# How to delete a Matrix in a 3D Matrix?

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Karoline Peters on 18 May 2022
Answered: Steven Lord on 19 May 2022
Hello,
I would like to delete a Matrix in my 3D Matrix if one element in that Matrix is NaN.
Size of T5 = 4x4x30
For example if Matrix 4x4x1 looks like this:
val(:,:,1) =
NaN NaN NaN NaN
NaN NaN NaN NaN
NaN NaN NaN NaN
0 0 0 1
I want to delete the whole Matrix.
My code does not work
[a b c] = size(T5)
for i = i:c
if isnan(T5(:,:,i))
T5(:,:,i) = [];
end
end
What am I doing wrong?
dpb on 18 May 2022
if any(isnan(T5(:,:,i)))
...
See the doc on explanation of how logical if works for arrays -- in short it returns true only if all elements are true.
You can do the above w/o any looping in MATLAB with logical addressing---
ip=find(isnan(x)); % the linear indices in x containing NaN
if ~isempty(ip)
[~,~,ip]=ind2sub(size(x),ip); % any plane with at least one NaN
x(:,:,unique(ip))=[]; % remove those planes
end

David Hill on 18 May 2022
idx=[];
for k=1:size(T5,3)
if any(isnan(T5(:,:,k)),'all')
idx=[idx,k];
end
end
T5(:,:,idx)=[];

Sulaymon Eshkabilov on 18 May 2022
You are trying to change the size of the matrix that can't be achieved in this way. If you need to find and substitute NaNs with some values, then it is much better to use this approach, e.g.:
clear A
A(:,:,1) = randi([-2, 2], 5);
A(:,:,2) = randi([-2, 2], 5);
A(:,:,3) = randi([-2, 2], 5);
A=A/0;
disp(A)
(:,:,1) = NaN -Inf Inf Inf NaN Inf Inf -Inf -Inf Inf NaN NaN -Inf Inf NaN -Inf -Inf NaN Inf Inf NaN -Inf Inf -Inf -Inf (:,:,2) = -Inf Inf Inf -Inf Inf -Inf Inf NaN Inf Inf -Inf -Inf Inf NaN Inf Inf Inf Inf Inf NaN -Inf Inf Inf NaN -Inf (:,:,3) = NaN -Inf NaN NaN NaN NaN -Inf NaN Inf Inf -Inf -Inf NaN -Inf NaN Inf Inf -Inf Inf -Inf Inf Inf Inf NaN Inf
%
[R C L] = size(A)
R = 5
C = 5
L = 3
for ii = 1:L
for jj = 1:R
for kk = 1:C
if isnan(A(jj,kk,ii))
A(jj,kk,ii) = 0;
else
A(jj,kk,ii) = 13;
end
end
end
end
disp(A)
(:,:,1) = 0 13 13 13 0 13 13 13 13 13 0 0 13 13 0 13 13 0 13 13 0 13 13 13 13 (:,:,2) = 13 13 13 13 13 13 13 0 13 13 13 13 13 0 13 13 13 13 13 0 13 13 13 0 13 (:,:,3) = 0 13 0 0 0 0 13 0 13 13 13 13 0 13 0 13 13 13 13 13 13 13 13 0 13
dpb on 18 May 2022
"... trying to change the size of the matrix that can't be achieved in this way..."
Certainly can delete a plane by setting it's values to [] by logical addressing.
The problem is with the tes, not the replacementt; it only will be true if all elements of the plane are NaN so that part of the code is never executed with the sample data.

Steven Lord on 19 May 2022
Let's make a sample 3-D array with some NaN values.
A = randi(10, 3, 3, 4);
A(randperm(numel(A), 3)) = NaN % Make 3 random values NaN
A =
A(:,:,1) = 1 9 6 7 NaN 7 7 3 NaN A(:,:,2) = 3 2 8 4 10 10 7 3 2 A(:,:,3) = 8 9 4 1 5 7 2 9 4 A(:,:,4) = 5 9 1 3 5 2 9 4 NaN
Identify the pages that contain NaN.
nanpages = any(isnan(A), [1 2]) % Compute any over dimensions 1 and 2
nanpages = 1×1×4 logical array
nanpages(:,:,1) = 1 nanpages(:,:,2) = 0 nanpages(:,:,3) = 0 nanpages(:,:,4) = 1
Now delete.
A(:, :, nanpages) = []
A =
A(:,:,1) = 3 2 8 4 10 10 7 3 2 A(:,:,2) = 8 9 4 1 5 7 2 9 4

R2020b

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