MATLAB: How to find the minimum number of row and column and concatenate to a matrix

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Hello.
When I used this code, I got the error.
This is because the cell of Phi_NuMax has different matrix.
Could you explain how to Find the minimum number of rows and column and concatenate it?
1 cell : 100 x 144
2 cell : 95 x 144
…….
I want to match the matrix as same size of rows and column.
        Phi_NuMax{i} = (U{i}(:, 1:r{i})*(S{i}(1:r{i}, 1:r{i}).^(1/2)))';
        
        % How to Find the minimum number of row and column of Phi_NuMax{i} to concatenate it ?
       
        % Each Phi_NuMax{i} has different size of rows and columns
        % Each data{i} has same size of rows and columns 
        Y{i} = Phi_NuMax{i} * data{i}; 
        Y{i} = reshape(Y{i},[],1);
    end
    K = cell2mat(Y);

Best Answer

Don't you need the minimum dimension in order to concatenate? In other words, you need to find the max dimensions. Untested code:
minRequiredRows = 0;
minRequiredColumns = 0;
for k = 1 : length(Phi_NuMax)
    thisMatrix = Phi_NuMax{k};
    [thisRows, thisColumns] = size(thisMatrix);
    fprintf('Cell #%d has %d rows and %d columns.\n', k, thisRows, thisColumns);
    % Find out how wide the matrix will need to be if they are all stacked on top of one another.
    minRequiredColumns = max(minRequiredColumns, thisColumns);
    % Sum up how many rows we'll need to hold all of them vertically.
    minRequiredRows = minRequiredRows + thisRows; 
end
fprintf('At a minimum, you will require %d rows and %d columns.\n', k, minRequiredRows, minRequiredColumns);
% Now instantiate a matrix for all the individual matrices stitched together vertically.
allMatrixes = zeros(minRequiredRows, minRequiredColumns);
% Now concatenate by inserting the cell into all Matrixes at the appropriate row number.
currentRow = 1;
for k = 1 : length(Phi_NuMax)
    thisMatrix = Phi_NuMax{k};
    [thisRows, thisColumns] = size(thisMatrix);
    fprintf('Inserting cell #%d with %d rows and %d columns.\n', k, thisRows, thisColumns);
    allMatrixes(currentRow : currentRow + thisRows - 1, 1:thisRows) = thisMatrix;
    % Move the pointer to the next place where we will insert the next matrix.
    currentRow = currentRow + thisRows;
end
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