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.

Use max with axis=1:

df = df.max(axis=1)
print (df)
0    2.0
1    3.2
2    8.8
3    7.8
dtype: float64

And if need new column:

df['max_value'] = df.max(axis=1)
print (df)
     a    b     c  max_value
0  1.2  2.0  0.10        2.0
1  2.1  1.1  3.20        3.2
2  0.2  1.9  8.80        8.8
3  3.3  7.8  0.12        7.8

Set display.max_rows:

pd.set_option('display.max_rows', 500)

For older versions of pandas (<=0.11.0) you need to change both display.height and display.max_rows.

pd.set_option('display.height', 500)
pd.set_option('display.max_rows', 500)

See also pd.describe_option('display').

You can set an option only temporarily for this one time like this:

from IPython.display import display
with pd.option_context('display.max_rows', 100, 'display.max_columns', 10):
    display(df) #need display to show the dataframe when using with in jupyter
    #some pandas stuff

You can also reset an option back to its default value like this:

pd.reset_option('display.max_rows')

And reset all of them back:

pd.reset_option('all')

Assuming df has a unique index, this gives the row with the maximum value:

In [34]: df.loc[df['Value'].idxmax()]
Out[34]: 
Country        US
Place      Kansas
Value         894
Name: 7

Note that idxmax returns index labels. So if the DataFrame has duplicates in the index, the label may not uniquely identify the row, so df.loc may return more than one row.

Therefore, if df does not have a unique index, you must make the index unique before proceeding as above. Depending on the DataFrame, sometimes you can use stack or set_index to make the index unique. Or, you can simply reset the index (so the rows become renumbered, starting at 0):

df = df.reset_index()

I would just forget the 'max_val_idx' column. I don't think it saves time and actually is more of a pain for syntax. Sample data:

df = pd.DataFrame({ 'x': range(3) }).applymap( lambda x: np.random.randn(3) )

                                                   x
0  [-1.17106202376, -1.61211460669, 0.0198122724315]
1    [0.806819945736, 1.49139051675, -0.21434675401]
2  [-0.427272615966, 0.0939459129359, 0.496474566...

You could extract the max like this:

df.applymap( lambda x: x.max() )

          x  
0  0.019812
1  1.491391
2  0.496475

But generally speaking, life is easier if you have one number per cell. If each cell has an array of length 3, you could rearrange like this:

for i, v in enumerate(list('abc')): df[v] = df.x.map( lambda x: x[i] )
df = df[list('abc')]

          a         b         c
0 -1.171062 -1.612115  0.019812
1  0.806820  1.491391 -0.214347
2 -0.427273  0.093946  0.496475

And then do a standard pandas operation:

df.apply( max, axis=1 )

          x  
0  0.019812
1  1.491391
2  0.496475

Admittedly, this is not much easier than above, but overall the data will be much easier to work with in this form.