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

user_8973620 0
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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.

numpy-ndarray numpy python
2 Answers

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.

user_36939220

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)]
user_88746960
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