# How to make a two-dimensional mask act on a three-dimensional array?

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Steve Francis il 25 Mag 2022
Modificato: Stephen23 il 16 Lug 2022
Here's a dummy three-dimensional array 'B' to illustrate the problem:
A = 9* ones (6,4,3); % create a 3 dimensional array
B = A;
B is a 6x4x3 array with two NaN elements.
Wherever a NaN appears in array B, I want a NaN to overwrite all elements in the corresponding first dimension. So, in the example above, I want to convert B so that my output is
B(:,:,1) =
9 NaN 9 9
9 NaN 9 9
9 NaN 9 9
9 NaN 9 9
9 NaN 9 9
9 NaN 9 9
B(:,:,2) =
9 9 NaN 9
9 9 NaN 9
9 9 NaN 9
9 9 NaN 9
9 9 NaN 9
9 9 NaN 9
B(:,:,3) =
9 9 9 9
9 9 9 9
9 9 9 9
9 9 9 9
9 9 9 9
9 9 9 9
The script below appears to work but is not very elegant. Is there a way to do it without the 'for' loop?
% Here is my attempt to get NaN to overwrite all elements along the corresponding FIRST dimension
C = isnan(B); % create a logical mask
D = any (C,1); % look for any NaNs along the first dimension
for k=1:(size(B,1))
E=squeeze(B(k,:,:));
E(squeeze(D))=NaN; % apply the 'any' mask
B(k,:,:)=E ;
end
There is a very similar question here: https://uk.mathworks.com/matlabcentral/answers/358514-replace-all-the-array-with-nan-if-any-of-the-value-is-nan but I couldn't successfully tailor Jan's solution to my question. Thank you.
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### Risposta accettata

Stephen23 il 26 Mag 2022
"I wondered if there was any neat trick to do it by array multipication..."
A = 9* ones (6,4,3); % create a 3 dimensional array
A(4,3,2) = NaN; % add a NaN
A(3,2,1) = NaN; % add another NaN;
B = A
B =
B(:,:,1) = 9 9 9 9 9 9 9 9 9 NaN 9 9 9 9 9 9 9 9 9 9 9 9 9 9 B(:,:,2) = 9 9 9 9 9 9 9 9 9 9 9 9 9 9 NaN 9 9 9 9 9 9 9 9 9 B(:,:,3) = 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
X = 0./~any(isnan(B),1);
B = B+X
B =
B(:,:,1) = 9 NaN 9 9 9 NaN 9 9 9 NaN 9 9 9 NaN 9 9 9 NaN 9 9 9 NaN 9 9 B(:,:,2) = 9 9 NaN 9 9 9 NaN 9 9 9 NaN 9 9 9 NaN 9 9 9 NaN 9 9 9 NaN 9 B(:,:,3) = 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
If you really want to use multilplication add 1 to X.
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Stephen23 il 16 Lug 2022
Modificato: Stephen23 il 16 Lug 2022
A = 9*ones(4,3,6); % create a 3 dimensional array
A(3,2,4) = NaN; % add a NaN
A(2,1,3) = NaN % add another NaN;
A =
A(:,:,1) = 9 9 9 9 9 9 9 9 9 9 9 9 A(:,:,2) = 9 9 9 9 9 9 9 9 9 9 9 9 A(:,:,3) = 9 9 9 NaN 9 9 9 9 9 9 9 9 A(:,:,4) = 9 9 9 9 9 9 9 NaN 9 9 9 9 A(:,:,5) = 9 9 9 9 9 9 9 9 9 9 9 9 A(:,:,6) = 9 9 9 9 9 9 9 9 9 9 9 9
B = A;
X = 0./~any(isnan(B),3);
B = B+X % ^ you just need to change this
B =
B(:,:,1) = 9 9 9 NaN 9 9 9 NaN 9 9 9 9 B(:,:,2) = 9 9 9 NaN 9 9 9 NaN 9 9 9 9 B(:,:,3) = 9 9 9 NaN 9 9 9 NaN 9 9 9 9 B(:,:,4) = 9 9 9 NaN 9 9 9 NaN 9 9 9 9 B(:,:,5) = 9 9 9 NaN 9 9 9 NaN 9 9 9 9 B(:,:,6) = 9 9 9 NaN 9 9 9 NaN 9 9 9 9

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### Più risposte (1)

Voss il 25 Mag 2022
Modificato: Voss il 25 Mag 2022
You can replicate your matrix D=any(isnan(B),1) in the first dimension using repmat, generating a 3D logical array the same size as B, and then use that 3D logical array as a logical index in B:
A = 9* ones (6,4,3); % create a 3 dimensional array
B = A;
B(repmat(any(isnan(B),1),[size(B,1) 1 1])) = NaN;
disp(B);
(:,:,1) = 9 NaN 9 9 9 NaN 9 9 9 NaN 9 9 9 NaN 9 9 9 NaN 9 9 9 NaN 9 9 (:,:,2) = 9 9 NaN 9 9 9 NaN 9 9 9 NaN 9 9 9 NaN 9 9 9 NaN 9 9 9 NaN 9 (:,:,3) = 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
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Steve Francis il 25 Mag 2022
Thanks very much for your quick response. I wondered if there was any neat trick to do it by array multipication but your way works well.
Steve Francis il 16 Lug 2022
I've changed the orientation of the matrix and was hoping that you could advise on the revised repmat statement, please?
A = 9* ones (4,3,6); % create a 3 dimensional array
B = A;
The desired output is now
val(:,:,1) =
9 9 9
NaN 9 9
9 NaN 9
9 9 9
val(:,:,2) =
9 9 9
NaN 9 9
9 NaN 9
9 9 9
val(:,:,3) =
9 9 9
NaN 9 9
9 NaN 9
9 9 9
val(:,:,4) =
9 9 9
NaN 9 9
9 NaN 9
9 9 9
val(:,:,5) =
9 9 9
NaN 9 9
9 NaN 9
9 9 9
val(:,:,6) =
9 9 9
NaN 9 9
9 NaN 9
9 9 9
The following seems to work:
B(repmat(any(isnan(B),3),[1 1 size(B,3)])) = NaN;
Is that correct?

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