CODEWITHSUNDEEP
pythonpandas
patricksurry
patricksurry

Reputation: 5878

Why doesn't dataframe groupby with a single group return a dataframe?

I suspect this is a simpler form of my question here. [Update: unfortunately not so.]

If you do something like this (in Pandas 0.11):

df = pd.DataFrame([[1,2],[1,3],[2,4]],columns='a b'.split())
print df
g = df.groupby('a').count()
print type(g)
print g

You get the expected:

   a  b
0  1  2
1  1  3
2  2  4
<class 'pandas.core.frame.DataFrame'>
   a  b
a      
1  2  2
2  1  1

But if there's only one resulting group, you get a very odd Series instead:

df = pd.DataFrame([[1,2],[1,3],[1,4]],columns='a b'.split())
...

   a  b
0  1  2
1  1  3
2  1  4
<class 'pandas.core.series.Series'>
a   
1  a    3
   b    3
Name: 1, dtype: int64

But I'd rather the result was a DataFrame equivalent to this:

print pd.DataFrame([[3,3]],index=pd.Index([1],name='a'),columns='a b'.split())

   a  b
a      
1  3  3

I'm stuck as to how to get that easily from the series (and not sure why I get that in the first place).

Upvotes: 3

Views: 2594

Answers (1)

Phillip Cloud
Phillip Cloud

Reputation: 25662

In pandas 0.12 this does exactly what you ask.

In [3]: df = pd.DataFrame([[1,2],[1,3],[1,4]],columns='a b'.split())

In [4]: df.groupby('a').count()
Out[4]:
   a  b
a
1  3  3

To replicate what you're seeing pass squeeze=True:

In [5]: df.groupby('a', squeeze=True).count()
Out[5]:
a
1  a    3
   b    3
Name: 1, dtype: int64

If you can't upgrade then do:

In [3]: df.groupby('a').count().unstack()
Out[3]:
   a  b
a
1  3  3

Upvotes: 3

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