How to find the common values of ith row and the rest of the rows in a matrix?

What I want to do is to find the common elements between ith row and the rest of the rows.

Assume that we have the following matrix; A = [1 5 6 7 8 9 ; 2 4 5 3 6 11; 3 4 1 12 11 6; 4 5 6 12 13 14];

The first values of the first column is the ID, and the rest of the line of the each column is the values of each ID.

For i = 1, for instance, I need to find the common values of (1,2), (1,3) and (1,4).

ComVal(1,2) = 5; ComVal(1,3) = 6 ComVal(1,4) = 5,6

By the same token, the common values of other IDs should be found, which are:

ComVal(2,1) = 5; ComVal(2,3) = 4,6; ComVal(2,4) = 5,6;

ComVal(3,1) = 6; ComVal(3,2) = 4,6 ComVal(3,4) = 6,12

ComVal(4,1) = 5,6 ComVal(4,2) = 5,6 ComVal(4,3) = 6,12

Is there any possibility that I can do it without using a for loop?

Thanks in advance for the answers. Samet

Answer 1

[Azzi Abdelmalek]

A = [1 5 6 7 8 9 ; 2 4 5 3 6 11; 3 4 1 12 11 6; 4 5 6 12 13 14];
n=size(A,1);
m=1:n;
for k=1:n
  p=setdiff(m,k);
  M=A(p,:);
  R=ismember(M,A(k,:));
  out{k}=arrayfun(@(x)  M(x,R(x,:)),(1:n-1)','un',0);
end 
celldisp(out)

[Sam]

It looks really good,with one minor problem. it also finds the first row, the IDs, as common values. We need to find the common values by comparing 2:size(A,1).

Answer 2

[Geoff Hayes]

If the idea is to find elements that are common to each row then you could try the following

 % get the number of rows of A
 [m] = size(A,1);

 % pre-allocate memory to the cell output matrix (which is symmetric)
 cellMtx = cell(m,m);

 for u=1:m
    for v=u+1:m
        % determine the intersection between the two rows
        cellMtx{u,v} = intersect(A(u,2:end),A(v,2:end));
        cellMtx{v,u} = cellMtx{u,v};
    end
 end

Note that elements along the diagonal of cellMtx are empty as the above code skips those comparisons.

[Sam]

My code is almost same as what you wrote here. The problem is the data I have is really big, and using two for loops makes it too heavy. Apart from that code works perfectly fine.