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我有一个类似于 的数据集iris，需要编写一个以下列方式处理异常值的函数：对于每个物种setosa、versicolor和virginica，在每个变量iris$Sepal.Length、iris$Sepal.Width、iris$Petal.Length和Petal.Width中，将 1.5*IQR 之外的值替换为IQR +/- 1.5*IQR（取决于它是高于还是低于 IQR）。我一直在使用下面的代码来实现这一点，但它非常重复、耗时且容易出错。此外，这样做会改变原始对象中的值。将参数合并到一个函数中会很好，该函数不仅可以实现这一点，还可以告诉我哪些值已更改并将所有输出保存到新数据框中，而不是更改原始数据集中的值。

data(iris)
#create separate objects containing the data for each species 
setosa <-
  iris%>%
  filter(Species == "setosa")

versicolor <- 
  iris%>%
  filter(Species == "versicolor")

virginica <- 
  iris%>%
  filter(Species == "virginica")
#for each variable within each species, do the following:
#create an object (qnt) that contains the 25th and 75th percentile 
#create an object (H) containing the value of 1.5 times the interquartile range(IQR)
#replace any number less than the 25th percentile minus H with the value of the
#25th percentile minus H
#replace any number greater than the 75th percentile plus H with the value of the 
#75th percentile plus H

qnt <- quantile(setosa$Sepal.Length, probs = c(.25, .75), na.rm = T)
H <- 1.5*IQR(setosa$Sepal.Length, na.rm = T)
setosa$Sepal.Length[setosa$Sepal.Length < (qnt[1] - H)] <- qnt[1]-H
setosa$Sepal.Length[setosa$Sepal.Length > (qnt[2] + H)] <- qnt[2]+H

qnt <- quantile(setosa$Sepal.Width, probs = c(.25, .75), na.rm = T)
H <- 1.5*IQR(setosa$Sepal.Width, na.rm = T)
setosa$Sepal.Width[setosa$Sepal.Width < (qnt[1] - H)] <- qnt[1]-H
setosa$Sepal.Width[setosa$Sepal.Width > (qnt[2] + H)] <- qnt[2]+H

qnt <- quantile(setosa$Petal.Length, probs = c(.25, .75), na.rm = T)
H <- 1.5*IQR(setosa$Petal.Length, na.rm = T)
setosa$Petal.Length[setosa$Petal.Length < (qnt[1] - H)] <- qnt[1]-H
setosa$Petal.Length[setosa$Petal.Length > (qnt[2] + H)] <- qnt[2]+H

qnt <- quantile(setosa$Petal.Width, probs = c(.25, .75), na.rm = T)
H <- 1.5*IQR(setosa$Petal.Width, na.rm = T)
setosa$Sepal.Width[setosa$Petal.Width < (qnt[1] - H)] <- qnt[1]-H
setosa$Sepal.Width[setosa$Petal.Width > (qnt[2] + H)] <- qnt[2]+H

#now do versicolor

qnt <- quantile(versicolor$Sepal.Length, probs = c(.25, .75), na.rm = T)
H <- 1.5*IQR(versicolor$Sepal.Length, na.rm = T)
versicolor$Sepal.Length[versicolor$Sepal.Length < (qnt[1] - H)] <- qnt[1]-H
versicolor$Sepal.Length[versicolor$Sepal.Length > (qnt[2] + H)] <- qnt[2]+H

qnt <- quantile(versicolor$Sepal.Width, probs = c(.25, .75), na.rm = T)
H <- 1.5*IQR(versicolor$Sepal.Width, na.rm = T)
versicolor$Sepal.Width[versicolor$Sepal.Width < (qnt[1] - H)] <- qnt[1]-H
versicolor$Sepal.Width[versicolor$Sepal.Width > (qnt[2] + H)] <- qnt[2]+H

qnt <- quantile(versicolor$Petal.Length, probs = c(.25, .75), na.rm = T)
H <- 1.5*IQR(versicolor$Petal.Length, na.rm = T)
versicolor$Petal.Length[versicolor$Petal.Length < (qnt[1] - H)] <- qnt[1]-H
versicolor$Petal.Length[versicolor$Petal.Length > (qnt[2] + H)] <- qnt[2]+H

qnt <- quantile(versicolor$Petal.Width, probs = c(.25, .75), na.rm = T)
H <- 1.5*IQR(versicolor$Petal.Width, na.rm = T)
versicolor$Sepal.Width[versicolor$Petal.Width < (qnt[1] - H)] <- qnt[1]-H
versicolor$Sepal.Width[versicolor$Petal.Width > (qnt[2] + H)] <- qnt[2]+H

#now do virginica

qnt <- quantile(virginica$Sepal.Length, probs = c(.25, .75), na.rm = T)
H <- 1.5*IQR(virginica$Sepal.Length, na.rm = T)
virginica$Sepal.Length[virginica$Sepal.Length < (qnt[1] - H)] <- qnt[1]-H
virginica$Sepal.Length[virginica$Sepal.Length > (qnt[2] + H)] <- qnt[2]+H

qnt <- quantile(virginica$Sepal.Width, probs = c(.25, .75), na.rm = T)
H <- 1.5*IQR(virginica$Sepal.Width, na.rm = T)
virginica$Sepal.Width[virginica$Sepal.Width < (qnt[1] - H)] <- qnt[1]-H
virginica$Sepal.Width[virginica$Sepal.Width > (qnt[2] + H)] <- qnt[2]+H

qnt <- quantile(virginica$Petal.Length, probs = c(.25, .75), na.rm = T)
H <- 1.5*IQR(virginica$Petal.Length, na.rm = T)
virginica$Petal.Length[virginica$Petal.Length < (qnt[1] - H)] <- qnt[1]-H
virginica$Petal.Length[virginica$Petal.Length > (qnt[2] + H)] <- qnt[2]+H

qnt <- quantile(virginica$Petal.Width, probs = c(.25, .75), na.rm = T)
H <- 1.5*IQR(virginica$Petal.Width, na.rm = T)
virginica$Sepal.Width[virginica$Petal.Width < (qnt[1] - H)] <- qnt[1]-H
virginica$Sepal.Width[virginica$Petal.Width > (qnt[2] + H)] <- qnt[2]+H
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1 回答 1

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创建一个函数，然后在按“物种”分组后应用于列，并将其分配给一个新对象会更容易。使用，除非我们使用特殊运算符 from (而不是) dplyr，否则原始数据集不会即时更改magrittr%<>%%>%

f1 <- function(x) {
       qnt <- quantile(x, probs = c(.25, .75), na.rm = TRUE)
       H <- 1.5*IQR(x, na.rm = TRUE)
       x[x< (qnt[1] - H)] <- qnt[1]-H
       x[x> (qnt[2] + H)] <- qnt[2]+H
       x
       }

library(dplyr)
iris1 <- iris %>% 
             group_by(Species) %>%
             mutate_at(vars(-group_cols()), f1)

此外，如果我们只需要应用数字列（如果数据集有其他不同类型的列）

iris1 <- iris %>%
            group_by(Species) %>%
            mutate_if(is.numeric, f1)
于 2019-11-30T17:41:21.317 回答