How to reset a DataFrame's indexes for all groups in one step?

Question

I\'ve tried to split my dataframe to groups

df = pd.DataFrame({\'A\' : [\'foo\', \'bar\', \'foo\', \'bar\',
                       \'foo\', \'bar\', \'foo\',         


        

Answer 1

Isn't this just grouped = grouped.apply(lambda x: x.reset_index()) ?

Answer 2

Something like this would work:

for group, index in grouped.indices.iteritems():
    grouped.indices[group] = range(0, len(index))

You could probably make it less verbose if you wanted to.

Answer 3

df=df.groupby('A').apply(lambda x: x.reset_index(drop=True)).drop('A',axis=1).reset_index()

Answer 4

Pass in as_index=False to the groupby, then you don't need to reset_index to make the groupby-d columns columns again:

In [11]: grouped = df.groupby('A', as_index=False)

In [12]: grouped.get_group('foo')
Out[12]:
     A  B
0  foo  1
2  foo  3
4  foo  5
6  foo  7
7  foo  8

Note: As pointed out (and seen in the above example) the index above is not [0, 1, 2, ...], I claim that this will never matter in practice - if it does you're going to have to just through some strange hoops - it's going to be more verbose, less readable and less efficient...

Answer 5

df = pd.DataFrame({'A' : ['foo', 'bar', 'foo', 'bar',
                       'foo', 'bar', 'foo', 'foo'],
                   'B' : ['1', '2', '3', '4',
                       '5', '6', '7', '8'],
                   })
grouped = df.groupby('A',as_index = False)

we get two groups

grouped_index = grouped.apply(lambda x: x.reset_index(drop = True)).reset_index()

Result in two new columns level_0 and level_1 getting added and the index is reset

 level_0level_1 A   B
0   0     0    bar  2
1   0     1    bar  4
2   0     2    bar  6
3   1     0    foo  1
4   1     1    foo  3
5   1     2    foo  5
6   1     3    foo  7
7   1     4    foo  8

result = grouped_index.drop('level_0',axis = 1).set_index('level_1')

Creates an index within each group of "A"

          A     B
level_1     
0        bar    2
1        bar    4
2        bar    6
0        foo    1
1        foo    3
2        foo    5
3        foo    7
4        foo    8