Question

说我有一些价值观：

raw = tibble(
    group = c("A", "B", "C", "A", "B", "C"),
    value = c(10, 20, 30, 40, 50, 60)
)

# A tibble: 6 x 2
  group value
  <chr> <dbl>
1     A    10
2     B    20
3     C    30
4     A    40
5     B    50
6     C    60

我想从我的tibble中的每个值中减去一定数量，具体取决于它属于哪个组。我需要减去的金额是另一个元素：

corrections = tibble(
    group = c("A", "B", "C"),
    corr = c(0, 1, 2)
)

# A tibble: 3 x 2
  group  corr
  <chr> <dbl>
1     A     0
2     B     1
3     C     2

实现这一目标的最优雅方式是什么？以下作品，但我觉得它很乱 - 当然还有另一种方式吗？

mutate(raw, corrected = value - as_vector(corrections[corrections["group"] == group, "corr"]))

# A tibble: 6 x 3
  group value corrected
  <chr> <dbl>     <dbl>
1     A    10        10
2     B    20        19
3     C    30        28
4     A    40        40
5     B    50        49
6     C    60        58

Answer 1

首先加入raw和corrections然后计算corrected怎么样？

library(dplyr)
left_join(raw, corrections, by = "group") %>% 
  mutate(corrected = value - corr) %>% 
  select(-corr) 

#> # A tibble: 6 x 3
#>   group value corrected
#>   <chr> <dbl>     <dbl>
#> 1     A    10        10
#> 2     B    20        19
#> 3     C    30        28
#> 4     A    40        40
#> 5     B    50        49
#> 6     C    60        58

基于另一个tibble从tibble列中减去值

1 个答案: