filtering a 3D numpy array according to 2D numpy array

Question

I have a 2D numpy array with the shape (3024, 4032).

I have a 3D numpy array with the shape (3024, 4032, 3).

2D numpy array is filled with 0s and 1s.

3D numpy array is filled with values between 0 and 255.

By looking at the 2D array values, I want to change the values in 3D array. If a value in 2D array is 0, I will change the all 3 pixel values in 3D array into 0 along the last axes. If a value in 2D array is 1, I won't change it.

I have checked this question, How to filter a numpy array with another array's values, but it applies for 2 arrays which have same dimensions. In my case, dimensions are different.

How the filtering is applied in two arrays, with same size on 2 dimensions, but not size on the last dimension?

I will change the pixel value in 3D array into 0 - All three values along the last axis? — Divakar
– Divakar, Commented Nov 21, 2017 at 8:34

Paul Panzer · Accepted Answer · 2017-11-21 09:46:51Z

7

Ok, I'll answer this to highlight one pecularity regarding "missing" dimensions. Lets' assume a.shape==(5,4,3) and b.shape==(5,4)

When indexing, existing dimensions are left aligned which is why @Divakar's solution a[b == 0] = 0 works.

When broadcasting, existing dimensions are right aligned which is why @InvaderZim's a*b does not work. What you need to do is a*b[..., None] which inserts a broadcastable dimension at the right

answered Nov 21, 2017 at 9:46

Paul Panzer

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Comments

InvaderZim · Accepted Answer · 2017-11-22 10:10:45Z

1

I think this one is very simple:

If a is a 3D array (a.shape == (5, 4, 3)) filled with values, and b is a 2D array (b.shape == (5, 4)) filled with 1 and 0, then reshape b and multiply them:

a = a * b.reshape(5, 4, 1)

Numpy will automatically expand the arrays as needed.

edited Nov 22, 2017 at 10:10

answered Nov 21, 2017 at 8:55

InvaderZim

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2 Comments

Paul Panzer Over a year ago

Did either you or OP who accepted this answer actually test it?

InvaderZim Over a year ago

Thought I checked, but seems like I did not. Edited. Thanks :)

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