How to put back a number-based index

Sat May 14 10:29:53 EDT 2016

David, it sounds like you'll need a thorough introduction to the basics of
Python.
Check out the tutorial: https://docs.python.org/3/tutorial/

On Sat, May 14, 2016 at 6:19 AM David Shi <davidgshi at yahoo.co.uk> wrote:

> Hello, Michael,
>
> I discovered that the problem is "two columns of data are put together"
> and "are recognised as one column".
>
> This is very strange.  I would like to understand the subject well.
>
> And, how many ways are there to investigate into the nature of objects
> dynamically?
>
> Some object types only get shown as an object.  Are there anything to be
> typed in Python, to reveal objects.
>
> Regards.
>
> David
>
>
> On Saturday, 14 May 2016, 4:30, Michael Selik <michael.selik at gmail.com>
> wrote:
>
>
> What were you hoping to get from ``df[0]``?
> When you say it "yields nothing" do you mean it raised an error? What was
> the error message?
>
> Have you tried a Google search for "pandas set index"?
>
> http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.set_index.html
>
> On Fri, May 13, 2016 at 11:18 PM David Shi <davidgshi at yahoo.co.uk> wrote:
>
> Hello, Michael,
>
> I tried to discover the problem.
>
> df[0]   yields nothing
> df[1]  yields nothing
> df[2] yields nothing
>
> However, df[3] gives the following:
>
> sid
> -9223372036854775808          NaN
>  1                      133738.70
>  4                      295256.11
>  5                      137733.09
>  6                      409413.58
>  8                      269600.97
>  9                       12852.94
>
>
> Can we split this back to normal?  or turn it into a dictionary, so that I can put values back properly.
>
>
> I like to use sid as index, some way.
>
>
> Regards.
>
>
> David
>
>
>
> On Friday, 13 May 2016, 22:58, Michael Selik <michael.selik at gmail.com>
> wrote:
>
>
> What have code you tried? What error message are you receiving?
>
> On Fri, May 13, 2016, 5:54 PM David Shi <davidgshi at yahoo.co.uk> wrote:
>
> Hello, Michael,
>
> How to convert a float type column into an integer or label or string type?
>
>
> On Friday, 13 May 2016, 22:02, Michael Selik <michael.selik at gmail.com>
> wrote:
>
>
> To clarify that you're specifying the index as a label, use df.iloc
>
>     >>> df = pd.DataFrame({'X': range(4)}, index=list('abcd'))
>     >>> df
>        X
>     a  0
>     b  1
>     c  2
>     d  3
>     >>> df.loc['a']
>     X    0
>     Name: a, dtype: int64
>     >>> df.iloc[0]
>     X    0
>     Name: a, dtype: int64
>
> On Fri, May 13, 2016 at 4:54 PM David Shi <davidgshi at yahoo.co.uk> wrote:
>
> Dear Michael,
>
> To avoid complication, I only groupby using one column.
>
> It is OK now.  But, how to refer to new row index?  How do I use floating
> index?
>
> Float64Index([ 1.0,  4.0,  5.0,  6.0,  8.0,  9.0, 10.0, 11.0, 12.0, 13.0, 16.0,
>               17.0, 18.0, 19.0, 20.0, 21.0, 22.0, 23.0, 24.0, 25.0, 26.0, 27.0,
>               28.0, 29.0, 30.0, 31.0, 32.0, 33.0, 34.0, 35.0, 36.0, 37.0, 38.0,
>               39.0, 40.0, 41.0, 42.0, 44.0, 45.0, 46.0, 47.0, 48.0, 49.0, 50.0,
>               51.0, 53.0, 54.0, 55.0, 56.0],
>              dtype='float64', name=u'StateFIPS')
>
>
> Regards.
>
>
> David
>
>
>
> On Friday, 13 May 2016, 21:43, Michael Selik <michael.selik at gmail.com>
> wrote:
>
>
> Here's an example.
>
>     >>> import pandas as pd
>     >>> df = pd.DataFrame({'group': list('AB') * 2, 'data': range(4)},
> index=list('wxyz'))
>     >>> df
>        data group
>     w     0     A
>     x     1     B
>     y     2     A
>     z     3     B
>     >>> df = df.reset_index()
>     >>> df
>       index  data group
>     0     w     0     A
>     1     x     1     B
>     2     y     2     A
>     3     z     3     B
>     >>> df.groupby('group').max()
>           index  data
>     group
>     A         y     2
>     B         z     3
>
> If that doesn't help, you'll need to explain what you're trying to
