Determine the number of "for" loops by the user

Hi everyone

I want to write a code that takes the number of uncertain parameters from the user and creates a "for" loop according to their number and performs the calculations.

for example:

The user enters the number 4 as the number of uncertainty parameters and the code puts 4 "for" loops in a row. This way :

for i=1:I
    for j=1:J
        for k=1:K
            for r=1:R
                "calculations"
            end
        end
    end
end

Is this possible in MatLab?

[Wan Ji]

Use while instead

[DGM]

How does while address the question? The number of iterations is known; the number of nested loops is not.

[f4r3in]

Let me explain the problem further:

In the problem, the values I, J, K, R are known (for example all are equal to 7). And the calculation that is done in "for" loops is as follows: (A1 : A4 are also known)

"calculations"
Scen(:,l) = [A1(:,i);A2(:,j);A3(:,k);A4(:,r)];
l = l+1;

The only unknown issue is the number of "for" loops.

How can we implement this?

[KSSV]

I think you can avoid using loops.....can you tell us what are those calculations?

I,J,K,R they will be always same?

Answer 1

[Chunru]

% The number of iterations
ni = [3 4 2 2];         % 4 iterations [I J K R]; i is innerest loop 
idx = ones(size(ni));   % first index 

k = cumprod(ni);
for i=1:prod(ni)
    % calculate index (ignore this part if calculation independent of index)
    ii = i;
    for j=length(ni):-1:2
        ir = rem(ii-1, k(j-1)) + 1;
        idx(j) = (ii - ir) /k(j-1) + 1;
        ii = ir;
    end
    idx(1) = ii;
    disp(idx)
    
    % do calculation depending on idx
end

     1     1     1     1

     2     1     1     1

     3     1     1     1

     1     2     1     1

     2     2     1     1

     3     2     1     1

     1     3     1     1

     2     3     1     1

     3     3     1     1

     1     4     1     1

     2     4     1     1

     3     4     1     1

     1     1     2     1

     2     1     2     1

     3     1     2     1

     1     2     2     1

     2     2     2     1

     3     2     2     1

     1     3     2     1

     2     3     2     1

     3     3     2     1

     1     4     2     1

     2     4     2     1

     3     4     2     1

     1     1     1     2

     2     1     1     2

     3     1     1     2

     1     2     1     2

     2     2     1     2

     3     2     1     2

     1     3     1     2

     2     3     1     2

     3     3     1     2

     1     4     1     2

     2     4     1     2

     3     4     1     2

     1     1     2     2

     2     1     2     2

     3     1     2     2

     1     2     2     2

     2     2     2     2

     3     2     2     2

     1     3     2     2

     2     3     2     2

     3     3     2     2

     1     4     2     2

     2     4     2     2

     3     4     2     2

[f4r3in]

Thank you very much for your answer.