numpy.where : how to delay evaluating parameters?

Question

I am using numpy.where, and I was wondering if there was a simply way to avoid calling the unused parameter. Example:

import numpy as np
z = np.array([-2, -1, 0, 1, -2])
np.where(z!=0, 1/z, 1)

returns:

array([-0.5, -1. ,  1. ,  1. , -0.5])

but I get a divide by zero warning because when z was 0, the code still evaluates 1/z even though it doesn't use it.

why don't you use z = z[np.nonzero(z)] then divide by z, but that just returns 1 so I guess I am not sure why — , Jul 09 '16 at 00:20
`1/z` is executed by the interpreter as part of calling `where`. That's not under the control of `where`. But you can do `out[ind]=1/z[ind]` where `ind` is the mask of ok values. There are other ways of dealing divide by zero. — hpaulj, Jul 09 '16 at 01:43

Mike Müller · Answer 1 · 2016-07-09T04:14:54.480

2

You can also turn off the warning and turn it back on after you are done using the context manager errstate:

with np.errstate(divide='ignore'):
    np.where(z!=0, 1/z, 1)

edited Jul 09 '16 at 04:14

answered Jul 09 '16 at 04:09

Mike Müller

score 1 · Answer 2 · answered Jul 09 '16 at 03:25

1

You can apply a mask:

out = numpy.ones_like(z)
mask = z != 0
out[mask] = 1/z[mask]

answered Jul 09 '16 at 03:25

Benjamin

score 0 · Answer 3 · answered Jul 11 '16 at 20:28

0

It's basically a matter of allocating the output array and using np.place (or putmask) to fill exactly what you need.

answered Jul 11 '16 at 20:28

ev-br

3 Answers3