> accomplish in detail -- what variables you started with, what
> transformations you want to do, and what variables you hope to have when
> finished.
>
> On Fri, May 13, 2016 at 4:36 PM David Shi <davidgshi at yahoo.co.uk> wrote:
>
> Hello, Michael,
>
> I changed groupby with one column.
>
> The index is different.
>
> Index([   u'AL',    u'AR',    u'AZ',    u'CA',    u'CO',    u'CT',    u'DC',
>           u'DE',    u'FL',    u'GA',    u'IA',    u'ID',    u'IL',    u'IN',
>           u'KS',    u'KY',    u'LA',    u'MA',    u'MD',    u'ME',    u'MI',
>           u'MN',    u'MO',    u'MS',    u'MT',    u'NC',    u'ND',    u'NE',
>           u'NH',    u'NJ',    u'NM',    u'NV',    u'NY',    u'OH',    u'OK',
>           u'OR',    u'PA',    u'RI',    u'SC',    u'SD', u'State',    u'TN',
>           u'TX',    u'UT',    u'VA',    u'VT',    u'WA',    u'WI',    u'WV',
>           u'WY'],
>       dtype='object', name=0)
>
>
> How to use this index?
>
>
> Regards.
>
>
> David
>
>
>
> On Friday, 13 May 2016, 21:19, David Shi <davidgshi at yahoo.co.uk> wrote:
>
>
> Hello, Michael,
>
> I typed in df.index
>
> I got the following
>
> MultiIndex(levels=[[1.0, 4.0, 5.0, 6.0, 8.0, 9.0, 10.0, 11.0, 12.0, 13.0, 16.0, 17.0, 18.0, 19.0, 20.0, 21.0, 22.0, 23.0, 24.0, 25.0, 26.0, 27.0, 28.0, 29.0, 30.0, 31.0, 32.0, 33.0, 34.0, 35.0, 36.0, 37.0, 38.0, 39.0, 40.0, 41.0, 42.0, 44.0, 45.0, 46.0, 47.0, 48.0, 49.0, 50.0, 51.0, 53.0, 54.0, 55.0, 56.0], [u'AL', u'AR', u'AZ', u'CA', u'CO', u'CT', u'DC', u'DE', u'FL', u'GA', u'IA', u'ID', u'IL', u'IN', u'KS', u'KY', u'LA', u'MA', u'MD', u'ME', u'MI', u'MN', u'MO', u'MS', u'MT', u'NC', u'ND', u'NE', u'NH', u'NJ', u'NM', u'NV', u'NY', u'OH', u'OK', u'OR', u'PA', u'RI', u'SC', u'SD', u'State', u'TN', u'TX', u'UT', u'VA', u'VT', u'WA', u'WI', u'WV', u'WY']],
>            labels=[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48], [0, 2, 1, 3, 4, 5, 7, 6, 8, 9, 11, 12, 13, 10, 14, 15, 16, 19, 18, 17, 20, 21, 23, 22, 24, 27, 31, 28, 29, 30, 32, 25, 26, 33, 34, 35, 36, 37, 38, 39, 41, 42, 43, 45, 44, 46, 48, 47, 49]],
>            names=[u'StateFIPS', 0])
>
> Regards.
>
>
> David
>
>
>
> On Friday, 13 May 2016, 21:11, David Shi <davidgshi at yahoo.co.uk> wrote:
>
>
> Dear Michael,
>
> I have done a number of operation in between.
>
> Providing that information does not help you
>
> How to reset index after grouping and various operations is of interest.
>
> How to type in a command to find out its current dataframe?
>
> Regards.
>
> David
>
>
> On Friday, 13 May 2016, 20:58, Michael Selik <michael.selik at gmail.com>
> wrote:
>
>
> Just in case I misunderstood, why don't you make a little example of
> before and after the grouping? This mailing list does not accept
> attachments, so you'll have to make do with pasting a few rows of
> comma-separated or tab-separated values.
>
> On Fri, May 13, 2016 at 3:56 PM Michael Selik <michael.selik at gmail.com>
> wrote:
>
> In order to preserve your index after the aggregation, you need to make
> sure it is considered a data column (via reset_index) and then choose how
> your aggregation will operate on that column.
>
> On Fri, May 13, 2016 at 3:29 PM David Shi <davidgshi at yahoo.co.uk> wrote:
>
> Hello, Michael,
>
> Why reset_index before grouping?
>
> Regards.
>
> David
>
>
> On Friday, 13 May 2016, 17:57, Michael Selik <michael.selik at gmail.com>
> wrote:
>
>
>
>
> On Fri, May 13, 2016 at 12:27 PM David Shi via Python-list <
> python-list at python.org> wrote:
>
> I lost my indexes after grouping in Pandas.
> I managed to rest_index and got back the index column.
> But How can I get back a index row?
>
>
> Was the grouping an aggregation? If so, the original indexes are
> meaningless. What you could do is reset_index before the grouping and when
> you aggregate decide how to handle the formerly-known-as-index column (min,
> max, mean, ?).
>
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