Rowwise numpy.isin for 2D arrays

Question

I have two arrays:

A = np.array([[3, 1], [4, 1], [1, 4]])
B = np.array([[0, 1, 5], [2, 4, 5], [2, 3, 5]])

Is it possible to use numpy.isin rowwise for 2D arrays? I want to check if A[i,j] is in B[i] and return this result into C[i,j]. At the end I would get the following C:

np.array([[False, True], [True, False], [False, False]])

It would be great, if this is also doable with the == operator, then I could use it also with PyTorch.

Edit: I also considered check for identical rows in different numpy arrays. The question is somehow similar, but I am not able to apply its solutions to this slightly different problem.

Please use Divakar's solution, not the accepted answer. The accepted answer makes some unnecessary temp arrays. — Mad Physicist, Jun 07 '21 at 11:47
@Christian. To the duplicate question: https://stackoverflow.com/a/51352806/2988730 — Mad Physicist, Jun 07 '21 at 14:31
@MadPhysicist what are the problems with the solutions here? — Christian, Jun 07 '21 at 14:46
@Christian. They are the same as the accepted one there. They are totally functional, but create massive and unnecessary temporary arrays, which can be eliminated by calling in1d with properly viewed data. — Mad Physicist, Jun 07 '21 at 14:58

score 3 · Accepted Answer · answered Jun 07 '21 at 11:21

Not sure that my code solves your problem perfectly. please run it on more test cases to confirm. but i would do smth like i've done in the following code taking advantage of numpys vector outer operations ability (similar to vector outer product). If it works as intended it should work with pytorch as well.

import numpy as np
A = np.array([[3, 1], [4, 1], [1, 4]])
B = np.array([[0, 1, 5], [2, 4, 5], [2, 3, 5]])

AA = A.reshape(3, 2, 1)
BB = B.reshape(3, 1, 3)
(AA == BB).sum(-1).astype(bool)

output:

array([[False,  True],
       [ True, False],
       [False, False]])

Thanks! I was looking for something like this. In contrast to the answer of Whole Brain, this code works also if `A` and `B` have the same dimension. You can reduce it to one line: `(A[:,:,None] == B[:,None,:).any(-1)` — Christian, Jun 07 '21 at 14:57

Whole Brain · Answer 2 · 2021-06-07T11:45:25.070

2

Updated answer

Here is a way to do this :

(A == B[..., None]).any(axis=1).astype(bool)
# > array([[False,  True],
#          [ True, False],
#          [False, False]])

Previous answer

You could also do it inside a list comprehension:

[np.isin(a, b) for a,b in zip(A, B)]
# > [array([False,  True]), array([ True, False]), array([False, False])]
np.array([np.isin(a, b) for a,b in zip(A, B)])
# > array([[False,  True],
#          [ True, False],
#          [False, False]])

But, as @alex said it defeats the purpose of numpy.

edited Jun 07 '21 at 11:45

answered Jun 07 '21 at 11:25

Whole Brain

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loops defeat the whole purpose of Numpy... – alex Jun 07 '21 at 11:35
1

Thanks @WholeBrain, I think I tried that, but unfortunately that did not work if `B` has the same dimension as `A` (in this case 2 columns). – Christian Jun 07 '21 at 13:49

Rowwise numpy.isin for 2D arrays

2 Answers2

Updated answer

Previous answer

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