# Any suggestion to replace the diagonal values of matrices in a 3D array?

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julian gaviria am 6 Nov. 2023
Kommentiert: Dyuman Joshi am 6 Nov. 2023
Any suggestion to replace the diagonal values of matrices in a 3D array?
in case1 to "0"
in case1 to "NaN"
e.g. for NaN
M_3D=randi(100, 4,4,3); %fake input
N=size(M_3D,1);
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Steven Lord am 6 Nov. 2023
One possibility is points at subscripts n*ones(1, ndimsOfArray) for n = 1:min(size(theArray)). Once you reach any "edge" of the array you stop.
So for an array of size (4, 3, 5) those would be the elements at subscripts (1, 1, 1), (2, 2, 2), and (3, 3, 3). That matches conceptually the behavior of the diag function on non-square matrices.
A = reshape(1:12, 3, 4)
A = 3×4
1 4 7 10 2 5 8 11 3 6 9 12
diag(A) % (1, 1), (2, 2), and (3, 3) since min(size(A)) is 3.
ans = 3×1
1 5 9
julian gaviria am 6 Nov. 2023
@Steven Lord, following you example, I want to replace the elements 1, 5, 9, 10, 14, 18, 19, 23, and 27 either by "0" or by "NaN" values. These are two different cases. Apologies for not providing all the info in the initial post. Sometimes my 3D matrices have NaN values in the diagonals, which I must replace by "0" value.s @Dyuman Joshi's suggestion worked out.
Thanks anyway.

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### Akzeptierte Antwort

Dyuman Joshi am 6 Nov. 2023
Replacing diagonal values of each 2D matrix of a 3D matrix with NaN
M = randi(100, 4,4,3); %fake input
N = eye(size(M,[1 2]));
M(M&N)=NaN
M =
M(:,:,1) = NaN 88 80 30 51 NaN 98 73 75 100 NaN 63 68 51 61 NaN M(:,:,2) = NaN 13 65 60 99 NaN 55 50 16 24 NaN 94 46 52 86 NaN M(:,:,3) = NaN 50 49 17 41 NaN 68 52 79 80 NaN 79 83 47 86 NaN
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julian gaviria am 6 Nov. 2023
It nicely worked! Thanks a lot @Dyuman Joshi
Dyuman Joshi am 6 Nov. 2023
You are welcome!

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