Difference between images of different sizes

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My problem is as follows. I have an image img0 (array shape (A,B,3)) and then a face img1 cut out from the middle of that image (by an algorithm I don't have access to: my input is only the whole image, and the face cut out from it), now an array shaped (C,D,3) where C<A and D<B. Now, I want to perform operations on the face (e.g., colour it differently) and then stick it back inside the original background (which is not coloured differently) -- these operations will not affect the shape of img1 array containing the face alone, it will remain (C,D,3). Something like img0-img1 doesn't work because of the shape mismatch.

I guess an approach like finding the starting coordinate of the face in img0 would work in the case that the face cut out is rectangular (which is possible for me to use, though not ideal), since it is guaranteed that the face is exactly identical in img1 and img0. That means, to get the background, we only need to find the starting coordinate of the img1 array in img0, cut out the subsequent elements (that correspond to img1) from img0, and we're left with the background. After I've done whatever I want to the face, I can use the new (C,D,3) array in place of the previous img1 part of the whole image (img0).

Is there a way to do this in Python? i.e., compute the difference between two images of different sizes, where one image is a 'subimage' of the other? Or, failing that, if we can find the starting coordinate of the rectangular portion of an image (img0) which corresponds to a rectangular cutout available to us (img1)?

opencv image-processing numpy image python
1 Answers

Or, failing that, if we can find the starting coordinate of the rectangular ?portion of an image (img0) which corresponds to a rectangular cutout available to us (img1)?

One easy way to do that would be to cross-correlate your zero-mean cut-out with the zero-mean original image. As you have no noise added to the image, any maximum of the cross-correlation is a possible candidate.

However:

(i) If you don't use faces but e.g. blocks, there will be multiple maxima and you don't have an unique solution. (ii) It is not exactly an elegant solution to your problem.

I modified the code example from [1] to make it clearer:

from scipy import signal, misc
import numpy as np

face = misc.face(gray=True)
face = face - np.mean(face)
face_cutout = np.copy(face[300:365, 670:750])
face_cutout = face_cutout - np.mean(face_cutout)

corr = signal.correlate2d(face, face_cutout, mode='valid')
y, x = np.unravel_index(np.argmax(corr), corr.shape)  # find the match

print(f'x: {x} y: {y}')

[1] https://docs.scipy.org/doc/scipy/reference/generated/scipy.signal.correlate2d.html

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