如何在分组的DataFrame上强制pandas.DataFrame.apply

Question

The behavior of pandas.DataFrame.apply(myfunc) is application of myfunc along columns. The behavior of pandas.core.groupby.DataFrameGroupBy.apply is more complicated. This difference shows up for functions myfunc such that frame.apply(myfunc) != myfunc(frame) .

I would like to group a DataFrame then apply myfunc along columns of each individual frame (in each group) and then paste together the results. There are hacky ways to do it, but I wonder it seems like there is some simple kwarg I'm missing.

Consider the example below:

In [22]: df = pd.DataFrame({'a':range(5), 'b': range(5, 10)})

In [23]: df
Out[23]: 
   a  b
0  0  5
1  1  6
2  2  7
3  3  8
4  4  9

In [24]: def myfunc(data):
             # Implements max in a funny way.
             # However, this is just an example of a function such that 
             # myfunc(frame) != frame.apply(myfunc)
             return data.values.ravel().max()

In [25]: df.apply(myfunc)
Out[25]: 
a    4
b    9

In [26]: df.groupby(df.a < 2).apply(myfunc)
Out[26]: 
a
False    9
True     6

As you can see, myfunc was called like myfunc(group) . This default behavior is reasonable, since myfunc takes in a DataFrame and returns a number, but this is not what I always want. Is there a canonical way to force myfunc to be applied along columns of each group, as in group.apply(myfunc) ? The best I can come up with is an awkward wrapper:

In [27]: def wrapped(frame):
   ....:     return frame.apply(myfunc)

In [28]: df.groupby(df.a < 2).apply(wrapped)
Out[28]: 
       a  b
a          
False  4  9
True   1  6

Answer 1

You can do this

In [25]: df.groupby(df.a<2).aggregate(myfunc)
Out[25]: 
       a  b
a          
False  4  9
True   1  6

[2 rows x 2 columns]

But this is simpler

In [26]: df.groupby(df.a<2).max()
Out[26]: 
       a  b
a          
False  4  9
True   1  6

[2 rows x 2 columns]