我有一个数据框，假设：
names<-c("a","a","a","a","a","b","b","b","b","b","c","c","c","c","c","c","c","c")
var1<-c(0.942999593,0.935507266,0.973589623,0.969415912,0.95230801,0.935507266,0.888740961,0.91750551,0.944482672,0.945468585,1.457579147,0.922206277,0.941511433,0.954724791,0.941014244,0.941511433,0.941511433,1.50511433)
var2<-c(-0.012678088,0.014313763,0.001138275,-0.020568206,0.012987126,0.001217192,0.03360358,0.009758172,0.015066932,-0.037879492,0.020471157,0.010738162,0.010952531,0.019377213,0.027140572,0.031116892,-0.018530676,-8.90E-05)
as.data.frame(cbind(names,var1,var2))->df
我想在列var1和var2中将异常值转换为Na。但是，我想为“名称”列中的每个类别独立计算离群值。因此var1中“a”的异常值将是var1中仅使用前5行的异常值。
我检测异常值的方式是分别低于或高于分位数0.25和0.75的所有值。
在R中有没有简单的方法呢？
非常感谢你。
蒂娜。
答案 0 :(得分：5)
以下是如何为var1执行此操作：
quantiles<-tapply(var1,names,quantile)
minq <- sapply(names, function(x) quantiles[[x]]["25%"])
maxq <- sapply(names, function(x) quantiles[[x]]["75%"])
var1[var1<minq | var1>maxq] <- NA
对var2（或df $ var2）重复相同的操作。