Numpy.where in a list of arrays, searching for an entire array

Asked Oct 16 '18 at 11:53

Active Oct 16 '18 at 11:53

Viewed 40 times

I have a list of arrays, and I am looking the position of a specific array:

import numpy as np
c = [np.array([0, 0]), np.array([-1,  0]), np.array([ 0, -1]), np.array([0, 1]), np.array([1, 0])]
n = np.array([0,-1])

np.where(c==n)
# array([[ True, False],
       [False, False],
       [ True,  True],
       [ True, False],
       [False, False]])

I guess it's comparing elemnet by element?
How do I return only the position where both x and y are True, without having to do another np.where to search [True, True]?

asked Oct 16 '18 at 11:53

SuperCiocia

1,823
6
23
40

`np.where(c==n` returns `(array([0, 2, 2, 3]), array([0, 0, 1, 0]))` for me – Mihai Alexandru-Ionut Oct 16 '18 at 12:01
1

You are looking for `np.equal(c, n).all(axis=1)` – Georgy Oct 16 '18 at 12:02
Great thanks @Georgy, if you write that an answer I 'll accept it – SuperCiocia Oct 16 '18 at 12:04
1

Most probably this question was asked before. I'll try to search for a duplicate, and if I won't find anything, I will write an answer. – Georgy Oct 16 '18 at 12:05

Numpy.where in a list of arrays, searching for an entire array

0 Answers0