## 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

**This answer is collected from stackoverflow and reviewed by FixPython community admins, is licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0 **