DOC: Dict of Dicts for renaming Groupby Aggregations

[TomAugspurger]

I didn't realize this was possible, and didn't see it in the docs.

df = pd.DataFrame({'A': [1, 2, 3], 'B': ['a', 'a', 'b'], 'C': [3, 4, 5]})
df.groupby('B').agg({'A': {'mean1': 'mean', 'med1': 'median'}, 'C': {'mean2': 'mean', 'med2': 'median'}})

[jreback]

xref is #8593 (which would replace / enhance this)

[aimboden]

Thanks for the tip. Didn't realize this was possible either, this will save me from building my multicolumns "by hand".

@jreback are you planning any API change for 0.16.0 on this? #8593 does not seem to interfere with this behaviour, but maybe a deeper change is planned?

I'd rather not rely on this if it's not tested atm. Or would you accept a test for this?

[jreback]

@Gimli510 this IS implemented. Its basically the same as the following (except the name determination is slightly different).

In [5]: df.groupby('B').agg({'A': ['mean','median'], 'C': ['mean','median']})
Out[5]: 
     A           C       
  mean median mean median
B                        
a  1.5    1.5  3.5    3.5
b  3.0    3.0  5.0    5.0

I haven't carefully looked thru, but I suspect their is at least 1 tests. Though would for sure accept a PR which makes these tests more prominent (e.g. test_agg_api or something).

pd.Summary will enhance this API, the existing will remain.

[jreback]

from mailing list

In [2]: df = pd.DataFrame({'A' : ['foo', 'bar', 'foo', 'bar',
   ...:                           'foo', 'bar', 'foo', 'foo'],
   ...:                    'B' : ['one', 'one', 'two', 'three',
   ...:                           'two', 'two', 'one', 'three'],
   ...:                    'C' : np.random.randn(8),
   ...:                    'D' : np.random.randn(8)})

In [3]: 

In [3]: grouped = df.groupby(['A', 'B'])

In [4]: grouped[['D','C']].agg({'r':np.sum, 'r2':np.mean})
Out[4]: 
                    r        r2
A   B                          
bar one   D -0.460078 -0.460078
          C  0.798220  0.798220
    three D  1.599986  1.599986
          C -0.554798 -0.554798
    two   D  0.124900  0.124900
          C  0.084758  0.084758
foo one   D -0.466082 -0.233041
          C -0.585512 -0.292756
    three D -0.184726 -0.184726
          C  0.130756  0.130756
    two   D -1.985586 -0.992793
          C  1.275138  0.637569

In [5]: grouped[['D','C']].agg({'r': { 'C' : np.sum }, 'r2' : { 'D' : np.mean }})
Out[5]: 
                    r        r2
                    C         D
A   B                          
bar one   D -0.460078 -0.460078
          C  0.798220  0.798220
    three D  1.599986  1.599986
          C -0.554798 -0.554798
    two   D  0.124900  0.124900
          C  0.084758  0.084758
foo one   D -0.466082 -0.233041
          C -0.585512 -0.292756
    three D -0.184726 -0.184726
          C  0.130756  0.130756
    two   D -1.985586 -0.992793
          C  1.275138  0.637569

In [6]: grouped[['D','C']].agg([np.sum, np.mean])
Out[6]: 
                  D                   C          
                sum      mean       sum      mean
A   B                                            
bar one   -0.460078 -0.460078  0.798220  0.798220
    three  1.599986  1.599986 -0.554798 -0.554798
    two    0.124900  0.124900  0.084758  0.084758
foo one   -0.466082 -0.233041 -0.585512 -0.292756
    three -0.184726 -0.184726  0.130756  0.130756
    two   -1.985586 -0.992793  1.275138  0.637569

with a trivial patch

diff --git a/pandas/core/groupby.py b/pandas/core/groupby.py
index add5080..b885b6f 100644
--- a/pandas/core/groupby.py
+++ b/pandas/core/groupby.py
@@ -2837,9 +2837,6 @@ class NDFrameGroupBy(GroupBy):
             keys = []
             if self._selection is not None:
                 subset = obj
-                if isinstance(subset, DataFrame):
-                    raise NotImplementedError("Aggregating on a DataFrame is "
-                                              "not supported")

                 for fname, agg_how in compat.iteritems(arg):
                     colg = SeriesGroupBy(subset, selection=self._selection,

of course need some tests......