我正在使用一种方法来删除单变量异常值。只有当向量包含异常值时,此方法才能工作。
如何将该方法推广到不带异常值的向量中。我试过用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发布于 2021-06-15 19:58:33
否定(!),而不是使用-,即使没有异常值也能工作
df3[!(df3$x %in% outliers3),]-output
# A tibble: 8 x 1
x
<dbl>
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8 8或者如果有异常值,它就移除。
df[!df$x %in% outliers,]
# A tibble: 7 x 1
x
<dbl>
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7 7https://stackoverflow.com/questions/67992709
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