You could combine these two lines into one, but I left them separate to make the code more readable:
m = [3;5;6;1;2;8;5;2;9;1;2;7;8;3;4;9;3];
r1 = [5;1;9;3];
r2 = [6;4];
r = [r1; r2];
ml = logical(prod(bsxfun(@ne, r', m),2));
mm = m(ml);
The bsxfun call creates matrices out of ‘r'’ (note transpose) and ‘m’ that are now the same size, then does element-by-element ‘not equal’ operations on them. This creates a logical matrix of (length(m) x length(r)). The prod call then treats the logical values as numeric, multiplies them row-wise to create a column vector with 1 values where the condition is met. It then converts this back to a logical vector to create ‘mm’. The setdiff function would not work here because you want repeated elements, and setdiff does not offer that option. This code does.
EDIT — You changed the question adding ‘r1’ and ‘r2’ (originally just ‘r’) while I was answering this. I updated my answer to reflect that. (The essential code didn’t change.)
