合并重复的特定记录并在 R 中保留一些已删除记录的值

Question

我有一个庞大的数据集，其中包含四列 user_id、action、start_time 和 end_time。我想合并连续的动作"o" 和start_time 将第一个start_time 和end_time 是最后一个合并记录的end_time。
假设df：

"user_id","action","start_time","end_time"
"11","o",23:25:27,23:25:49
"11","o",23:25:28,23:25:28
"11","o",23:25:48,23:26:50
"11","v",23:25:49,23:25:49
"11","v",23:25:49,23:25:50
"11","o",23:28:24,00:22:33
"11","o",00:10:48,00:23:44
"22","o",00:11:52,00:22:33
"22","o",00:22:32,00:27:44
"22","v",00:22:42,00:22:42
"22","o",00:22:42,00:22:42
"22","z",00:22:42,00:22:43

我想合并第 1 行、第 2 行和第 3 行，因为它们都有动作 "o" 并且合并有第一行的 start_time 和第二行的 end_time。这同样适用于行号 6 和 7 以及行号 8 和 9。
所以想要的输出：

    "user_id","action","start_time","end_time"
    "11","o",23:25:27,23:26:50
    "11","v",23:25:49,23:25:49
    "11","v",23:25:49,23:25:50
    "11","o",23:28:24,00:23:44
    "22","o",00:11:52,00:27:44
    "22","v",00:22:42,00:22:42
    "22","o",00:22:42,00:22:42
    "22","z",00:22:42,00:22:43   

我如何在 R 中做到这一点？ 谢谢

Answer 1

如果您不介意data.table 解决方案，

library(data.table)
setDT(df)
df[, {
    if (action[1L]=="o") {
        .(start_time=start_time[1L], end_time=end_time[.N])
    } else {
        .(start_time, end_time)
    }
}, by=.(rleid(action), user_id, action)][, -1L]

#   user_id action start_time end_time
#1:      11      o   23:25:27 23:26:50
#2:      11      v   23:25:49 23:25:49
#3:      11      v   23:25:49 23:25:50
#4:      11      o   23:28:24 00:23:44
#5:      22      o   00:11:52 00:27:44
#6:      22      v   00:22:42 00:22:42
#7:      22      o   00:22:42 00:22:42
#8:      22      z   00:22:42 00:22:43

数据：

df <- read.csv(text='"user_id","action","start_time","end_time"
"11","o",23:25:27,23:25:49
"11","o",23:25:28,23:25:28
"11","o",23:25:48,23:26:50
"11","v",23:25:49,23:25:49
"11","v",23:25:49,23:25:50
"11","o",23:28:24,00:22:33
"11","o",00:10:48,00:23:44
"22","o",00:11:52,00:22:33
"22","o",00:22:32,00:27:44
"22","v",00:22:42,00:22:42
"22","o",00:22:42,00:22:42
"22","z",00:22:42,00:22:43')

Answer 2

我的流程：首先我们添加一个带有rleid 的运行长度编码ID，这将允许我们将操作视为单独的组。接下来我们添加两个临时列st 和et，表示每个组的开始和结束时间。然后，我们filter 并得到所有不是“o”的动作，或者如果它是“o”，我们采取第一个动作。然后在动作为“o”的组中，我们希望用临时列替换开始时间和结束时间。最后我们select 只列出了您想要用于决赛桌的列。

这应该适用于“o”动作的多次长度运行。我确信有更好的方法来进行最后一次变异，但我想把它放在那里。

library(data.table)
library(dplyr)
df  %>% 
  mutate(rlid = rleid(user_id,action)) %>% 
  group_by(rlid) %>% 
  mutate(st = start_time[row_number()==1], et = end_time[row_number = n()]) %>%
  filter(action!="o" | row_number()==1) %>% 
  mutate(start_time = case_when(action=="o"~st,
                                action!="o"~start_time),
         end_time = case_when(action=="o"~et,
                              action!="o"~end_time)) %>% 
  ungroup() %>% 
  select(user_id:end_time)

# # A tibble: 8 x 4
#   user_id action start_time end_time
#     <int> <fct>  <fct>      <fct>   
# 1      11 o      23:25:27   23:26:50
# 2      11 v      23:25:49   23:25:49
# 3      11 v      23:25:49   23:25:50
# 4      11 o      23:28:24   00:23:44
# 5      22 o      00:11:52   00:27:44
# 6      22 v      00:22:42   00:22:42
# 7      22 o      00:22:42   00:22:42
# 8      22 z      00:22:42   00:22:43

Answer 3

感谢@jasbner 建议data.table::rleid 的 99% tidyverse 解决方案：

使用data.table::rleid，我们可以给每个顺序组一个唯一的ID。然后就是简单的按rlid分组，然后用summarize找出最早的start_time和最新的end_time。默认情况下，summary 会删除所有其他变量，因此您必须如下所示明确保留它们。最后，我们删除了 rlid 变量，使其与您的示例相匹配，但这可能会为将来保留。

library(dplyr)
library(data.table)

df  %>% 
    mutate(rlid = data.table::rleid(user_id,action)) %>% 
    group_by(rlid) %>%
    summarize(user_id = user_id[1],
              action = action[1],
              start_time = min(start_time),
              end_time = max(end_time)) %>%
    select(-rlid)

  user_id action start_time end_time
    <int> <chr>  <chr>      <chr>   
1      11 o      23:25:27   23:26:50
2      11 v      23:25:49   23:25:50
3      11 o      00:10:48   00:23:44
4      22 o      00:11:52   00:27:44
5      22 v      00:22:42   00:22:42
6      22 o      00:22:42   00:22:42
7      22 z      00:22:42   00:22:43

此方法将折叠任意数量的重复行，并且（我认为）比纯 data.table 方法更易于理解。