Comments (2)
感觉这样可能不太对吧。是不是可以在sigmoid里面加一个权重? sigmoid( a * x)
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请教下,时长的loss=-[(t/1+t)*log(p) + (1/1+t)log(1-p)] ,这个应怎么理解呢?
我想到的2种理解方式:
1.未点击的样本,t=0,走后面这部分 (1/1+t)log(1-p) = 1log(1-p)
有点击的样本,走前面这部分(t/1+t)log(p) < 1log(p) => 带来的问题是负样本的loss权重均高于正样本,且正样本间的loss权重区分性不大
2.未点击的样本,t=0,带入上述完整公式推出 = 1log(1-p)
点击的样本,t>0,走上述完整公式,那正样本受2部分loss的影响
请问是哪种呢理解?
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