For each row in one array: is it in another array?

Question

How can check for each row in a numpy array if it can be found in another array? So if I have

import numpy as np
a = np.array([[1, 2.6], [3, 4], [2.6, 1]])
b = np.array([[0, 0], [3, 4], [1, 2.6], [4.3, 4], [5, 5]])

I would like to get

array([True, True, False])

Obvioulsly

np.isin(a,b)
array([[ True,  True],
       [ True,  True],
       [ True,  True]])

is not an answer and of course I can write something like

return_ = np.zeros(a.shape[0], dtype=bool)
for index, loc in enumerate(a):
    for loc2 in b:
        if np.allclose(loc, loc2):
            return_[index] = True
            break

but this is slow and looks horrible. I would prefer using proper numpy commands.

gofvonx · Accepted Answer · 2021-03-17 11:54:53Z

2

You could try creating a boolean index with appropriate broadcasting and then checking in the result for a True subarray

np.any(np.all(b == a[:, None, :], axis=-1), axis=1)

answered Mar 17, 2021 at 11:54

gofvonx

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1 Comment

Make42 Over a year ago

Using np.argwhere instead of np.any allows for getting the indices and using np.isclose instead of == deals with floating point issues. Perfect!

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