仅当异常值存在于 R 中时,如何使用通用方法删除异常值
How to use a generic method to remove outliers only if they exist in R
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才有效。
如何推广此方法以适用于没有异常值的向量。我尝试 ifelse
但没有成功。
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
否定 (!
) 而不是使用 -
即使没有异常值也可以工作
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
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才有效。
如何推广此方法以适用于没有异常值的向量。我尝试 ifelse
但没有成功。
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
否定 (!
) 而不是使用 -
即使没有异常值也可以工作
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