how to remove the zeros inside a 3D matrix?

AA,

I have a 3D matrix

Xroot(z,w,k) (10x33x402).

The problem is, for example Xroot(1,1,:) it contians 400 zeros and just 2 needed values.

and this for most of Xroot(z,w,:).

Xroot(1,1,:)

ans(:,:,1) =
   69.8033
ans(:,:,2) =
   83.5743
ans(:,:,3) =
     0
ans(:,:,4) =
     0
ans(:,:,5) =
     0
     
     etc...
     ______________ % another example 
     Xroot(10,15,:)
     ...
ans(:,:,363) =
     0
ans(:,:,364) =
     0
ans(:,:,365) =
   33.0133
ans(:,:,366) =
   84.2195
ans(:,:,367) =
     0
         etc...

I want to extract the 2 needed values and Xroot to be (10x33x2).

Notes:

[some of Xroot(z,w) contains fully of zeros, as that means there is no solution at that z,w] so, in this case i want them to be two zeros as Xroot (10x33x2).

[Note that, not all Xroot(:,:,1:2) are the needed values, they are can be Xroot(:,:,365:366) like Xroot(10,15,:), but there are always come Consecutivly ]

Answer 1

[Voss]

Here's one way:

% a (slightly) smaller example, to demonstrate
Xroot = zeros(7,15,402);
% with a few non-zero elements in consecutive "pages" (i.e., 3rd dim)
Xroot(1,1,[11 12]) = [5 6];
Xroot(1,2,[201 202]) = [7 8];
Xroot(5,1,[301 302]) = [-9 -10];

[m,n,p] = size(Xroot);
% page_idx is the index in the 3rd dimension where the first
% non-zero element occurs in each (row,column) of Xroot
[~,page_idx] = max(logical(Xroot),[],3)

page_idx = 7×15

11 201 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 301 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

% convert those locations to linear indices
[r,c] = ndgrid(1:m,1:n);
linear_idx = sub2ind([m n p],r(:),c(:),page_idx(:))

linear_idx = 105×1

1051 2 3 4 31505 6 7 21008 9 10

% get those elements and the ones next to them in 
% the next page, m*n elements later;
% reshape result into 2 columns
temp = reshape(Xroot(linear_idx+[0 m*n]),[],2)

temp = 105×2

5 6 0 0 0 0 0 0 -9 -10 0 0 0 0 7 8 0 0 0 0

% reshape to m-by-n-by-2 for final result
Xroot = reshape(temp,m,n,[]);

format compact
disp(Xroot);

(:,:,1) =
     5     7     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
    -9     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
(:,:,2) =
     6     8     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
   -10     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0

[Anas Zh]

thank you a lot

[Voss]

You're welcome!

Regarding your message, please post a new question, and someone may be able to help.

Answer 2

[Chunru]

% Xroot(z,w,k) (10x33x402). 
m1 = 10; m2 = 33; m3 = 402;
m1 = 4; m2 = 5; m3 = 6;

% Generate data
Xroot = zeros(m1, m2, m3); 
for i=1:m1
    for j=1:m2
        Xroot(i, j, randperm(m3, 2)) = rand(1,2);  % 2 nonzeros
    end
end

% Get the index
idx = zeros(m1, m2, 2);
val = zeros(m1, m2, 2);
for i=1:m1
    for j=1:m2
        [v1, i1] = sort(Xroot(i,j,:), 'descend');
        idx(i,j,:) = i1(1:2);
        val(i,j,:) = v1(1:2);
        fprintf('i=%3d  j=%3d   idx: %3d  %3d;   val: %f   %f\n', i, j, idx(i,j,:), val(i,j,:));
    end
end

i=  1  j=  1   idx:   5    2;   val: 0.670144   0.208733
i=  1  j=  2   idx:   2    4;   val: 0.729268   0.418120
i=  1  j=  3   idx:   4    5;   val: 0.644506   0.643362
i=  1  j=  4   idx:   3    6;   val: 0.603974   0.202120
i=  1  j=  5   idx:   6    1;   val: 0.746606   0.455431
i=  2  j=  1   idx:   3    5;   val: 0.530988   0.385374
i=  2  j=  2   idx:   4    5;   val: 0.745114   0.379642
i=  2  j=  3   idx:   1    2;   val: 0.813252   0.653010
i=  2  j=  4   idx:   6    2;   val: 0.880486   0.027366
i=  2  j=  5   idx:   4    6;   val: 0.960762   0.034169
i=  3  j=  1   idx:   2    4;   val: 0.395492   0.358227
i=  3  j=  2   idx:   4    6;   val: 0.206305   0.127420
i=  3  j=  3   idx:   2    5;   val: 0.985977   0.252404
i=  3  j=  4   idx:   2    1;   val: 0.454050   0.113484
i=  3  j=  5   idx:   1    6;   val: 0.848944   0.591683
i=  4  j=  1   idx:   4    6;   val: 0.600984   0.133792
i=  4  j=  2   idx:   3    1;   val: 0.971629   0.004523
i=  4  j=  3   idx:   3    2;   val: 0.840334   0.108039
i=  4  j=  4   idx:   3    4;   val: 0.374167   0.316749
i=  4  j=  5   idx:   4    3;   val: 0.864410   0.842234

[Anas Zh]

Thank you too

Answer 3

[DGM]

Assuming that each dim3 vector contains either 2 or 0 nonzero elements:

% an example array
A = cat(3,[0 0 0; 0 1 0; 5 0 0], ...
    [0 3 0; 0 2 0; 6 0 0], ...
    [0 4 0; 0 0 0; 0 0 0])

A = 
A(:,:,1) =

     0     0     0
     0     1     0
     5     0     0


A(:,:,2) =

     0     3     0
     0     2     0
     6     0     0


A(:,:,3) =

     0     4     0
     0     0     0
     0     0     0

% get nonzero elements, rearrange
Ar = nonzeros(permute(A,[3 1 2]));
Ar = reshape(Ar,2,[]).'

Ar = 3×2

5 6 3 4 1 2

% construct output array
hasroots = repmat(any(A,3),1,1,2);
B = zeros([size(A,1) size(A,2) 2]);
B(hasroots) = Ar

B = 
B(:,:,1) =

     0     3     0
     0     1     0
     5     0     0


B(:,:,2) =

     0     4     0
     0     2     0
     6     0     0

There are probably other ways as well.