a function with increasing number of its arguments in a loop

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mehdi J 2018-10-27

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https://ww2.mathworks.cn/matlabcentral/answers/426392-a-function-with-increasing-number-of-its-arguments-in-a-loop

编辑： Stephen23 2018-10-27

hi dears,

I'm using a function for generating Cartesian products of two or more vectors and I want the number of its arguments increases in each iteration of a loop (the vectors as its arguments are same). For instance in the first iteration of the loop the function calculates the Cartesian(A,A), in the second iteration calculates Cartesian(A,A,A), in the next iteration calculates Cartesian(A,A,A,A), and so on. How could I do it? Thanks

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Kevin Chng 2018-10-27

Have you tried out any code? Do you mind to share to us?

mehdi J 2018-10-27

在 MATLAB Online 中打开

a code has been shared on this link and it is as below. I want to use it in a loop and increase its size in each iteration of the loop

------
function C = cartesian(varargin)
  args = varargin;
  n = nargin;
    [F{1:n}] = ndgrid(args{:});
    for i=n:-1:1
        G(:,i) = F{i}(:);
    end
    C = unique(G , 'rows');
end

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Answer 1

Stephen23 2018-10-27

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https://ww2.mathworks.cn/matlabcentral/answers/426392-a-function-with-increasing-number-of-its-arguments-in-a-loop#answer_343709

编辑：Stephen23 2018-10-27

在 MATLAB Online 中打开

Use a comma-separated list:

N = ... number of iterations
C = cell(1,N); % output array
for k = 1:N
    T = repmat({A},1,1+k);
    C{k} = Cartesian(T{:});
end

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mehdi J 2018-10-27

Thanks, but how could I put all of them (all of the cells) in one matrix (below each other)?

Stephen23 2018-10-27

编辑：Stephen23 2018-10-27

在 MATLAB Online 中打开

@Mhedi J: each of the Cartesian products has a different number of rows and columns: this means there is no simple way to concatenate them into one matrix. The simplest solution is to leave them in a cell array, otherwise you would have to pad them with NaN or some other value/s and then concatenate them into one matrix.

Here is a simple example of how to use a cell array:

A = 1:3;
N = 4;
C = cell(1,N);
for k = 1:N;
    T = repmat({A},1,1+k);
    [T{:}] = ndgrid(T{:});
    T = cellfun(@(v)v(:),T,'uni',0);
    C{k} = [T{:}];
end