How to find a row-wise intersection of 2d numpy arrays?

Question

I look for an efficient way to get a row-wise intersection of two two-dimensional numpy ndarrays. There is only one intersection per row. For example:

[[1, 2], ∩ [[0, 1], -> [1,
 [3, 4]]    [0, 3]]     3]

In the best case zeros should be ignored:

[[1, 2, 0], ∩ [[0, 1, 0], -> [1,
 [3, 4, 0]]    [0, 3, 0]]     3]

My solution:

import numpy as np

arr1 = np.array([[1, 2],
                 [3, 4]])
arr2 = np.array([[0, 1],
                 [0, 3]])
arr3 = np.empty(len(arr1))

for i in range(len(arr1)):
    arr3[i] = np.intersect1d(arr1[i], arr2[i])

print(arr3)
# [ 1.  3.]

I have about 1 million rows, so the vectorized operations are most preferred. You are welcome to use other python packages.

If you found a solution (which looks right, in my humble opinion, and also vectorized), please post it as a solution and "accept" it so it's visible that the question has an accepted answer. Also, the pandas and scipy tags aren't relevant here. — Itamar Mushkin
– Itamar Mushkin, Commented Jul 2, 2019 at 10:34
You don't have to use a loop here, you can just transpose the arrays: np.intersect1d(arr1.transpose(),arr2.transpose()).transpose() — Itamar Mushkin
– Itamar Mushkin, Commented Jul 2, 2019 at 10:36
@ItamarMushkin Vectorization allows to execute similar operations simultaneously on a bunch of data. Your solution has the for loop so it's executed line by line. I accept solution which use scipy and pandas packages as well. Try to read documentation for the intersect1d function. It is a coincidence that you get the same result. Try these arrays [[1,2],[3,4]] and [[3,4],[1,2]]. — Mykola Zotko
– Mykola Zotko, Commented Jul 2, 2019 at 13:13

user3693922 · Accepted Answer · 2020-12-03 09:40:11Z

You can use np.apply_along_axis. I wrote a solution that pads to the size of the arr1. Didn't test the efficiency.

    import numpy as np

    def intersect1d_padded(x):
        x, y = np.split(x, 2)
        padded_intersection = -1 * np.ones(x.shape, dtype=np.int)
        intersection = np.intersect1d(x, y)
        padded_intersection[:intersection.shape[0]] = intersection
        return padded_intersection

    def rowwise_intersection(a, b):
        return np.apply_along_axis(intersect1d_padded,
                        1, np.concatenate((a, b), axis=1))

    result = rowwise_intersection(arr1,arr2)

    >>> array([[ 1, -1],
               [ 3, -1]])

if you know you have only one element in the intersection you can use

    result = rowwise_intersection(arr1,arr2)[:,0]

    >>> array([1, 3])

You can also modify intersect1d_padded to return a scalar with the intersection value.

Itamar Mushkin · Accepted Answer · 2019-07-02 09:31:06Z

0

I don't know of an elegant way to do it in numpy, but a simple list comprehension can do the trick:

[list(set.intersection(set(_x),set(_y)).difference({0})) for _x,_y in zip(x,y)]

answered Jul 2, 2019 at 9:31

Itamar Mushkin

2,9232 gold badges19 silver badges34 bronze badges

Collectives™ on Stack Overflow

How to find a row-wise intersection of 2d numpy arrays?

2 Answers 2

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related