Remove subarray from a 2d array based on their existence in another 2d array (return a new 2d array)

Question

I have two 2d array, A and B, like the following.

A=array([[1, 1, 1],
       [2, 2, 2],
       [3, 3, 3],
       [4, 4, 4],
       [5, 5, 5],
       [6, 6, 6],
       [7, 7, 7],
       [8, 8, 8],
       [9, 9, 9]])
B=array([[1, 1, 1],
       [3, 3, 3],
       [8, 8, 8]])

I want to remove subarrays of A if they exist in B. Then return a new 2d array C like the following:

C=array([[2, 2, 2],
       [4, 4, 4],
       [5, 5, 5],
       [6, 6, 6],
       [7, 7, 7],
       [9, 9, 9]])

Currently I have tried the np.isin function but the result is no longer a 2d array.

mask = np.isin(A, B, invert=True)
A[mask]
>>array([2, 2, 2, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 9, 9, 9])

mozway · Accepted Answer · 2021-09-08 07:34:41Z

1

You need to aggregate the booleans on axis 1:

C = A[~np.isin(A,B).all(1)]

output:

array([[2, 2, 2],
       [4, 4, 4],
       [5, 5, 5],
       [6, 6, 6],
       [7, 7, 7],
       [9, 9, 9]])

answered Sep 8, 2021 at 7:34

mozway

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U13-Forward · Accepted Answer · 2021-09-08 07:36:55Z

0

Or in1d:

C = A[~np.in1d(A, B).all(axis=1)]

And now:

print(C)

Output:

[[2 2 2]
 [4 4 4]
 [5 5 5]
 [6 6 6]
 [7 7 7]
 [9 9 9]]

edited Sep 8, 2021 at 7:36

answered Sep 8, 2021 at 7:35

U13-Forward

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3 Comments

mozway Over a year ago

This is incorrect, that would match only the first column

U13-Forward Over a year ago

@mozway Edited mine

mozway Over a year ago

numpy.in1d is semi-deprecated. From the docs: We recommend using :func:isin instead of in1d for new code. ;)

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