Replace 2D numpy array elements based on 2D indexes [duplicate]

Question

This question has probably been asked somewhere before, but I couldn't find any after some search, hence posting here.

Say I have an array A and an index array idx. Let both the arrays be 2D for the time being.

import numpy as np

A = np.array([[3,3,4],
              [4,5,4],
              [3,4,5]])

idx = np.array([[1,1],
                [2,1],
                [1,0],
                [0,0]])

Now I want to replace the corresponding elements in A based on the indexes in idx to 0. Basically, I want to do A[idx]=0, which doesn't work.

How do I do this efficiently without running a loop?

Preferably, the proposed solution should be scalable to a higher dimensional (3D and above) arrays.

Ehsan · Accepted Answer · 2020-04-28 22:39:07Z

9

You can try:

A[idx[:,0], idx[:,1]]=0

output:

[[0 3 4]
 [0 0 4]
 [3 0 5]]

If you have excessive number of dimensions that you cannot hard code it, you can use:

A[tuple(idx.T)]=0

edited Apr 28, 2020 at 22:39

answered Apr 28, 2020 at 22:26

Ehsan

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