R - 创建具有连续数字的列并使用 mutate 基于其他列重置答案

【问题标题】：R - Create a column with consecutive numbers and reset based on other column using mutateR - 创建具有连续数字的列并使用 mutate 基于其他列重置
【发布时间】：2020-11-25 13:15:50
【问题描述】：

我正在尝试使用mutate 在我的df 中创建一个新变量game_plus，该变量计算自在另一列game 中发生事件的天数并重置每次事件发生。例如，我的df 中的变量game 是二进制的，可以采用"Game" 或"Training"。我正在考虑使用嵌套的ifelse 语句来生成以下输出：

game         game_plus
Game             0
Training         1
Training         2
Training         3
Game             0
Training         1
Training         2
Game             0
Training         1
Training         2
Training         3
Training         4

我还想在对面列game_minus，它基本上计算事件之前的天数，如下所示。

game         game_plus      game_minus
Game             0              0
Training         1              3
Training         2              2
Training         3              1
Game             0              0
Training         1              2
Training         2              1 
Game             0              0
Training         1              4
Training         2              3
Training         3              2
Training         4              1

有人可以帮忙吗？我知道我可以使用ifelse(game == "Game", 0,，但我正在努力弄清楚如何将 - 事件发生之前或之后- 元素合并到这个 ifelse 语句中。任何帮助将不胜感激！

【问题讨论】：

标签： r if-statement multiple-columns reset dplyr

【解决方案1】：

library(tibble)
library(dplyr)

game_tbl <- 
tibble(game = c("Game",rep("Training", 3), "Game",rep("Training", 2),"Game",rep("Training", 4)))


game_tbl  %>% 
  mutate(period = cumsum(game == "Game")) %>% ## which rows belong to one game period 
  group_by(period, game) %>% 
  mutate(game_plus = case_when(game == "Game" ~ 0L , TRUE ~ row_number())) %>%
  group_by(period) %>%  
  mutate(units = n() ) %>% ## how many rows per  game period
  mutate(game_minus =  case_when(game == "Game" ~ 0L, TRUE ~  units - row_number() + 1L )) %>%
  ungroup() %>%
  select(game, game_plus, game_minus)

基本上您需要使用 group_by 和 row_number。 row_number 添加每行的编号。与 group_by 结合使用，它会遍历每个组的行。我添加了辅助变量 period 来识别属于一个游戏周期（从游戏到最后一次训练）的所有行。因此，row_number 将对 game == "Game" 和 game == "Training" 的每个游戏周期计算行数。我还添加了辅助变量 units 来计算每个游戏周期的行数。

【讨论】：

【解决方案2】：

这是一个使用data.table 的解决方案，这里的要点是使用n_games 将数据分解为多个部分。然后data.table 有一个内置的方法来基本上使用.I 获取行号，因此如果我们将数据分解为每个部分，我们可以反向获取行号以倒计时。剩下的最后一件事是为每个比赛日分配一个值 0，而不是距离下一场比赛的天数。

library(data.table)

dt = data.table(game = c("Game",rep("Training", 3), "Game",rep("Training", 2),"Game",rep("Training", 4)))
## Create an indicator for if a day has a game
dt[,game_ind := ifelse(game == 'Game', 1, 0)]
## Use the indicators to break up the data into groups by taking the cumulative sum of games
dt[,n_games := cumsum(game_ind)]
## .SD[,.I] gets the row number for each group of n_games, rev makes it so that it's 
## counting down instead of up
dt[,game_minus := rev(unlist(.SD[,.I])), by = n_games]
## Set game days to 0
dt[game == 'Game', game_minus := 0]

dt
#>         game game_ind n_games game_minus
#>  1:     Game        1       1          0
#>  2: Training        0       1          3
#>  3: Training        0       1          2
#>  4: Training        0       1          1
#>  5:     Game        1       2          0
#>  6: Training        0       2          2
#>  7: Training        0       2          1
#>  8:     Game        1       3          0
#>  9: Training        0       3          4
#> 10: Training        0       3          3
#> 11: Training        0       3          2
#> 12: Training        0       3          1

## If you want to clean up
dt[,c('game_ind', 'n_games') := NULL]

head(dt)
#>        game game_minus
#> 1:     Game          0
#> 2: Training          3
#> 3: Training          2
#> 4: Training          1
#> 5:     Game          0
#> 6: Training          2

^{由reprex package (v0.3.0) 于 2020 年 11 月 25 日创建}

【讨论】：