Efficient way to find mirror entries

Hello,

I have a matrix A with N rows. Any arbitrary row i=1,...,N looks like [a b xi], where a,b are positive integers.

For my purposes, for any two rows i,j, if xi=xj, then [a b xi] is equivalent to [a b xj] and [b a xj] and I want to keep only one of them. Note that the in the last example a and b is 'flipped'.

I can easily screen out the cases [a b xi], [a b xj]

[~,loc,~] = unique(A,'rows');
A = A(loc,:);

This is very quick.

However, then I take care of [a b xi], [b a xj] by

k=0;
temp = zeros(size(A));
for i = 1:size(A,1)
    pass = 1;
    %discard nodes at the same x, but reverse orientation
    for j=1:i       
        if j~=i && ...
                ~isequal(A(i,3),A(j,3)) && ...
                isequal(A(i,1:2),fliplr(A(j,1:2)))
            pass = 0;
        end
    end
    if pass ==1
        k = k+1;
        temp(k,:) = A(i,:);        
    end
end
A = temp(1:k,:);

Which is painfully slow (A is quite large).

Can someone suggest a quicker way to do this?

Thank you very much!

Answer 1

[Guillaume]

A = unique(A, 'rows')

would be even faster than going through loc.

Since the order of a and b does not matter, then the simplest way to remove all duplicates is probably:

unique([sort(A(:, [1 2]), 2), A(:, 3)], 'rows')

[Adam]

Excellent!