Why isn't my Pandas 'apply' function referencing multiple columns working?

Question:

I have some problems with the Pandas apply function, when using multiple columns with the following dataframe

df = DataFrame ({'a' : np.random.randn(6),
                 'b' : ['foo', 'bar'] * 3,
                 'c' : np.random.randn(6)})

and the following function

def my_test(a, b):
    return a % b

When I try to apply this function with :

df['Value'] = df.apply(lambda row: my_test(row[a], row[c]), axis=1)

I get the error message:

NameError: ("global name 'a' is not defined", u'occurred at index 0')

I do not understand this message, I defined the name properly.

I would highly appreciate any help on this issue

Update

Thanks for your help. I made indeed some syntax mistakes with the code, the index should be put ”. However I still get the same issue using a more complex function such as:

def my_test(a):
    cum_diff = 0
    for ix in df.index():
        cum_diff = cum_diff + (a - df['a'][ix])
    return cum_diff
Asked By: Andy

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Source

Answers:

If you just want to compute (column a) % (column b), you don’t need apply, just do it directly:

In [7]: df['a'] % df['c']                                                                                                                                                        
Out[7]: 
0   -1.132022                                                                                                                                                                    
1   -0.939493                                                                                                                                                                    
2    0.201931                                                                                                                                                                    
3    0.511374                                                                                                                                                                    
4   -0.694647                                                                                                                                                                    
5   -0.023486                                                                                                                                                                    
Name: a
Answered By: herrfz

Seems you forgot the '' of your string.

In [43]: df['Value'] = df.apply(lambda row: my_test(row['a'], row['c']), axis=1)

In [44]: df
Out[44]:
                    a    b         c     Value
          0 -1.674308  foo  0.343801  0.044698
          1 -2.163236  bar -2.046438 -0.116798
          2 -0.199115  foo -0.458050 -0.199115
          3  0.918646  bar -0.007185 -0.001006
          4  1.336830  foo  0.534292  0.268245
          5  0.976844  bar -0.773630 -0.570417

BTW, in my opinion, following way is more elegant:

In [53]: def my_test2(row):
....:     return row['a'] % row['c']
....:     

In [54]: df['Value'] = df.apply(my_test2, axis=1)
Answered By: waitingkuo

Let’s say we want to apply a function add5 to columns ‘a’ and ‘b’ of DataFrame df

def add5(x):
    return x+5

df[['a', 'b']].apply(add5)
Answered By: Mir_Murtaza

All of the suggestions above work, but if you want your computations to by more efficient, you should take advantage of numpy vector operations (as pointed out here).

import pandas as pd
import numpy as np


df = pd.DataFrame ({'a' : np.random.randn(6),
             'b' : ['foo', 'bar'] * 3,
             'c' : np.random.randn(6)})

Example 1: looping with pandas.apply():

%%timeit
def my_test2(row):
    return row['a'] % row['c']

df['Value'] = df.apply(my_test2, axis=1)

The slowest run took 7.49 times longer than the fastest. This could
mean that an intermediate result is being cached. 1000 loops, best of
3: 481 µs per loop

Example 2: vectorize using pandas.apply():

%%timeit
df['a'] % df['c']

The slowest run took 458.85 times longer than the fastest. This could
mean that an intermediate result is being cached. 10000 loops, best of
3: 70.9 µs per loop

Example 3: vectorize using numpy arrays:

%%timeit
df['a'].values % df['c'].values

The slowest run took 7.98 times longer than the fastest. This could
mean that an intermediate result is being cached. 100000 loops, best
of 3: 6.39 µs per loop

So vectorizing using numpy arrays improved the speed by almost two orders of magnitude.

Answered By: Blane

This is same as the previous solution but I have defined the function in df.apply itself:

df['Value'] = df.apply(lambda row: row['a']%row['c'], axis=1)
Answered By: shaurya airi

I have given the comparison of all three discussed above.

Using values

%timeit df['value'] = df['a'].values % df['c'].values

139 µs ± 1.91 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

Without values

%timeit df['value'] = df['a']%df['c']

216 µs ± 1.86 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Apply function

%timeit df['Value'] = df.apply(lambda row: row['a']%row['c'], axis=1)

474 µs ± 5.07 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Answered By: Gursewak Singh