How to loop through an unknown number of matrix dimensions?

I would like to loop through all dimensions of a n-dimensional matrix, but the matrix dimensions are unknown beforehand. Below is an example of generating a matrix of unknown dimensions, if it was a 2D matrix looping over those 2 dimensions. But in fact it could have 1-5 dimensions. Can you think of a good solution?

inputMatrixOfUnknownSize = ones(factor(round(rand(1)*500))); % Example matrix that could have 1 to many dimensions
dimList = size(inputMatrixOfUnknownSize) % List the dimensions
% If it was 2D, here's how we'd loop over those two dimensions...but it isn't necessarily 2D!
for iDim1 = 1:dimList(1)
  for iDim2 = 1:dimList(2)
     % Operate on each element. In reality the operation will not be the same for all dimensions. 
     outputMatrixOfUnknownSize(iDim1,iDim2) = outputMatrixOfUnknownSize(iDim1,iDim2)^2;
  end  
end

In my real code, the operation that is performed on each matrix element will depend on the dimension size, so I can't eliminate the loop in the example above.

Answer 1

[Jan]

You can use linear indexing, if you want to operate on all elements:

X = rand(2,3,4);
Y = zeros(size(X));
for k = 1:numel(X)
  Y(k) = X(k) ^ 2;
end

Of course Y = X .^ 2 would be easier without a loop, but this is thought as demonstration only.

If you want to operate on submatrices, you can move them to the first dimensions temporarily:

X = rand(2,3,4,5,6);
operateOn = [3,5];  % Operate on submatrix along 3rd and 5th dimension

v   = 1:ndims(X);
XX  = permute(X, [operateOn, setdiff(v, operateOn)]);
sXX = size(XX);
XX  = reshape(XX, [sXX(1:2), prod(sXX(3:end)]);
for k = 1:size(XX, 3)
   submatrix = XX(:, :, k);
   ... do what you want
end

Another example is using a cell vector for indexing. If you want a subarray with indexing the the last two dimensions - independent from knowing how many dimensions the input has:

X = rand(2,3,4,5,6);
n = ndims(X);
index        = cell(1, n);
index(:)     = {':'};
index{end-1} = 3;
index{end}   = 4;
Y = X(index{:});  % Equivalent to: X(:, :, :, 3, 4)

You can replace a bunch of nested loops by a single loop also using an index vector instead of scalar indices, see Answers: 333926-recursive-function-for-replacing-multiple-for-loops :

X = rand(2,3,4,5,6);

nv   = ndims(X);
v    = ones(1, nv);
vLim = size(X);

ready = false;
while ~ready
  % Do what you need with X and the index vector v
  ...

    % Update the index vector:
    ready = true;       % Assume that the WHILE loop is ready
    for k = 1:nv
      v(k) = v(k) + 1;
      if v(k) <= vLim(k)
        ready = false;  % No, WHILE loop is not ready now
        break;          % v(k) increased successfully, leave "for k" loop
      end
      v(k) = 1;         % Reset v(k), proceed to next k
    end
  end

This is equivalent to the following without the need to know the number of dimensions before:

for i1 = 1:size(X, 1)
  for i2 = 1:size(X, 2)
    for i3 = 1:size(X, 3)
      for i4 = 1:size(X, 4)
        for i5 = 1:size(X, 5)
          v = [i1, i2, i3, i4, i5];
          % Do what you need with X and the index vector v
          ...
        end
      end
    end
  end
end

See an example: https://www.mathworks.com/matlabcentral/answers/447469-looping-through-a-matrix-of-unknown-dimensions

[Stephen23]

+1 very good use of permute

[KAE]

Very educational. Thanks.