Use one array to filter out another array

Question

How to use one array filter out another array with non-zero value?

from numpy import array

a = array([[ 0,  1,  2,  3,  4],
           [ 5,  6,  7,  8,  9],
           [10, 11, 12, 13, 14],
           [15, 16, 17, 18, 19],
           [20, 21, 22, 23, 24]])

b = array([[0, 0, 1, 0, 0],
           [0, 0, 2, 0, 0],
           [0, 0, 3, 0, 0],
           [0, 0, 4, 0, 0],
           [0, 0, 5, 0, 0]])

Expected result:

array([[ 0, 0, 2,  0, 0],
       [ 0, 0, 7,  0, 0],
       [ 0, 0, 12, 0, 0],
       [ 0, 0, 17, 0, 0],
       [ 0, 0, 22, 0, 0]])

Thank you

MSeifert · Accepted Answer · 2017-09-30 11:30:11Z

1

The easiest way if you want a new array would be np.where with 3 arguments:

>>> import numpy as np
>>> np.where(b, a, 0)
array([[ 0,  0,  2,  0,  0],
       [ 0,  0,  7,  0,  0],
       [ 0,  0, 12,  0,  0],
       [ 0,  0, 17,  0,  0],
       [ 0,  0, 22,  0,  0]])

If you want to change a in-place you could instead use boolean indexing based on b:

>>> a[b == 0] = 0
>>> a
array([[ 0,  0,  2,  0,  0],
       [ 0,  0,  7,  0,  0],
       [ 0,  0, 12,  0,  0],
       [ 0,  0, 17,  0,  0],
       [ 0,  0, 22,  0,  0]])

edited Sep 30, 2017 at 11:30

answered Sep 30, 2017 at 11:24

MSeifert

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Comments

Sraw · Accepted Answer · 2017-09-30 10:03:13Z

1

One line solution:

a * (b != 0)

answered Sep 30, 2017 at 10:03

Sraw

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