How to return 0 with divide by zero

Question

In numpy v1.7+, you can take advantage of the "where" option for ufuncs. You can do things in one line and you don't have to deal with the errstate context manager.

>>> a = np.array([-1, 0, 1, 2, 3], dtype=float)
>>> b = np.array([ 0, 0, 0, 2, 2], dtype=float)

# If you don't pass `out` the indices where (b == 0) will be uninitialized!
>>> c = np.divide(a, b, out=np.zeros_like(a), where=b!=0)
>>> print(c)
[ 0.   0.   0.   1.   1.5]

In this case, it does the divide calculation anywhere 'where' b does not equal zero. When b does equal zero, then it remains unchanged from whatever value you originally gave it in the 'out' argument.

Building on @Franck Dernoncourt's answer, fixing -1 / 0 and my bug on scalars:

def div0( a, b, fill=np.nan ):
    """ a / b, divide by 0 -> `fill`
        div0( [-1, 0, 1], 0, fill=np.nan) -> [nan nan nan]
        div0( 1, 0, fill=np.inf ) -> inf
    """
    with np.errstate(divide='ignore', invalid='ignore'):
        c = np.true_divide( a, b )
    if np.isscalar( c ):
        return c if np.isfinite( c ) \
            else fill
    else:
        c[ ~ np.isfinite( c )] = fill
        return c

Building on the other answers, and improving on:

• 0/0 handling by adding invalid='ignore' to numpy.errstate()
• introducing numpy.nan_to_num() to convert np.nan to 0.

Code:

import numpy as np

a = np.array([0,0,1,1,2], dtype='float')
b = np.array([0,1,0,1,3], dtype='float')

with np.errstate(divide='ignore', invalid='ignore'):
    c = np.true_divide(a,b)
    c[c == np.inf] = 0
    c = np.nan_to_num(c)

print('c: {0}'.format(c))

Output:

c: [ 0.          0.          0.          1.          0.66666667]

DEPRECATED (PYTHON 2 SOLUTION):

One-liner (throws warning)

np.nan_to_num(array1 / array2)