How to delete a Matrix in a 3D Matrix?

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]

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

Answer 1

[David Hill]

idx=[];
for k=1:size(T5,3)
    if any(isnan(T5(:,:,k)),'all')
        idx=[idx,k];
    end
end
T5(:,:,idx)=[];

Answer 2

[Sulaymon Eshkabilov]

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]

"... 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.

Answer 3

[Steven Lord]

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