How to build on an existing struct 4x4xn matrix?

I have an struct, for example...

A = struct();

where we have a 4x4 matrix

A.G = rand(4,4);

I want to build on values within the (4x4) matrix but stack new values in a n'th dimension.

How do I build on the maxtrix in the n'th dimesnion if I don't know how many non-zero values exist in respective field A.G(1,1,n), A.G(1,2,n), A.G(2,2.n) and A.G(4,4,n). the n't value can stack randomly. I wan't to know how many non-zero values lies behind for example A.G.(1,2,n) where n is a unkown value depending how many values i already put into A.G(1,2,n), which i don't keep track on.

[Rik]

To you have a struct array, or is your variable simply stored in A.G?

And what you want to know? Neither your input nor your desired output is clear to me.

[Happy PhD]

I have an 2x4x3 matrix for example

A.G(:,:,1) = [ 1 2 0 1 ; 4 9 5 6 ];

A.G(:,:,2) = [ 0 1 0 0; 0 6 0 0];

A.G(:,:,3) = [0 5 0 0; 0 0 0 0 ];

I want to add an value to for example A.G(2,2,n) or A.G(1,1,n) or A.G(1,2,n) but I don't know how many values we have in respective X and Y position in n'th dimension, where coordinate system is the 1'st dimension is A.G(X,Y,1) = [ (x1,y1) (x1,y2) (x1,y3) (x1,y4); (x2, y1) (x2,y2) (x2,y3) (x2,y4)]; if A.G(:,:,1) is the 1st dimension, A.G(:,:,2) is the 2nd dimesnion and A.G(:,:,3) is the 3rd dimesnion etc... So that means ...A.G(1,2,1) = 2 and A.G(2,2,1) =9 and A.G(2,2,2) = 6.. etc...

Hopefully I am making sense now. Note 0 is a non-value here.

I want to for example add a value to A.G(1,2,4) = 8 or A.G(1,1,2) = 8 .. but I don't know how many values are in A.G(1,2,n) or A.G(1,1,n) where the n-value is unkown. How do I add next value there. Note 0 is a non-value here.

[Mrutyunjaya Hiremath]

Hello Happy PhD,

do you want to add next value where there are 0's are there?

[Happy PhD]

That is correct! I want to add values after all non-zero values. And increase the matrix size in the third dimension when needed.

Answer 1

[Mrutyunjaya Hiremath]

Hello Happy PhD,

Try this one...

clc;
clear all;

A.G(:,:,1) = [1 2 1 1 ; 4 9 5 6 ];
A.G(:,:,2) = [0 1 0 0; 0 6 0 0];
A.G(:,:,3) = [0 5 0 0; 0 0 0 0];

N = 10; % random number 

idx = find(A.G == 0);
if ~isempty(idx)
    A.G(idx(1)) = N;
else
    n = size(A.G,3);
    A.G(1,1,n+1) = N;
end
disp(A.G);

Answer 2

[Rik]

It sounds like you can just assign that value. Matlab will expand the array and fill empty positions with 0.

A.G=cat(3,[ 1 2 0 1 ; 4 9 5 6 ], [ 0 1 0 0; 0 6 0 0],[0 5 0 0; 0 0 0 0 ]);
A.G(1,2,4)=8;%Matlab fills the 4th page with zeros
disp(A.G)

[Happy PhD]

In my application I don’t know how my matrix looks like, so I don’t know I can add a value in A.G(1,2,4) -where 4 is unknown to me. But I want add a number in A.G(1,2,n) where n is the next available position.

It might be zeroes here .. I want to add the next value after the next non-zero value in that dimensional ‘column‘.

[Rik]

You can find the first empty page like this:

[r,c,val]=deal(1,2,8);

A.G=cat(3,[ 1 2 0 1 ; 4 9 5 6 ], [ 0 1 0 0; 0 6 0 0],[0 5 0 0; 0 0 0 0 ]);
N_by_rc=sum(logical(A.G),3)+1;
n=N_by_rc(r,c);
A.G(r,c,n)=val;

clc,disp(A.G)