Numpy: find index of elements in one array that occur in another array

Question

I have two 1D arrays and I want to find out if an element in one array occurs in another array or not.

For example:

import numpy as np
A = np.array([ 1, 48, 50, 78, 85, 97])
B = np.array([38, 43, 50, 62, 78, 85])

I want:

C = [2,3,4] # since 50 in second array occurs in first array at index 2, 
            # similarly 78 in second array occurs in first array in index 3,
            # similarly for 85, it is index 4

I tried:

accuracy = np.searchsorted(A, B)

But it gives me undesirable results.

This question is similar to: Getting indices of elements that are in another list in numpy. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. — ClimateUnboxed
– ClimateUnboxed, Commented Sep 16, 2024 at 6:40

Alex Riley · Accepted Answer · 2015-03-06 14:55:50Z

15

You could use np.where and np.in1d:

>>> np.where(np.in1d(A, B))[0]
array([2, 3, 4])

np.in1d(A, B) returns a boolean array indicating whether each value of A is found in B. np.where returns the indices of the True values. (This will also work if your arrays are not sorted.)

answered Mar 6, 2015 at 14:55

Alex Riley

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perimosocordiae · Accepted Answer · 2015-03-06 14:53:09Z

7

You should start with np.intersect1d, which finds the set intersection (common elements) between arrays.

In [5]: np.intersect1d(A, B)
Out[5]: array([50, 78, 85])

To get the desired output from your question, you can then use np.searchsorted with just those items:

In [7]: np.searchsorted(A, np.intersect1d(A, B))
Out[7]: array([2, 3, 4])

answered Mar 6, 2015 at 14:53

perimosocordiae

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ClimateUnboxed · Accepted Answer · 2024-09-16 06:38:38Z

0

Since version 2.0 of numpy function in1d is deprecated. Recommendation is to use isin instead:

np.where(np.isin(A, B))[0]
(array([2, 3, 4]),)

answered Sep 16, 2024 at 6:38

ClimateUnboxed

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