Anne Anne - 4 months ago 41
Python Question

Pandas groupby: How to get a union of strings

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?

Thanks,
Anne

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 (that the applicatino 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 a whole frame 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}
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