I tried to come up with a vectorised solution but could not.
My non-vectorised solution:
A=[2 3 5 7; 1 4 8 9; 3 4 9 7];
B=[2; 8; 11];
C = zeros(size(B));
for k1 = 1:size(B,1)
T = min(A(k1,(find(A(k1,:) > B(k1)))));
if ~isempty(T)
C(k1) = T;
end
end
The ‘C’ (result) vector is the output of the script.
It produces:
A vectorised solution is not always more efficient than a loop. This one is a bit more efficient than your posted solution (it eliminates one loop and vectorises the comparison step), so it will speed things up a bit.
Best Answer