Numpy generate binary array with at most N ones

Question

I can easily create a random binary array in Numpy by doing this

random_mask = np.random.randint(0,2, (r, c))

But what what I actually want is to set the maximum number of 1s that can be in the array.

For example, if I want a 5,5 binary matrix, I want there to be at most 10 ones randomly placed throughout the matrix, and the rest are 0s.

I was thinking of an approach where I generate the random array like normal, count the number of 1s that are currently placed, and somehow subtracting off the ones I don't need.

I'm wondering if there's already a way to do this in numpy

user8408080 · Accepted Answer · 2019-01-27 22:01:29Z

1

This is the most basic approach I could think of:

import numpy as np

def binary_mask_random(r, c, n):
    a = np.zeros((r,c)).flatten()

    for i in range(np.random.randint(0, n+1)):
        x = np.random.randint(0, r*c)
        a[x] = 1

    return a.reshape((r,c))

It creates a 1xr*c array of zeros and fills it with up to n 1s at random positions. Returns a rxc array.

edited Jan 27, 2019 at 22:01

answered Jan 27, 2019 at 21:54

user8408080

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4 Comments

Carpetfizz Over a year ago

hm yeah this would do the trick - I guess there isn't a built in way to do this in numpy? was trying to avoid loops

user8408080 Over a year ago

I don't think, that there is a built-in solution, but it may be possible to find a fully vectorized solution

Carpetfizz Over a year ago

yeah I’ll play around with it - thanks for pointing me in the right direction

user8408080 Over a year ago

@Carpetfizz Maybe you could also try np.random.choice([0, 1], (r, c), p=[0.9, 0.1]), but i'm not exactly sure, if this will always give you the exact results you need, because after all, it's just a probability

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