代码之家 › 专栏 › 技术社区 › Emily Kothe

计算组中行之间的差异分数

dplyr r

1

Emily Kothe · 技术社区 · 7 年前

我可以得到第二队的得分差(在diff列中),但是我不知道如何计算第一团队。它应该是第二个团队的目标差的倒数(即在样本数据集中“种植者”应该拥有 1 -1

library(dplyr)

dat <-
  structure(
    list(
      Match = c(1, 1, 2, 2, 3, 3),
      Team = c("Growlers",
               "Rollers", "Strike", "Bandits", "Cats", "Blues"),
      Goals = c(1,0, 0, 1, 1, 2)
    ),
    row.names = c(NA,-6L),
    groups = structure(
      list(
        Match = c(895825, 895826, 895827),
        .rows = list(1:2, 3:4,
                     5:6)
      ),
      row.names = c(NA,-3L),
      class = c("tbl_df", "tbl",
                "data.frame"),
      .drop = TRUE
    ),
    class = c("grouped_df", "tbl_df",
              "tbl", "data.frame")
  )

dat %>% 
    group_by(Match) %>% 
    mutate(diff = Goals - lag(Goals))
#> # A tibble: 6 x 4
#> # Groups:   Match [3]
#>   Match Team     Goals  diff
#>   <dbl> <chr>    <dbl> <dbl>
#> 1     1 Growlers     1    NA
#> 2     1 Rollers      0    -1
#> 3     2 Strike       0    NA
#> 4     2 Bandits      1     1
#> 5     3 Cats         1    NA
#> 6     3 Blues        2     1

创建于2019-02-26 reprex package

0 回复 | 直到 7 年前

1

Andrie 7 年前

一种快速而肮脏的方法是显式计算团队1和团队2的分数,如下所示:

dat %>% 
  group_by(Match) %>% 
  mutate(
    diff = c(
      Goals[1] - Goals[2],
      Goals[2] - Goals[1] 
    )
  )

#> # A tibble: 6 x 4
#> # Groups:   Match [3]
#>   Match Team     Goals  diff
#>   <dbl> <chr>    <dbl> <dbl>
#> 1     1 Growlers     1     1
#> 2     1 Rollers      0    -1
#> 3     2 Strike       0    -1
#> 4     2 Bandits      1     1