How do I store the values of a for loop in a matrix array?

I am in need to use a delta function and to preserve the output values into different matrix arrays.

I defined the delta function, , for a 64 x 64 matrix as:

function [d] = delta2(u, u0, v, v0)
d=0; 
if u - u0 ==0 && v -v0 ==0 
    d=1;
end 

The intention is for the programme to analyse every (u,v) point for a centre point (u0, v0). This means, to start at (u,v) = (0,0) and move throught the square image until finishing at (u,v) =(63,63). Once this is done, the programme should restart for a different (u0,v0) value until reaching (u0,v0) = (63,63).

The purpose is to obtain a different matrix of 0s and a single 1 for every (u0,v0) value.

My attempt so far was:

for u0 = 1:64
    for v0 = 1:64
        for u = 1:64
            for v = 1:64
                mat = delta2(u, u0, v, v0);
            end
        end
    end
end

The problem is that the values obtained in the loop are not being stored, because the output for mat is always a single value of 1 or 0, and as stated above I wanted a matrix per (u0,v0) value.

How can I make the code run for every (u0,v0) value in order (I mean start at (0,0) , (0,1), .... and end up at (64,64))?

How do I store the results in a 64 x 64 matrix array?

Answer 1

[Jan]

One solution might be indexing the output

...
    mat(u, u0, v, v0) = delta2(u, u0, v, v0);
...    

But this can be simplified. At first the function delta():

function d = delta2(u, u0, v, v0)
d = double((u == u0) && (v == v0));
end

But the main loops can be omitted also:

v = 1:64;
m = (v == reshape(v, [1, 1, 1, 64]) & ...
     v.' == reshape(v, [1, 1, 64]));

As side effect this needs 0.05 seconds on my computer (i5, Matlab R2018b) instead of 1.08 seconds of the original loop.

[Goncalo Costa]

But this won't help me with the problem of creating an individual 64 x 64 matrix for every (u0,v0) value, at least it hasn't for me....how would you do that?

Thank you so much for your help so far. Have a great day.

[Jan]

"problem of creating an individual 64 x 64 matrix for every (u0,v0) value"

what exactly is your problem? My code produces the same output as yours.

"The problem is that the values obtained in the loop are not being stored."

And with my codes the values are stored. So please explain on a short example with e.g. 1x4 vectors, what you want to achieve instead.

[Goncalo Costa]

Thanks for the help, what I intend to attain is, in the case of a 4 x 4 matrix (more suitable than 1x4 for the example). If (u0, v0) = (0,0) , the matrix would look something like:

I intend to get a matrix for every different (u0,v0) value. I hope this makes more sense. Thanks once more.

[Jan]

@Goncalo Costa: You do get a set of such matrices with my code:

v = 1:4;
m = (v == reshape(v, [1, 1, 1, 4]) &  v.' == reshape(v, [1, 1, 4]));
% Test:
m(:, :, 1, 1)

ans = 4×4 logical array
   1   0   0   0
   0   0   0   0
   0   0   0   0
   0   0   0   0

m(:, :, 1, 2)

ans = 4×4 logical array
   0   1   0   0
   0   0   0   0
   0   0   0   0
   0   0   0   0

and so on. These matrices are stored in a 4D array efficiently. An alterative, which takes more RAM is to store them in a cell array:

n = 4;   % Or 64, as you want
v = 1:n;
m = (v == reshape(v, [1, 1, 1, 4]) &  v.' == reshape(v, [1, 1, 4]));
C = squeeze(num2cell(m, [1,2]));
C{1,1}
C{1,2}  % Same as above

[Goncalo Costa]

Thank you, I will give this ago again, I struggled to understand you omitted the main loop and so I struggled to comprehend its use. I will give this a second go. Thank you so much.