'Find the column name which has the maximum value for each row
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
Solution 1:[1]
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)
).
Solution 2:[2]
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)
Solution 3:[3]
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
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
---|---|
Solution 1 | |
Solution 2 | user1718097 |
Solution 3 |