How can I write zeros to a 2D numpy array by both row and column indices

Question

I have a large (90k x 90k) numpy ndarray and I need to zero out a block of it. I have a list of about 30k indices that indicate which rows and columns need to be zero. The indices aren't necessarily contiguous, so a[min:max, min:max] style slicing isn't possible.

As a toy example, I can start with a 2D array of non-zero values, but I can't seem to write zeros the way I expect.

import numpy as np

a = np.ones((6, 8))
indices = [2, 3, 5]

# I thought this would work, but it does not.
# It correctly writes to (2,2), (3,3), and (5,5), but not all
# combinations of (2, 3), (2, 5), (3, 2), (3, 5), (5, 2), or (5, 3)
a[indices, indices] = 0.0
print(a)

[[1. 1. 1. 1. 1. 1. 1. 1.]
 [1. 1. 1. 1. 1. 1. 1. 1.]
 [1. 1. 0. 1. 1. 1. 1. 1.]
 [1. 1. 1. 0. 1. 1. 1. 1.]
 [1. 1. 1. 1. 1. 1. 1. 1.]
 [1. 1. 1. 1. 1. 0. 1. 1.]]

# I thought this would fix that problem, but it doesn't change the array.
a[indices, :][:, indices] = 0.0
print(a)

[[1. 1. 1. 1. 1. 1. 1. 1.]
 [1. 1. 1. 1. 1. 1. 1. 1.]
 [1. 1. 1. 1. 1. 1. 1. 1.]
 [1. 1. 1. 1. 1. 1. 1. 1.]
 [1. 1. 1. 1. 1. 1. 1. 1.]
 [1. 1. 1. 1. 1. 1. 1. 1.]]

In this toy example, I'm hoping for this result.

[[1. 1. 1. 1. 1. 1. 1. 1.]
 [1. 1. 1. 1. 1. 1. 1. 1.]
 [1. 1. 0. 0. 1. 0. 1. 1.]
 [1. 1. 0. 0. 1. 0. 1. 1.]
 [1. 1. 1. 1. 1. 1. 1. 1.]
 [1. 1. 0. 0. 1. 0. 1. 1.]]

I could probably write a cumbersome loop or build some combinatorically huge list of indices to do this, but it seems intuitive that this must be supported in a cleaner way, I just can't find the syntax to make it happen. Any ideas?

@hpaulj, I had never seen this function, and wouldn't have known to search for it. And it's exactly what I need. Thanks!! I wrote an answer, but if you'd rather take the credit, I'll delete mine. — mightypile
– mightypile, Commented Jan 16 at 23:19

mightypile · Accepted Answer · 2025-01-16 23:28:57Z

3

Based on hpaulj's comment, I came up with this, which works perfectly on the toy example.

a[np.ix_(indices, indices)] = 0.0
print(a)

[[1. 1. 1. 1. 1. 1. 1. 1.]
 [1. 1. 1. 1. 1. 1. 1. 1.]
 [1. 1. 0. 0. 1. 0. 1. 1.]
 [1. 1. 0. 0. 1. 0. 1. 1.]
 [1. 1. 1. 1. 1. 1. 1. 1.]
 [1. 1. 0. 0. 1. 0. 1. 1.]]

It also worked beautifully on the real data. It was faster than I expected and didn't noticeably increase memory consumption. Exhausting memory has been a constant concern with these giant arrays.

edited Jan 16 at 23:28

answered Jan 16 at 23:18

mightypile

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1 Comment

hpaulj Jan 17 at 6:49

Look at the result of np.ix_(indices, indices). It is easy to create the same two arrays without ix_. Ir's a convenience function, not a necessity. Those 2 arrays are broadcastable.

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