如何根据条件删除数据帧的N行

Question

我的问题来自How to find tail rows of a data frame that satisfy set criteria?，因此，我的(更新的)示例数据的结构如下：

Individ <- data.frame(Participant = c("Bill", "Bill", "Bill", "Bill", "Bill", "Jane", "Jane", "Jane", "Jane", "Jane", "Jane", "Jane", "Jane", "Jane", "Jane", "Jane", "Jane", "Bill", "Bill", "Bill", "Bill"),  
                      Time = c(1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4),
                      Condition = c("Placebo", "Placebo", "Placebo", "Placebo", "Placebo", "Expr", "Expr", "Expr", "Expr", "Expr", "Expr", "Placebo", "Placebo", "Placebo", "Placebo", "Placebo", "Placebo", "Expr", "Expr", "Expr", "Expr"),
                      Location = c("Home", "Home", "Home", "Home", "Home", "Home", "Home", "Home", "Home", "Home", "Home", "Home", "Home", "Home", "Home", "Home", "Home", "Away", "Away", "Away", "Away"),
                      Power = c(400, 250, 180, 500, 300, 600, 512, 300, 500, 450, 200, 402, 210, 130, 520, 310, 451, 608, 582, 390, 570))

我已经学会了根据每个Participant在不同的Condition + Location上最后出现的Power来找到它们的尾行。现在，我希望从每个Participant中删除每个Condition和Location的最后3行。但是，为每个Participant和Condition收集的Participant不同，因此我不能完全基于标准化的Time删除行。

如何快速遍历每个Participant及其各自的Condition+ Location并删除最后3行？我的实际数据框架是400万行+，有50多个参与者，所以理想情况下，在每个Participant和Condition上迭代的解决方案是可取的。

我的预期产出如下：

Output <- data.frame(Participant = c("Bill", "Bill", "Jane", "Jane", "Jane", "Jane", "Jane", "Jane", "Bill"),
                     Time = c(1, 2, 1, 2, 3, 1, 2, 3, 1),
                     Condition = c("Placebo", "Placebo", "Expr", "Expr", "Expr", "Placebo", "Placebo", "Placebo", "Expr"),
                     Location = c("Home", "Home", "Home", "Home", "Home", "Home", "Home", "Home", "Away"),
                     Power = c(400, 250, 600, 512, 300, 402, 210, 130, 608))

Answer 1

如果你用dplyr和row_number()和n()..。

library(dplyr)
Individ %>%
  group_by(Participant, Condition, Location) %>%
  filter(row_number() < n() - 2)

返回

Source: local data frame [9 x 5]
Groups: Participant, Condition, Location [4]

  Participant  Time Condition Location Power
       (fctr) (dbl)    (fctr)   (fctr) (dbl)
1        Bill     1   Placebo     Home   400
2        Bill     2   Placebo     Home   250
3        Jane     1      Expr     Home   600
4        Jane     2      Expr     Home   512
5        Jane     3      Expr     Home   300
6        Jane     1   Placebo     Home   402
7        Jane     2   Placebo     Home   210
8        Jane     3   Placebo     Home   130
9        Bill     1      Expr     Away   608

Answer 2

使用data.table的选项。我们将“data.frame”转换为“data.table”(setDT(Individ))，按“参与者”、“条件”和“位置”分组，使用head删除每个组合的最后3个观察值。

library(data.table)
setDT(Individ)[, head(.SD, -3) , .(Participant, Condition, Location)]
#   Participant Condition Location Time Power
#1:        Bill   Placebo     Home    1   400
#2:        Bill   Placebo     Home    2   250
#3:        Jane      Expr     Home    1   600
#4:        Jane      Expr     Home    2   512
#5:        Jane      Expr     Home    3   300
#6:        Jane   Placebo     Home    1   402
#7:        Jane   Placebo     Home    2   210
#8:        Jane   Placebo     Home    3   130
#9:        Bill      Expr     Away    1   608

dplyr中的等效选项是

library(dplyr)
Individ %>% 
     group_by(Participant, Condition, Location) %>% 
     do(head(., -3))
#  Participant  Time Condition Location Power
#       (fctr) (dbl)    (fctr)   (fctr) (dbl)
#1        Bill     1      Expr     Away   608
#2        Bill     1   Placebo     Home   400
#3        Bill     2   Placebo     Home   250
#4        Jane     1      Expr     Home   600
#5        Jane     2      Expr     Home   512
#6        Jane     3      Expr     Home   300
#7        Jane     1   Placebo     Home   402
#8        Jane     2   Placebo     Home   210
#9        Jane     3   Placebo     Home   130