首页 > 解决方案 > 仅当 R 中存在异常值时,如何使用通用方法删除异常值

问题描述

我正在使用一种方法来删除单变量异常值。此方法仅在向量包含异常值时有效。

如何将这种方法推广到也适用于没有异常值的向量。我试过ifelse没有成功。

library(tidyverse)

df <- tibble(x = c(1,2,3,4,5,6,7,80))

outliers <- boxplot(df$x, plot=FALSE)$out
print(outliers)
#> [1] 80

# This removes the outliers
df2 <- df[-which(df$x %in% outliers),]

# a new tibble withou outliers
df3 <- tibble(x = c(1,2,3,4,5,6,7,8))

outliers3 <- boxplot(df3$x, plot=FALSE)$out
print(outliers3) # no outliers
#> numeric(0)

# if I try to use the same expression to remove 0 outliers
df4 <- df[-which(df3$x %in% outliers),]

# boxplot gives an error because df4 has 0 observations
# when I was expecting 8 observations
boxplot(df4$x)
#> Warning in min(x): no non-missing arguments to min; returning Inf
#> Warning in max(x): no non-missing arguments to max; returning -Inf
#> Error in plot.window(xlim = xlim, ylim = ylim, log = log, yaxs = pars$yaxs): need finite 'ylim' values

标签: rboxplotoutliers

解决方案


即使没有异常值,也可以使用否定( !) 而不是使用which-

df3[!(df3$x %in% outliers3),]

-输出

# A tibble: 8 x 1
      x
  <dbl>
1     1
2     2
3     3
4     4
5     5
6     6
7     7
8     8

或者如果有异常值,它会删除

df[!df$x %in% outliers,]
# A tibble: 7 x 1
      x
  <dbl>
1     1
2     2
3     3
4     4
5     5
6     6
7     7

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