如何在 MATLAB 中为矩阵的列向量生成所有可能的组合？

Question

If B=[1; 2] B=[1; 2] and A=[BB B...(n times B)] , how to obtain the matrix C corresponding to all the possible combinations between the column vectors of A .ie I want to get the combinations between n copies of the same vector.

For example, for n=3:

A =
     1     1     1
     2     2     2

So, C can be obtained using the function from File Exchange 'allcomb(varargin)':

C=allcomb(A(:,1),A(:,2),A(:,3))
C =
     1     1     1
     1     1     2
     1     2     1
     1     2     2
     2     1     1
     2     1     2
     2     2     1
     2     2     2

In my case n is variable. How to obtain C for any value of n?

Answer 1

You can put the repetitions in a cell, and use the {:} syntax to put all cell elements as inputs to allcomb

n = 3;
B = [1,2];
A = repmat( {B}, n, 1 );
C = allcomb( A{:} ); % allcomb  is FileExchange.
                     % combvec  is a documented alternative.

Output:

C =
 1     1     1
 1     1     2
 1     2     1
 1     2     2
 2     1     1
 2     1     2
 2     2     1
 2     2     2

Answer 2

Since the alphabets for each of the places is the same, this is really a base conversion. MATLAB only accepts integer bases, but we can use that integer as an index into the alphabet B :

B=[1; 2];
n = 3;
b = numel(B);

for k = 0:(b^n-1)   % loop over all possible combinations
   C(k+1,:) = dec2base(k, b, n);
end
C = C - '0' + 1;   % convert 0..b-1 (in chars) into 1..b (in ints) for indexing
C = B(C);          % index into alphabet B

Results:

>> C
C =

   1   1   1
   1   1   2
   1   2   1
   1   2   2
   2   1   1
   2   1   2
   2   2   1
   2   2   2

The last line of the script doesn't appear to do much in this case because the alphabet happens to be the same range as our indices, but changing the alphabet to B = [7; 14] B = [7; 14] will correctly result in:

C =

    7    7    7
    7    7   14
    7   14    7
    7   14   14
   14    7    7
   14    7   14
   14   14    7
   14   14   14

Answer 3

Funnily enough, allcomb from MATLAB File Exchange seems to be what you want.

allcomb([1; 2],[1; 2], [1; 2])

ans =

     1     1     1
     1     1     2
     1     2     1
     1     2     2
     2     1     1
     2     1     2
     2     2     1
     2     2     2

To do it for any n , simply construct the matrix with:

>> n = 3;
>> repmat(B, 1, n)

ans =

     1     1     1
     2     2     2