如果我们选择常量预测logloss，那么二元分类器的f=0.5是多少？
这个等式的发展是否正确？
logloss = (1/N) * sum(-y*log(f)-(1-y)*log(1-f))
logloss(y=0) = -(1-0)*log(1-f) = -log(f)
logloss(y=1) = -1*log(f) = -log(f)
// That is either ground true y is 0 or 1 the cost is -log(f)
logloss = (1/N) * N * (-log(f)) = -log(f) = -log(5) = 0.69315