If ‘M’ is your matrix, this will work:
idx = any(bsxfun(@eq, M(:,1), (M(M(:,3)==-1,1))'),2);
Mout = M(~idx,:);
The ‘M(M(:,3)==-1,1)’ checks column 3 for -1 values, and returns the corresponding values in column 1. The bsxfun call searches for elements of column 1 that are equal to those values, and returns a matrix of column vectors of logical indices for each equality. The any call will be 1 if any of the column vectors are 1 in all rows, creating a single column logical vector from the matrix of logical vectors. Then the ‘Mout’ assignment returns all the rows that are not 1, producing the desired output.
Also, you can avoid retyping the ‘idx’ assignment with your actual array name by creating an anonymous function from it, so that if you call your array ‘A’:
idxfcn = @(M) ~any(bsxfun(@eq, M(:,1), (M(M(:,3)==-1,1))'),2);
idx = idxfcn(A);
Result = A(~idx,:);
producing the desired result.
