Covert a cell to multidimensional matrix

I have a 50 X 1 cell array, A, and each cell contains an m x 100 matrix, where m is variable and not the same for each of the cells. I would like to convert this cell array to a 3D matrix, Anew(M,100,50), where M is the first dimension of the matrix with biggest rows, and pad zero for the small matrices. I tried cell2mat but it concatenated the cell into a 2D matrix ([...],100). I also tried:

for ii=1:50
Anew(:,:,ii)=A{ii};
end

but I got

Subscripted assignment dimension mismatch.

error. Does anyone know how it can be done?

Thanks

This is the first 6 cells of A:

A = 

      [ 44x100 double]
      [ 80x100 double]
      [103x100 double]
      [ 96x100 double]
      [ 94x100 double]
      [ 92x100 double]
  .
  .
  .

Answer 1

[Azzi Abdelmalek]

%------------Example---------------------------
s=arrayfun(@(x) randi(5,randi(5),100),1:5,'un',0)
%---------------the code-------------------------
n=max(cellfun(@(x) size(x,1),s))
m=size(s{1},2)
a=cellfun(@(x) [zeros(n-size(x,1),m);x],s,'un',0)
out=cat(3,a)

[Orion]

I followed this up to

a=cellfun(@(x) [zeros(...

which makes sense. 'out' is still a 1x5 cell, but I want it to be 4x100x5 matrix. I guess I can figure out how to that. Thanks

[Guillaume]

Azzi made a mistake in the last line, it should be

out = cat(3, a{:});

Please use good variable names in your code, not m, n, a, etc. It makes the code so much easier to maintain:

%input cell array: c
maxheight = max(cellfun(@(m) size(m, 1), c);
resizedmatrices = cellfun(@(m) [m; zeros(maxheight - size(m, 1), size(m, 2)], c, 'UniformOutput', false);
out = cat(3, resizedmatrices{:});

[Orion]

thanks

Answer 2

[Image Analyst]

Try this (untested) intuitive looping method:

Anew = zeros(m, 100, 50); % Initialize
for thisCellIndex = 1 : 50
    % Extract contents of this particular cell.
     % It will be a 1 x 100 row vector.
    thismx100Array = A{thisCellIndex};
    finalZ = min([size(Anew, 3), length(thismx100Array )]);
    % Put this row vector into a column vector along the "Z" dimension.
    for z = 1 : finalZ 
        Anew(1:m, 1:100, ii) = thismx100Array(z);
    end
end

It's hard to do much more without your actual data.

[Orion]

m is changing in each cell. but I think to use your suggestion, it should be the biggest first index of the matrices.