146
votes

I have a DataFrame like this one:

In [7]:
frame.head()
Out[7]:
Communications and Search   Business    General Lifestyle
0   0.745763    0.050847    0.118644    0.084746
0   0.333333    0.000000    0.583333    0.083333
0   0.617021    0.042553    0.297872    0.042553
0   0.435897    0.000000    0.410256    0.153846
0   0.358974    0.076923    0.410256    0.153846

In here, I want to ask how to get column name which has maximum value for each row, the desired output is like this:

In [7]:
    frame.head()
    Out[7]:
    Communications and Search   Business    General Lifestyle   Max
    0   0.745763    0.050847    0.118644    0.084746           Communications 
    0   0.333333    0.000000    0.583333    0.083333           Business  
    0   0.617021    0.042553    0.297872    0.042553           Communications 
    0   0.435897    0.000000    0.410256    0.153846           Communications 
    0   0.358974    0.076923    0.410256    0.153846           Business 
3

3 Answers

212
votes

You can use idxmax with axis=1 to find the column with the greatest value on each row:

>>> df.idxmax(axis=1)
0    Communications
1          Business
2    Communications
3    Communications
4          Business
dtype: object

To create the new column 'Max', use df['Max'] = df.idxmax(axis=1).

To find the row index at which the maximum value occurs in each column, use df.idxmax() (or equivalently df.idxmax(axis=0)).

38
votes

And if you want to produce a column containing the name of the column with the maximum value but considering only a subset of columns then you use a variation of @ajcr's answer:

df['Max'] = df[['Communications','Business']].idxmax(axis=1)
11
votes

You could apply on dataframe and get argmax() of each row via axis=1

In [144]: df.apply(lambda x: x.argmax(), axis=1)
Out[144]:
0    Communications
1          Business
2    Communications
3    Communications
4          Business
dtype: object

Here's a benchmark to compare how slow apply method is to idxmax() for len(df) ~ 20K

In [146]: %timeit df.apply(lambda x: x.argmax(), axis=1)
1 loops, best of 3: 479 ms per loop

In [147]: %timeit df.idxmax(axis=1)
10 loops, best of 3: 47.3 ms per loop