代码中有两个case_when()，条件是相同的。有什么办法把它简化吗？

编程语言 2026-07-11

在下方的代码中，有两个 case_when()，条件相同。有没有办法把它简化？

library(tidyverse)
diamonds %>%
  mutate(kpi_a = case_when(
    color %in% c('E','I') ~ "A",
    color %in% c('J','H') ~ "B",
    TRUE ~ "C"
  ),
    kpi_b = case_when(
    color %in% c('E','I') ~ 1.1,
    color %in% c('J','H') ~ 1.2,
    TRUE ~ 2
  )
  )

解决方案

也许你可以把tibble和case_when一起使用

# library(tidyverse)

library(dplyr)
library(tibble)

diamonds %>%
  mutate(case_when(
    color %in% c('E','I') ~ tibble(kpi_a = "A", kpi_b = 1.1),
    color %in% c('J','H') ~ tibble(kpi_a = "B", kpi_b = 1.2),
    TRUE ~ tibble(kpi_a = "C", kpi_b = 2)
  ))

就会得到

# A tibble: 53,940 × 12
   carat cut       color clarity depth table price     x     y     z kpi_a kpi_b
   <dbl> <ord>     <ord> <ord>   <dbl> <dbl> <int> <dbl> <dbl> <dbl> <chr> <dbl>
 1  0.23 Ideal     E     SI2      61.5    55   326  3.95  3.98  2.43 A       1.1
 2  0.21 Premium   E     SI1      59.8    61   326  3.89  3.84  2.31 A       1.1
 3  0.23 Good      E     VS1      56.9    65   327  4.05  4.07  2.31 A       1.1
 4  0.29 Premium   I     VS2      62.4    58   334  4.2   4.23  2.63 A       1.1
 5  0.31 Good      J     SI2      63.3    58   335  4.34  4.35  2.75 B       1.2
 6  0.24 Very Good J     VVS2     62.8    57   336  3.94  3.96  2.48 B       1.2
 7  0.24 Very Good I     VVS1     62.3    57   336  3.95  3.98  2.47 A       1.1
 8  0.26 Very Good H     SI1      61.9    55   337  4.07  4.11  2.53 B       1.2
 9  0.22 Fair      E     VS2      65.1    61   337  3.87  3.78  2.49 A       1.1
10  0.23 Very Good H     VS1      59.4    61   338  4     4.05  2.39 B       1.2
# ℹ 53,930 more rows

duplicate