Use the pandas idxmax function. It's straightforward:

>>> import pandas
>>> import numpy as np
>>> df = pandas.DataFrame(np.random.randn(5,3),columns=['A','B','C'])
>>> df
          A         B         C
0  1.232853 -1.979459 -0.573626
1  0.140767  0.394940  1.068890
2  0.742023  1.343977 -0.579745
3  2.125299 -0.649328 -0.211692
4 -0.187253  1.908618 -1.862934
>>> df['A'].idxmax()
3
>>> df['B'].idxmax()
4
>>> df['C'].idxmax()
1

• Alternatively you could also use numpy.argmax, such as numpy.argmax(df['A']) -- it provides the same thing, and appears at least as fast as idxmax in cursory observations.

• idxmax() returns indices labels, not integers.

• Example': if you have string values as your index labels, like rows 'a' through 'e', you might want to know that the max occurs in row 4 (not row 'd').

• if you want the integer position of that label within the Index you have to get it manually (which can be tricky now that duplicate row labels are allowed).

HISTORICAL NOTES:

• idxmax() used to be called argmax() prior to 0.11
• argmax was deprecated prior to 1.0.0 and removed entirely in 1.0.0
• back as of Pandas 0.16, argmax used to exist and perform the same function (though appeared to run more slowly than idxmax).
• argmax function returned the integer position within the index of the row location of the maximum element.
• pandas moved to using row labels instead of integer indices. Positional integer indices used to be very common, more common than labels, especially in applications where duplicate row labels are common.

For example, consider this toy DataFrame with a duplicate row label:

In [19]: dfrm
Out[19]: 
          A         B         C
a  0.143693  0.653810  0.586007
b  0.623582  0.312903  0.919076
c  0.165438  0.889809  0.000967
d  0.308245  0.787776  0.571195
e  0.870068  0.935626  0.606911
f  0.037602  0.855193  0.728495
g  0.605366  0.338105  0.696460
h  0.000000  0.090814  0.963927
i  0.688343  0.188468  0.352213
i  0.879000  0.105039  0.900260

In [20]: dfrm['A'].idxmax()
Out[20]: 'i'

In [21]: dfrm.iloc[dfrm['A'].idxmax()]  # .ix instead of .iloc in older versions of pandas
Out[21]: 
          A         B         C
i  0.688343  0.188468  0.352213
i  0.879000  0.105039  0.900260

So here a naive use of idxmax is not sufficient, whereas the old form of argmax would correctly provide the positional location of the max row (in this case, position 9).

This is exactly one of those nasty kinds of bug-prone behaviors in dynamically typed languages that makes this sort of thing so unfortunate, and worth beating a dead horse over. If you are writing systems code and your system suddenly gets used on some data sets that are not cleaned properly before being joined, it's very easy to end up with duplicate row labels, especially string labels like a CUSIP or SEDOL identifier for financial assets. You can't easily use the type system to help you out, and you may not be able to enforce uniqueness on the index without running into unexpectedly missing data.

So you're left with hoping that your unit tests covered everything (they didn't, or more likely no one wrote any tests) -- otherwise (most likely) you're just left waiting to see if you happen to smack into this error at runtime, in which case you probably have to go drop many hours worth of work from the database you were outputting results to, bang your head against the wall in IPython trying to manually reproduce the problem, finally figuring out that it's because idxmax can only report the label of the max row, and then being disappointed that no standard function automatically gets the positions of the max row for you, writing a buggy implementation yourself, editing the code, and praying you don't run into the problem again.

This is more of a programming than a data science question, and would therefore be better suited for stackoverflow, but this can be achieved relatively easily using a combination of sorting and grouping:

(
    df
    # sort such that the first row is within each group is the one you want
    .sort_values(["A", "B", "C"], ascending=[True, False, False])
    # group based on column A
    .groupby("A")
    # select the first row within each group
    .first()
    # reset the index such that A is a column instead of the index
    .reset_index()
)

Which gives the following result:

This would be another approach a little bit shorter:

df.assign(resultado = lambda x: x.rolling(2).max())

EDIT:

For your comment try:

def idx(x):
    return x.index.values[np.argmax(x.values)]

df.rolling(2).agg(['max', idx])

will return both, the pairwise maximum and the index that corresponds to that value.