Can pandas groupby aggregate into a list, rather than sum, mean, etc?

Question

I've had success using the groupby function to sum or average a given variable by groups, but is there a way to aggregate into a list of values, rather than to get a single result? (And would this still be called aggregation?)

I am not entirely sure this is the approach I should be taking anyhow, so below is an example of the transformation I'd like to make, with toy data.

That is, if the data look something like this:

    A    B    C  

    1    10   22

    1    12   20

    1    11   8

    1    10   10

    2    11   13

    2    12   10 

    3    14   0

What I am trying to end up with is something like the following. I am not totally sure whether this can be done through groupby aggregating into lists, and am rather lost as to where to go from here.

Hypothetical output:

     A    B    C  New1  New2  New3  New4  New5  New6

    1    10   22  12    20    11    8     10    10

    2    11   13  12    10 

    3    14   0

Perhaps I should be pursuing pivots instead? The order by which the data are put into columns does not matter - all columns B through New6 in this example are equivalent. All suggestions/corrections are much appreciated.

Answer 1

Use the below line of code that aggregates values to lists.

import pandas as pd

df = pd.DataFrame( {'A' : [1, 1, 1, 1, 2, 2, 3], 'B' : [10, 12, 11, 10, 11, 12, 14], 'C' : [22, 20,     8, 10, 13, 10, 0]})

print df

# Old version

# df2=df.groupby(['A']).apply(lambda tdf: pd.Series(  dict([[vv,tdf[vv].unique().tolist()] for vv in tdf if vv not in ['A']])  )) 

df2 = df.groupby('A').aggregate(lambda tdf: tdf.unique().tolist())

print df2

Output:

In [3]: run tmp

   A   B   C

0  1  10  22

1  1  12  20

2  1  11   8

3  1  10  10

4  2  11  13

5  2  12  10

6  3  14   0

[7 rows x 3 columns]

              B                C

A                               

1  [10, 12, 11]  [22, 20, 8, 10]

2      [11, 12]         [13, 10]

3          [14]              [0]

[3 rows x 2 columns]