How to check if a list of one numpy array contains another numpy array

Question

For example, I want, given the list given_list = [np.array([1,44,21,2])] Check if this list contains another numpy array, for example np.array([21,4,1,21])

I tried this but it checks if each element of the numpy array is the equal to the new numpy array

import numpy as np
given_list = [np.array([1,44,21,2])]
new_elem = np.array([21,4,1,21])

print([new_elem==elem for elem in given_list])

I would like to receive '[False]', but instead I get '[array([False, False, False, False])]'.

I don't understand why given_list is identified as a numpy array and not as a list of one numpy array.

numpy is too smart, if you have a == operator it will compare by elements. I just executed print(np.ones(4) == np.arange(4)) and it gives [False True False False] — Giovanni Tardini
– Giovanni Tardini, Commented Jul 26, 2022 at 13:01
You are iterating on given_list and comparing arrays. Nothing is treating the list as an array. — hpaulj
– hpaulj, Commented Jul 26, 2022 at 14:16

mozway · Accepted Answer · 2022-07-26 12:58:56Z

1

You can use numpy.array_equal:

[np.array_equal(new_elem, elem) for elem in given_list]

or numpy.allclose if you want to have some tolerance:

[np.allclose(new_elem, elem) for elem in given_list]

output: [False]

answered Jul 26, 2022 at 12:58

mozway

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Salvatore Daniele Bianco · Accepted Answer · 2022-07-26 13:28:47Z

1

If you are sure that your arrays have the same length you can do this task without cycling the elements of given_list. For example:

given_list = [np.array([1,44,21,2]), np.array([21,4,1,21])]
new_elem = np.array([21,4,1,21])

(np.array(given_list)==new_elem).all(axis=1) #array([False,  True])

Otherwise, you can use the @mozway 's [np.array_equal(new_elem, elem) for elem in given_list]

answered Jul 26, 2022 at 13:28

Salvatore Daniele Bianco

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