Fix Python – Pandas groupby: How to get a union of strings

Question

Asked By – Anne

I have a dataframe like this:

   A         B       C
0  1  0.749065    This
1  2  0.301084      is
2  3  0.463468       a
3  4  0.643961  random
4  1  0.866521  string
5  2  0.120737       !

Calling

In [10]: print df.groupby("A")["B"].sum()

will return

A
1    1.615586
2    0.421821
3    0.463468
4    0.643961

Now I would like to do “the same” for column “C”. Because that column contains strings, sum() doesn’t work (although you might think that it would concatenate the strings). What I would really like to see is a list or set of the strings for each group, i.e.

A
1    {This, string}
2    {is, !}
3    {a}
4    {random}

I have been trying to find ways to do this.

Series.unique() (http://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.unique.html) doesn’t work, although

df.groupby("A")["B"]

is a

pandas.core.groupby.SeriesGroupBy object

so I was hoping any Series method would work. Any ideas?

Now we will see solution for issue: Pandas groupby: How to get a union of strings


Answer

In [4]: df = read_csv(StringIO(data),sep='\s+')

In [5]: df
Out[5]: 
   A         B       C
0  1  0.749065    This
1  2  0.301084      is
2  3  0.463468       a
3  4  0.643961  random
4  1  0.866521  string
5  2  0.120737       !

In [6]: df.dtypes
Out[6]: 
A      int64
B    float64
C     object
dtype: object

When you apply your own function, there is not automatic exclusions of non-numeric columns. This is slower, though, than the application of .sum() to the groupby

In [8]: df.groupby('A').apply(lambda x: x.sum())
Out[8]: 
   A         B           C
A                         
1  2  1.615586  Thisstring
2  4  0.421821         is!
3  3  0.463468           a
4  4  0.643961      random

sum by default concatenates

In [9]: df.groupby('A')['C'].apply(lambda x: x.sum())
Out[9]: 
A
1    Thisstring
2           is!
3             a
4        random
dtype: object

You can do pretty much what you want

In [11]: df.groupby('A')['C'].apply(lambda x: "{%s}" % ', '.join(x))
Out[11]: 
A
1    {This, string}
2           {is, !}
3               {a}
4          {random}
dtype: object

Doing this on a whole frame, one group at a time. Key is to return a Series

def f(x):
     return Series(dict(A = x['A'].sum(), 
                        B = x['B'].sum(), 
                        C = "{%s}" % ', '.join(x['C'])))

In [14]: df.groupby('A').apply(f)
Out[14]: 
   A         B               C
A                             
1  2  1.615586  {This, string}
2  4  0.421821         {is, !}
3  3  0.463468             {a}
4  4  0.643961        {random}

This question is answered By – Jeff

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