robot520 opened this issue 8 years ago · 0 comments
Your code：z = y .* (res.x3 - 1); My codes： idx_pos = (pos==1); idx_neg = (neg==1); z = max(0,1-res.x3.*idx_pos)+max(0,res.x3.*idx_neg); My Question: Is res.x3-1 suitable for computing the loss of negative examples?