Loop over combinations of elements of several vectors

In the simplest case, if I have vectors , I want to loop over all the possible combinations of their elements, that is, over I wat to do the same thing in a general case where I have and arbitrary number of vectors with respective dimensions , with denoting the k-th component of vector ; that is, I want to loop over all the combinations of elements . How to do this for a general number of vectors? For now, the only thing I came up with is to make nested loops for each vector, but I want some code whose structure does not depend on the number of vectors.

Thanks for your time!

Answer 1

[Dyuman Joshi]

Use ndgrid -

x1 = 1:5;
x2 = [2 3 5 7 11];
x3 = [2 4 6 8 10];

%Store vectors in a cell array
x = {x1,x2,x3};

n = numel(x);
%Preallocate
C = cell(1,n);

%Reverse order to get the proper order when concatenating
[C{end:-1:1}] = ndgrid(x{end:-1:1});

%Concatenate and reshape the data corresponding to number of vectors
out = reshape(cat(n,C{:}),[],n);
disp(out)

     1     2     2
     1     2     4
     1     2     6
     1     2     8
     1     2    10
     1     3     2
     1     3     4
     1     3     6
     1     3     8
     1     3    10
     1     5     2
     1     5     4
     1     5     6
     1     5     8
     1     5    10
     1     7     2
     1     7     4
     1     7     6
     1     7     8
     1     7    10
     1    11     2
     1    11     4
     1    11     6
     1    11     8
     1    11    10
     2     2     2
     2     2     4
     2     2     6
     2     2     8
     2     2    10
     2     3     2
     2     3     4
     2     3     6
     2     3     8
     2     3    10
     2     5     2
     2     5     4
     2     5     6
     2     5     8
     2     5    10
     2     7     2
     2     7     4
     2     7     6
     2     7     8
     2     7    10
     2    11     2
     2    11     4
     2    11     6
     2    11     8
     2    11    10
     3     2     2
     3     2     4
     3     2     6
     3     2     8
     3     2    10
     3     3     2
     3     3     4
     3     3     6
     3     3     8
     3     3    10
     3     5     2
     3     5     4
     3     5     6
     3     5     8
     3     5    10
     3     7     2
     3     7     4
     3     7     6
     3     7     8
     3     7    10
     3    11     2
     3    11     4
     3    11     6
     3    11     8
     3    11    10
     4     2     2
     4     2     4
     4     2     6
     4     2     8
     4     2    10
     4     3     2
     4     3     4
     4     3     6
     4     3     8
     4     3    10
     4     5     2
     4     5     4
     4     5     6
     4     5     8
     4     5    10
     4     7     2
     4     7     4
     4     7     6
     4     7     8
     4     7    10
     4    11     2
     4    11     4
     4    11     6
     4    11     8
     4    11    10
     5     2     2
     5     2     4
     5     2     6
     5     2     8
     5     2    10
     5     3     2
     5     3     4
     5     3     6
     5     3     8
     5     3    10
     5     5     2
     5     5     4
     5     5     6
     5     5     8
     5     5    10
     5     7     2
     5     7     4
     5     7     6
     5     7     8
     5     7    10
     5    11     2
     5    11     4
     5    11     6
     5    11     8
     5    11    10

Answer 2

[Chunru]

You can use combinations.

% arbitrary number of vectors with arbitrary size
v{1}=rand(3,1);
v{2}=rand(2,1);
v{3}=rand(2,1);

p =table2array(combinations(v{:}))

p = 12×3

0.0085 0.6301 0.2202 0.0085 0.6301 0.6508 0.0085 0.7299 0.2202 0.0085 0.7299 0.6508 0.8834 0.6301 0.2202 0.8834 0.6301 0.6508 0.8834 0.7299 0.2202 0.8834 0.7299 0.6508 0.4095 0.6301 0.2202 0.4095 0.6301 0.6508

for i=1:size(p,1)       % only one loop instead of nested loop
    p(i, :)  % this is the combination and do whatever you want in the loop
end

ans = 1×3

0.0085 0.6301 0.2202

ans = 1×3

0.0085 0.6301 0.6508

ans = 1×3

0.0085 0.7299 0.2202

ans = 1×3

0.0085 0.7299 0.6508

ans = 1×3

0.8834 0.6301 0.2202

ans = 1×3

0.8834 0.6301 0.6508

ans = 1×3

0.8834 0.7299 0.2202

ans = 1×3

0.8834 0.7299 0.6508

ans = 1×3

0.4095 0.6301 0.2202

ans = 1×3

0.4095 0.6301 0.6508

ans = 1×3

0.4095 0.7299 0.2202

ans = 1×3

0.4095 0.7299 0.6508

[Dyuman Joshi]

Just a note - combinations is available from R2023a onwards.