Counting number of conflicting signs between two vectors

Question

I have two vectors (actually 1xN matrices) that have numbers between [-1, 1]. I want to find the number of instances where the sign of two corresponding elements is not the same (sign(A[k]) ~= sign(B[k])). Is there a way to do this that's more efficient than just iterating over the two vectors?

@LuisMendo - One more for the hell of it. This is rather obfuscated: `A = (A + 1)/2; B = (B + 1) / 2; C = accumarray([A(:)+1 B(:)+1], 1); out = C(2) + C(3);` Nowhere near as efficient as your's and Divakar's, but just another way of completely blowing up this comments stream. — rayryeng, Feb 19 '15 at 22:23
@rayryeng With `accumarray`! Hahaha, it would have never occured to me :-) — Luis Mendo, Feb 19 '15 at 22:26
@LuisMendo - hehehe I know. Basically what I'm doing is I'm tallying up combinations of `[-1,1], [1,-1], [-1,-1]` and `[1,1]` between `A` and `B` then summing up the occurrences of `[-1,1]` and `[1,-1]` together. — rayryeng, Feb 19 '15 at 22:28
@rayryeng when you have a big hammer, everything looks like a nail! — David, Feb 19 '15 at 22:43
@David - ahahaha that's really funny. That was my laugh of the day. Thank you :) — rayryeng, Feb 19 '15 at 22:46

score 0 · Accepted Answer · edited Jun 20 '20 at 09:12

Both solutions posted in the comments work equally well for non-time-critical computations:

sum(sign(A) ~= sign(B)) -- Divakar

If you're feeling playful: nnz(A.*B<0) -- Luis Mendo

But the non-playful version sum(sign(A) ~= sign(B)) appears to be faster, presumably because it avoids multiplications involved in A.*B. I compared them like this:

Reps = 1e7; 
n = 10;
tic;
for i=1:Reps
  A = rand(1,n)-.5;
  B = rand(1,n)-.5;
  differ = sum(sign(A) ~= sign(B));   % or nnz(A.*B<0);
end
toc;

For the sum-of-signs version, computation time was 18.95 seconds; for the A.*B version, it was 47.70seconds.

Counting number of conflicting signs between two vectors

1 Answers1