How to put back a number-based index

Fri May 13 23:46:51 EDT 2016

Hello, Michael,
I do not understand this.
I tried list =df[3]
it worked.  But, it does not behave like a list.
list[0] nothinglist[1] a valuelist[2] nothing
list[4] a value
It behaves like a dictionary. 

    On Saturday, 14 May 2016, 4:27, David Shi <davidgshi at yahoo.co.uk> wrote:

 Hello, Michael,
This is very weird.
55                     145340.20
 56                     253333.43
Name: 3, dtype: float64
It looks like two columns, but it shows one single object.
Any clue? 

    On Saturday, 14 May 2016, 4:15, David Shi <davidgshi at yahoo.co.uk> wrote:

 Hello, Michael,
I tried to discover the problem.
df[0]   yields nothingdf[1]  yields nothingdf[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 followingMultiIndex(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, ?).