Fix Python – How to access pandas groupby dataframe by key

Question

Asked By – beardc

How do I access the corresponding groupby dataframe in a groupby object by the key?

With the following groupby:

rand = np.random.RandomState(1)
df = pd.DataFrame({'A': ['foo', 'bar'] * 3,
                   'B': rand.randn(6),
                   'C': rand.randint(0, 20, 6)})
gb = df.groupby(['A'])

I can iterate through it to get the keys and groups:

In [11]: for k, gp in gb:
             print 'key=' + str(k)
             print gp
key=bar
     A         B   C
1  bar -0.611756  18
3  bar -1.072969  10
5  bar -2.301539  18
key=foo
     A         B   C
0  foo  1.624345   5
2  foo -0.528172  11
4  foo  0.865408  14

I would like to be able to access a group by its key:

In [12]: gb['foo']
Out[12]:  
     A         B   C
0  foo  1.624345   5
2  foo -0.528172  11
4  foo  0.865408  14

But when I try doing that with gb[('foo',)] I get this weird pandas.core.groupby.DataFrameGroupBy object thing which doesn’t seem to have any methods that correspond to the DataFrame I want.

The best I could think of is:

In [13]: def gb_df_key(gb, key, orig_df):
             ix = gb.indices[key]
             return orig_df.ix[ix]

         gb_df_key(gb, 'foo', df)
Out[13]:
     A         B   C
0  foo  1.624345   5
2  foo -0.528172  11
4  foo  0.865408  14  

but this is kind of nasty, considering how nice pandas usually is at these things.
What’s the built-in way of doing this?

Now we will see solution for issue: How to access pandas groupby dataframe by key


Answer

You can use the get_group method:

In [21]: gb.get_group('foo')
Out[21]: 
     A         B   C
0  foo  1.624345   5
2  foo -0.528172  11
4  foo  0.865408  14

Note: This doesn’t require creating an intermediary dictionary / copy of every subdataframe for every group, so will be much more memory-efficient than creating the naive dictionary with dict(iter(gb)). This is because it uses data-structures already available in the groupby object.


You can select different columns using the groupby slicing:

In [22]: gb[["A", "B"]].get_group("foo")
Out[22]:
     A         B
0  foo  1.624345
2  foo -0.528172
4  foo  0.865408

In [23]: gb["C"].get_group("foo")
Out[23]:
0     5
2    11
4    14
Name: C, dtype: int64

This question is answered By – Andy Hayden

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