Numpy using where with varying array sizes

Question

So I have two PIL images of RGBA. What I want to do is find all locations where the RGB values are the same and alpha is 255. It looks like this:

from PIL import Image
import numpy as np
img1 = np.array(Image.open(/path/to/img1).convert('RGBA'), float).reshape(32,32,4)
img2 = np.array(Image.open(/path/to/img2).convert('RGBA'), float).reshape(32,32,4)

# Checks to see if RGB of img1 == RGB of img2 in all locations that A=255
np.where((img1[:,:,:-1] == img2[:,:,:-1]) &\ # RGB check
         (img1[:,:,3] == 255)) # Alpha check

But this results in operands could not be broadcast together with shapes (32,32,3) (32,32).
I didn't think I was trying to broadcast them together, I just wanted to find the indeces, which I guess in turn broadcasts them in this statement. Is there another way to do this, or a way to not broadcast unequal shapes?

img1[:, :, :-1] results in an array of shape 32, 32, 3. img1[:, :, 3] results in an array of shape 32, 32. — mgilson
– mgilson, Commented May 20, 2016 at 19:12
@mgilson yes I know. But is what I'm trying to accomplish clear? I thought the above would work because I thought np.where would allow for multiple "where" statements, not broadcast them together — Chrispresso
– Chrispresso, Commented May 20, 2016 at 19:14
And, to point out a bit of style, the \ for line continuation is unnecessary here. Python concatenates lines that are in unenclosed braces, brackets or parenthesis (as is the case here). In fact, PEP 8 (the "official" style guide) recommends using parenthesis to continue lines and never using the \ for line continuation. — mgilson
– mgilson, Commented May 20, 2016 at 19:33

Community · Accepted Answer · 2017-05-23 11:44:07Z

3

Use .all(axis=-1) to find locations where all three RGB values are equal:

np.where((img1[..., :-1] == img2[..., :-1]).all(axis=-1) 
         & (img1[..., 3] == 255))

As mgilson points out, (img1[..., :-1] == img2[..., :-1]) has shape (32, 32, 3). Calling .all(axis-1) reduces the last axis to a scalar boolean value, so

(img1[..., :-1] == img2[..., :-1]).all(axis=-1)

has shape (32, 32). This matches the shape of (img1[..., 3] == 255), so these boolean arrays can be combined with the bitwise-and operator, &.

edited May 23, 2017 at 11:44

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answered May 20, 2016 at 19:23

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3 Comments

mgilson Over a year ago

+1 I think the reduction along the last axis is what OP is asking for. I wonder if OP wants to make sure that the alpha values are the same too ... (In which case it simplifies to (img1 == img2).all(axis=-1) without the ellipsis)

Chrispresso Over a year ago

Awesome! This is exactly what I wanted. And yes @mgilson I was able to add yours too to get the alpha check. Thanks!

mgilson Over a year ago

I'll also point out that & is not the logical-and operator (though it's frequently used as such). It's the binary & operator which is special cased to return an array of logicals if both the right-hand and left-hand sides are logical arrays. For logical and, I've always recommended using np.logical_and (which doesn't have the precedence issues that the & operator has that sometimes bite people trying to use it)...

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