cut 将切割级别的最小值和最大值提取为 data.frame 中的列

Question

我有一个数值向量，我想将其转换为五个数值级别。我可以使用 cut

获得五个级别

dx <- data.frame(x=1:100)
dx$cut <- cut(dx$x,5)

但我现在在提取级别的下限和上限时遇到问题。例如 (0.901,20.8] 在 dx$min 中为 0.901，在 dx$max 中为 20.8。

我试过了：

dx$min <- pmin(dx$cut)
dx$max <- pmax(dx$cut)
dx

但这不起作用。

Answer 1

你可以尝试根据逗号拆分标签（事先转换为character并修改以抑制除,和.之外的标点符号）然后创建2列：

min_max <- unlist(strsplit(gsub("(?![,.])[[:punct:]]", "", as.character(dx$cut), perl=TRUE), ",")) # here, the regex ask to replace every punctuation mark except a . or a , by an empty string

dx$min <- min_max[seq(1, length(min_max), by=2)]
dx$max <- min_max[seq(2, length(min_max), by=2)]

head(dx)
#  x          cut   min  max
#1 1 (0.901,20.8] 0.901 20.8
#2 2 (0.901,20.8] 0.901 20.8
#3 3 (0.901,20.8] 0.901 20.8
#4 4 (0.901,20.8] 0.901 20.8
#5 5 (0.901,20.8] 0.901 20.8
#6 6 (0.901,20.8] 0.901 20.8

Answer 2

下面是 tidyverse 风格的解决方案。

library(tidyverse)

tibble(x = seq(-1000, 1000, length.out = 10),
       x_cut = cut(x, 5)) %>% 
  mutate(x_tmp = str_sub(x_cut, 2, -2)) %>% 
  separate(x_tmp, c("min", "max"), sep = ",") %>% 
  mutate_at(c("min", "max"), as.double)
#> # A tibble: 10 x 4
#>         x x_cut           min   max
#>     <dbl> <fct>         <dbl> <dbl>
#>  1 -1000  (-1e+03,-600] -1000  -600
#>  2  -778. (-1e+03,-600] -1000  -600
#>  3  -556. (-600,-200]    -600  -200
#>  4  -333. (-600,-200]    -600  -200
#>  5  -111. (-200,200]     -200   200
#>  6   111. (-200,200]     -200   200
#>  7   333. (200,600]       200   600
#>  8   556. (200,600]       200   600
#>  9   778. (600,1e+03]     600  1000
#> 10  1000  (600,1e+03]     600  1000

^{由 reprex package (v0.2.1)}

于 2019-01-10 创建

cut 将切割级别的最小值和最大值提取为 data.frame 中的列

cut extracting the minimum and maximum of cut levels as columns in data.frame

r

cut