roc.p=pROC::roc(label, pred,ci =T)
if (roc.p$auc>0.5){
cutoff=roc.p$thresholds[which.max(roc.p$sensitivities+roc.p$specificities)]
sensitivity=roc.p$sensitivities[which.max(roc.p$sensitivities+roc.p$specificities)]
specificity=roc.p$specificities[which.max(roc.p$sensitivities+roc.p$specificities)]
ci = roc.p$ci
df=data.frame(type='positive classification',
auc=round(roc.p$auc,3),cutoff=cutoff,
sensitivity=sensitivity,specificity=specificity,ci_lower = ci[1],ci_upper=ci[3])
return(df)
}
else{
cutoff=roc.p$thresholds[which.min(roc.p$sensitivities+roc.p$specificities)]
sensitivity=roc.p$sensitivities[which.min(roc.p$sensitivities+roc.p$specificities)]
specificity=roc.p$specificities[which.min(roc.p$sensitivities+roc.p$specificities)]
ci = roc.p$ci
df=data.frame(type='negative classification',
auc=1-round(roc.p$auc,3),cutoff=cutoff,
sensitivity=1-sensitivity,specificity=1-specificity,ci_lower = ci[1],ci_upper=ci[3])
return(df)
}
}
不需要排序。直接用quantile就可以。你看看是不是跟你的情况相似set.seed(1234)
d <- sample(1:10000,1000,replace = F)
quantile(d, probs = seq(0,1,0.25))
由已知条件可知μ=100,δ=5，我们需要把符合正态分布的随即变量x根据公式u=(x-μ)/δ转化为符合标准正态分布的常量u。把区间[85，115]带入公式，可得u的常量区间为[-3,3],然后查标准正态分布表
得u<3的区间概率为0.9987，u<-3的概率为1-0.9987，so,[-3,3]区间的概率为0.9987-（1-0.9987）=0.9974
