Finding location of subarray in numpy array

Question

I want to find the location of a subarray in an numpy array.
For example, if the large array is of shape (10000, 10, 10, 10), namely 10000 arrays of shape (10,10,10) and I want to find the location of the subarray of shape (3,3,3) as follows:

subarray = array(  
        [[[0., 0., 0.],
        [0., 0., 0.],
        [0., 0., 0.]],

       [[0., 0., 0.],
        [0., 0., 0.],
        [0., 0., 0.]],

       [[0., 0., 0.],
        [0., 1., 0.],
        [0., 0., 0.]]])

So I will know that at the location of n, i, j, k I will get:
largearray[n, i:i+subarray.shape[0], j:j+subarray.shape[1], k:k+subarray.shape[2]] == subarray

Daniel F · Accepted Answer · 2022-04-13 06:55:09Z

Long ago I made a numpy-only function for this, but skimage.util.view_as_windows is a bit easier to implement.

import numpy as np
from skimage.util import view_as_windows as winview

test = np.random.choice((0, 1), p = (.75, .25), size = (5,10,10,10))  #smaller test array
subarray = np.zeros((3,3,3))                                          #what we're looking for
view = winview(test, (1,3,3,3))                                       #a view over 3x3x3 windows 
out = np.all((view == subarray[None, ...]), (-4, -3, -2, -1))         #out=1 where the windows match
print(np.nonzero(out))                                                #output coordinates

Out[]: array([0, 1], dtype=int64), array([2, 7], dtype=int64), array([4, 4], dtype=int64), array([1, 1], dtype=int64))

That gives coordinates for 2 locations with the 3x3x3 zeros box in my random input.

Why do this as windows? Because looping takes forever in python and making the windows as a copy will kill your RAM. view is just another way to address memory in test that is more useful for this type of comparison.

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