How to replace diagonal of square matrices within a cell array?

I have a 940X1 cell array where each element is a 298X298 square matrix. I want to replace the main diagonals of each square matrix with zero. Is there a way to do it?

 Accepted Answer

The easiest way is to subtract the diagonal of each matrix from the matrix —
C = {rand(4); rand(4); rand(4)}
C = 3×1 cell array
{4×4 double} {4×4 double} {4×4 double}
Cr = cellfun(@(x) x-diag(x), C, 'Unif',0)
Cr = 3×1 cell array
{4×4 double} {4×4 double} {4×4 double}
Cr{1}
ans = 4×4
0 0.5651 0.1597 0.4081 0.0049 0 -0.5032 0.3582 0.4049 0.8147 0 0.2985 -0.4793 -0.4366 -0.5270 0
Cr{3}
ans = 4×4
0 0.2540 -0.2275 -0.0101 0.1521 0 -0.7862 -0.3234 0.5575 -0.0462 0 -0.1965 -0.4953 -0.3632 -0.5305 0
Make appropriate changes to get different results.
.

8 Comments

As always, my plesaure!
Replacing the diagonals with different values is simply an extension of that approach —
C = {rand(4); rand(4); rand(4)}
C = 3×1 cell array
{4×4 double} {4×4 double} {4×4 double}
Cr = cellfun(@(x) x-diag(x)+diag([1 2 3 4]), C, 'Unif',0)
Cr = 3×1 cell array
{4×4 double} {4×4 double} {4×4 double}
Cr{1}
ans = 4×4
1.0000 0.1342 -0.2432 0.2293 -0.6770 2.0000 -0.3277 0.1889 0.6493 0.5733 3.0000 0.7478 0.1705 0.4593 0.5354 4.0000
Cr{3}
ans = 4×4
1.0000 -0.7256 0.2266 -0.5666 0.0518 2.0000 -0.5614 -0.2324 -0.0021 0.7514 3.0000 0.3976 -0.5289 -0.0793 -0.1245 4.0000
.
Well it gets the job done. And this is easier.
Hi Star Strider.
Sorry to revive this post again. I just realized that by using your method although the diagonal is replaced by zeros, but the values around the diagonal element also changed. Why is that? The other values should remain the same.
It is because I have not used diag in a while, and assumed that it only changes the diagonal.
Change ‘Cr’ to:
Cr = cellfun(@(x) x-diag(diag(x)), C, 'Unif',0);
and it works the way I intend it to (and the way I thought it originally did).
My apologies for the inconvenience.
.
Cheers! It is all fixed now. Really appreciate all your help.

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