.

I think you just said it backwards: You want scientific ntoation, not decimal notation.

Cheers,   Jody





On Oct 29, 2017, at  19:29 PM, Ardeal LIANG <ardeal.liang at philips.com<mailto:ardeal.liang at philips.com>> wrote:

Hi Jody,

Please check the red ellipse in the picture.

I want the value 1.85092e+06 to be 1850922 in the red ellipse in the following picture.


fmt_no_sci = ScalarFormatter(useOffset=False)
fmt_no_sci.set_scientific(False)
# ax.fmt_xaxis = fmt_no_sci

fig = plt.figure()
ax = fig.add_subplot(111)
ax.plot(testdata)
ax.fmt_xaxis = fmt_no_sci
plt.show(0)

The x axis value displayed at the left-bottom corner of the figure is still scientific notation, but not decimal notation.
Please check the x axis value displayed at the left-bottom corner of the figure in the red circle.
<image001.jpg>




Best Regards,
Ardeal Liang

From: Jody Klymak [mailto:jklymak at uvic.ca]
Sent: 2017?10?30? 10:24
To: Ardeal LIANG <ardeal.liang at philips.com<mailto:ardeal.liang at philips.com>>
Cc: Firing Eric <efiring at hawaii.edu<mailto:efiring at hawaii.edu>>; matplotlib-users at python.org<mailto:matplotlib-users at python.org>
Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure







On Oct 29, 2017, at  17:46 PM, Ardeal LIANG <ardeal.liang at philips.com<mailto:ardeal.liang at philips.com>> wrote:

HI Eric,

Did you review the picture attached in my former email?
I suspect you go to the wrong direction about what I want to ask.

Best Regards,
Ardeal Liang

Hi Ardeal,

Its not clear to me what you want.  Do you want 3.2 x 10^3 instead of 0.0032?  The former is scientific and the latter decimal.

Thanks,  Jody








-----Original Message-----
From: Eric Firing [mailto:efiring at hawaii.edu]
Sent: 2017?10?28? 1:23
To: Ardeal LIANG <ardeal.liang at philips.com<mailto:ardeal.liang at philips.com>>; matplotlib-users at python.org<mailto:matplotlib-users at python.org>
Subject: Re: [Matplotlib-users] Scientific notation digits at left-bottom corner of figure

I'm getting a different result, but on the basic point we now agree:
the ScalarFormatter is not the right tool for this.  What you need is a function that takes a number and returns the kind of string you want.

Examples:

def format_as_int(num):
    return str(int(num))

def format_float_no_sci(num):
    return "%f" % num

Eric

On 2017/10/26 10:07 PM, Ardeal LIANG wrote:




Hi Eric,

I added your code to my code:

*from *matplotlib.ticker *import *ScalarFormatter

fmt_no_sci = ScalarFormatter(useOffset=False)
fmt_no_sci.set_scientific(False)

/# ax.fmt_xaxis = fmt_no_sci

/prow = 4
pcol = 1
pidx = 0
fig = plt.figure()
pidx += 1
axx = fig.add_subplot(prow, pcol, pidx, ylabel=*'rawdata'*)
axx.plot(np.abs(testdata))

axx.fmt_xaxis = fmt_no_sci

The *x axis value displayed at the left-bottom corner of the figure*is
still scientific notation, but not decimal notation.

Best Regards,

Ardeal Liang

-----Original Message-----
From: Matplotlib-users
[mailto:matplotlib-users-bounces+ardeal.liang=philips.com at python.org]
On Behalf Of Eric Firing
Sent: 2017?10?27? 16:00
To: matplotlib-users at python.org<mailto:matplotlib-users at python.org>
Subject: Re: [Matplotlib-users] Scientific notation digits at
left-bottom corner of figure

On 2017/10/26 9:50 PM, Ardeal LIANG wrote:





Hi,











If I put the mouse on the figure, x and y axis are displayed on the





left-bottom corner of the figure.





but the x axis is displayed as scientific notation, I want it to be





displayed as decimal notation.











how to make x axis or y axis values displayed as decimal notation
on





the left-bottom corner?











*Best Regards,*











*Ardeal Liang*

Something like this, assuming your Axes is ax:

from matplotlib.ticker import ScalarFormatter

fmt_no_sci = ScalarFormatter(useOffset=False)

fmt_no_sci.set_scientific(False)

ax.fmt_xaxis = fmt_no_sci

Eric

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