How to make zero by zero devision result in zero in Python pandas?

Question

I want to divide one dataframe by another in Pandas to eventually represent a percentage change. Both dataframes values contain NaN and 0. Now, when I divide one dataframe by the other, the result where the value from both dataframes was zero is NaN. I know why 0/0 is set to np.nan, but from a percentage change perspective, I need 0/0 to be 0.

What is the cleanest way to accomplish this?

Reproducing the problem:

import pandas as pd
import numpy as np

data_with_zeros = pd.DataFrame({'a': [2, np.nan, 0, 3], 'b': [np.nan, 2, 0, 6]})

data_with_zeros['a'].div(data_with_zeros['b'], fill_value=0)

Result:

0         inf
1         0.0
2         NaN
3         0.5
dtype: float64

Answer 1

Here's an approach with dataframe.where method -

mask = (data_with_zeros[['a','b']].values == [0,0]).all(1)
data_with_zeros['a'].div(data_with_zeros['b'], fill_value=0).where(~mask,0)

Alternatively mask could be created in a bit more intuitive way, like so -

mask = (data_with_zeros.a == 0) & (data_with_zeros.b == 0)

Sample runs -

Case #1:

In [66]: data_with_zeros
Out[66]: 
     a    b
0  2.0  NaN
1  NaN  2.0
2  0.0  0.0
3  3.0  6.0

In [67]: mask = (data_with_zeros.a == 0) & (data_with_zeros.b == 0)

In [68]: data_with_zeros['a'].div(data_with_zeros['b'], fill_value=0).where(~mask,0)
Out[68]: 
0         inf
1    0.000000
2    0.000000
3    0.500000
dtype: float64

Case #2:

In [70]: data_with_zeros
Out[70]: 
     a    b
0  2.0  0.0
1  NaN  2.0
2  0.0  0.0
3  3.0  6.0

In [71]: mask = (data_with_zeros.a == 0) & (data_with_zeros.b == 0)

In [72]: data_with_zeros['a'].div(data_with_zeros['b'], fill_value=0).where(~mask,0)
Out[72]: 
0         inf
1    0.000000
2    0.000000
3    0.500000
dtype: float64

Answer 2

In addition to the solution provided by Divakar, this function uses that method to do a dataframe/dataframe division:

def divide(a, other, fill_value=None):

    serie_name = a.name
    mask = ((a == 0) & (other[serie_name] == 0))
    result_with_zeros = a.div(other[serie_name], fill_value=fill_value)
    result_filled = result_with_zeros.where(~mask,0)

    return result_filled

data_with_zeros.apply(divide, args=(data_with_zeros,))

Result:

    a   b
0   1.0 0.0
1   NaN 1.0
2   0.0 0.0
3   1.0 1.0