logical_and with nested boolean arrays

Question

I have one (very long) boolean array a with k True entries, and one boolean array b of length k. I would like to get a boolean array c that is True if and only if a "and" b are True:

import numpy

a = numpy.array([False, False, True, False, True, False])
b = numpy.array([True, False])

assert numpy.sum(a) == len(b)  # guaranteed

c = numpy.zeros(len(a), dtype=bool)
idx_b = 0
for k in range(len(a)):
    if a[k]:
        if b[idx_b]:
            c[k] = True
        idx_b += 1

print(c)

[False False  True False False False]

This here uses a loop, but I'm thinking there must a faster way with boolean indexing, but I can't quite get to it.

Any hints?

So out of bounds is treated as False?

Ted Brownlow
– Ted Brownlow

2020-11-10 19:01:18 +00:00
Commented Nov 10, 2020 at 19:01 — Ted Brownlow
– Ted Brownlow, Commented Nov 10, 2020 at 19:01
out-of-bounds how?

Nico Schlömer
– Nico Schlömer

2020-11-10 19:01:59 +00:00
Commented Nov 10, 2020 at 19:01 — Nico Schlömer
– Nico Schlömer, Commented Nov 10, 2020 at 19:01

Divakar · Accepted Answer · 2020-11-10 19:20:05Z

6

Simply mask input array with itself (self-masking?) and assign -

a[a] = b

If you need output in a new array, copy the input array and perform masking on the same.

answered Nov 10, 2020 at 19:20

Divakar

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Comments

Michael Szczesny · Accepted Answer · 2020-11-10 19:29:49Z

1

If the assertion holds true, you can use np.flatnonzero

import numpy as np

a = np.array([False, False, True, False, True, False])
b = np.array([True, False])

assert np.sum(a) == len(b)

c = np.copy(a)
idx = np.flatnonzero(c) 
c[idx] = b

print(c)

Out:

[False, False,  True, False, False, False]

edited Nov 10, 2020 at 19:29

answered Nov 10, 2020 at 19:04

Michael Szczesny

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