notilas notilas - 1 year ago 168
Python Question

Python Pandas: Groupby and Apply multi-column operation

df1 is DataFrame with 4 columns.

I want to created a new DataFrame (df2) by grouping df1 with Column 'A' with multi-column operation on column 'C' and 'D'

Column 'AA' = mean(C)+mean(D)

Column 'BB' = std(D)

df1= 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)})

0 foo one 1.652675 -1.983378
1 bar one 0.926656 -0.598756
2 foo two 0.131381 0.604803
3 bar three -0.436376 -1.186363
4 foo two 0.487161 -0.650876
5 bar two 0.358007 0.249967
6 foo one -1.150428 2.275528
7 foo three 0.202677 -1.408699

def fun1(gg): # this does not work
return pd.DataFrame({'AA':C.mean()+gg.C.std(), 'BB':gg.C.std()})

dg1 = df1.groupby('A')
df2 = dg1.apply(fun1)

This does not work. It seems like aggregation() only works for Series and multi-column operation is not possible.
And apply() only produce Series output with multi-column operation.
Is there any other way to produce multi-column output (DataFrame) with multi-column operation?

Answer Source

Do you have a typo in your f function? Should AA be C.mean() + C.std() or C.mean() + D.mean()

In this first case, AA = C.mean() + C.std(),

In [91]: df = df1.groupby('A').agg({'C': lambda x: x.mean() + x.std(),
                                    'D': lambda x  x.std()})

In [92]: df
            C         D
bar  1.255506  0.588981
foo  1.775945  0.442724

For the second one C.mean() + D.mean(), things aren't quite as nice. When you give the .agg function on groupby objects a dict, I don't think there's a way to get values from two columns.

In [108]: g = df1.groupby('A')

In [109]: df = pd.DataFrame({"AA": g.mean()['C'] + g.mean()['D'], "BB": g.std()['D']})

In [110]: df
           AA        BB
bar  0.532263  0.721351
foo  0.427608  0.494980

You may want to assign g.mean() and g.std() to temporary variables to avoid calculating them twice.

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