如何根据条件删除数据框的 N 行答案

【问题标题】：How to remove N rows of a data frame, according to conditions如何根据条件删除数据框的 N 行
【发布时间】：2016-02-22 00:56:00
【问题描述】：

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）样本数据的结构如下：

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))

我已经学会根据Power 的最后一次出现在不同的Condition 加上Location 中找到每个Participant 的尾行。我现在希望为每个Condition 和Location 从每个Participant 中删除最后3 行。但是，为每个Participant 和Condition 收集的Time 不同，因此我不能纯粹基于标准化的Time 删除行。

如何快速遍历每个 Participant 和它们各自的 Conditionplus Location 并删除最后 3 行？我的实际数据框是 400 万行 + 超过 50 名参与者，因此理想情况下，需要迭代每个 Participant 和 Condition 的解决方案。

我的预期输出是：

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))

【问题讨论】：

标签： r

【解决方案1】：

如果您使用dplyr，与row_number() 和n()...

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

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

【讨论】：

谁知道会这么容易？！感谢您提供dplyr 解决方案。

【解决方案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，我们使用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

【讨论】：