how to exploit parallelism in a numpy matrix operation

Question

the following code generates a kind of 'skyline' profile where the initial values are 0's and the profile is made with 1's. I implemented using a for loop and I wonder if there is a more efficient way to obtain the same result.

#https://app.codility.com/programmers/task/max_square_on_matrix/
import numpy as np
np.random.seed(2)
A = np.zeros((5,5), dtype=int)
X = np.random.randint(low=1, high=5, size = 5)

print('A ')
print(A)
print('----------------------')
print('X ')
print(X)
#
for j in range(len(A)):
        A[5-X[j]:5,j] = 1
print('----------------------')
print('A modified ')
print(A)

If you were simply setting one value A[5-X, np.arange(5)] =1. But with slice to the end, that approach doesn't work. We need to generate a mask array with something like X[:,None]<np.arange(5). But details need to worked out in an answer. — hpaulj
– hpaulj, Commented May 26, 2021 at 18:42
Regarding the input size, parallelism will not help. The input should be quite big for parallelism to be useful. — Jérôme Richard
– Jérôme Richard, Commented May 26, 2021 at 20:13

Alexander S. Brunmayr · Accepted Answer · 2021-05-26 23:27:19Z

1

Unless you want to use matrix A for some computation which result in numbers other than 1 or 0, you can simply treat 1 and 0 like True and False:

N = 5
X = np.random.randint(low=1, high=N, size = (N,1) )
A = (X >= range(N,0,-1)).T

EDIT:

Oooop, I responded too quickly. Turns out my method is not "more efficient" than the for loop for N=5.

Also, the following works slightly better than my original answer, especially for large N:

a = np.array(range(N,0,-1),ndmin=2).T
X = np.random.randint(low=1, high=N, size=N)
A = X>=a

On my computer, this is a tiny bit slower than the for loop when N=5. However, when N=100, it is 2 times faster than doing the for loop.

edited May 26, 2021 at 23:27

answered May 26, 2021 at 22:42

Alexander S. Brunmayr

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1 Comment

Alexander S. Brunmayr Over a year ago

Also, if you really need the A.dtype == int instead of bool, then do A = np.empty((N,N), dtype=int) and A[:] = X>=a.

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