CODEWITHSUNDEEP
pythonarraysfunctionnumpy
Denver Dang
Denver Dang

Reputation: 2615

How do I accept "divide by zero" as zero? (Python)

So I have the following code:

import numpy as np

array1 = np.array([[[[2, 2, 3], [0, 2, 0], [2, 0, 0]],
[[1, 2, 2], [2, 2, 0], [0, 2, 3]],
[[0, 4, 2], [2, 2, 2], [2, 2, 3]]],
[[[2, 3, 0], [3, 2, 0], [2, 0, 3]],
[[0, 2, 2], [2, 2, 0], [2, 2, 3]],
[[1, 0, 2], [2, 2, 2], [2, 2, 0]]],
[[[2, 0, 0], [0, 2, 0], [2, 0, 0]],
[[2, 2, 2], [0, 2, 0], [2, 2, 0]],
[[0, 2, 2], [2, 2, 2], [2, 2, 0]]]])

array2 = np.array([[[[2, 2, 3], [0, 2, 0], [2, 0, 0]],
[[1, 2, 2], [2, 2, 0], [0, 2, 3]],
[[0, 4, 2], [2, 2, 2], [2, 2, 3]]],
[[[2, 3, 0], [3, 2, 0], [2, 0, 3]],
[[0, 2, 2], [2, 10, 0], [2, 2, 3]],
[[1, 0, 2], [2, 2, 2], [2, 2, 0]]],
[[[2, 0, 0], [0, 2, 0], [2, 0, 0]],
[[2, 2, 2], [0, 2, 0], [2, 2, 0]],
[[0, 2, 2], [2, 2, 2], [2, 2, 0]]]])

def calc(x, y):
    result = y/x
    return result

final_result = []
for x, y in zip(array1, array2):
    final_result.append(calc(np.array(x), np.array(y)))

So all in all I have two lists that include some 3D arrays, and then I have defined a function. The last part is where I use each 3D array in the function, and I ultimately end up with a list (final_result) of some other 3D arrays where the function has been used on each entry from array1 and array2.

However, as you can see, array1 which ultimately gives the x values in the function does have 0 values in some of the entries. And yes, mathematically, this is no good. But in this case, I really just need the entries that does have a zero x-entry to be zero. So it doesn't need to run the function whenever that happens, but just skip it, and leave that entry as zero.

Can this be done?

Upvotes: 0

Views: 215

Answers (1)

nico
nico

Reputation: 2121

This question has been answered here. Numpy has a specific way to catch such errors:

def calc( a, b ):
    """ ignore / 0, div0( [-1, 0, 1], 0 ) -> [0, 0, 0] """
    with np.errstate(divide='ignore', invalid='ignore'):
        c = np.true_divide( a, b )
        c[ ~ np.isfinite( c )] = 0  # -inf inf NaN
    return c

Upvotes: 3

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