r - 查找每个工作日的平均事件

Question

我有一个数据集如下：

+----+-------+---------------------+
| ID | SUBID |        date         |
+----+-------+---------------------+
| A  |     1 | 2021-01-01 12:00:00 |
| A  |     1 | 2021-01-02 01:00:00 |
| A  |     1 | 2021-01-02 02:00:00 |
| A  |     1 | 2021-01-03 03:00:00 |
| A  |     2 | 2021-01-05 16:00:00 |
| A  |     2 | 2021-01-06 13:00:00 |
| A  |     2 | 2021-01-07 06:00:00 |
| A  |     2 | 2021-01-08 08:00:00 |
| A  |     2 | 2021-01-08 10:00:00 |
| A  |     2 | 2021-01-08 11:00:00 |
| A  |     3 | 2021-01-09 09:00:00 |
| A  |     3 | 2021-01-10 19:00:00 |
| A  |     3 | 2021-01-11 20:00:00 |
| A  |     3 | 2021-01-12 22:00:00 |
| B  |     1 | 2021-02-01 23:00:00 |
| B  |     1 | 2021-02-02 15:00:00 |
| B  |     1 | 2021-02-03 06:00:00 |
| B  |     1 | 2021-02-04 08:00:00 |
| B  |     2 | 2021-02-05 18:00:00 |
| B  |     2 | 2021-02-05 19:00:00 |
| B  |     2 | 2021-02-06 22:00:00 |
| B  |     2 | 2021-02-06 23:00:00 |
| B  |     2 | 2021-02-07 04:00:00 |
| B  |     2 | 2021-02-08 02:00:00 |
| B  |     3 | 2021-02-09 01:00:00 |
| B  |     3 | 2021-02-10 03:00:00 |
| B  |     3 | 2021-02-11 13:00:00 |
| B  |     3 | 2021-02-12 14:00:00 |
+----+-------+---------------------+

我希望能够以小时为单位获得每个 ID 和 SUBID 组之间的时差，最好是在营业时间方面，其中每个出现在周末或联邦假日的日期都可以移动到最近的工作日（之前或成功），时间为 23:59:59，如下所示：

+----+-------+---------------------+------------------------------------------------------------------+
| ID | SUBID |        date         | timediff (hours) with preceding date for each group (ID, SUBID) |
+----+-------+---------------------+------------------------------------------------------------------+
| A  |     1 | 2021-01-01 12:00:00 |                                                                0 |
| A  |     1 | 2021-01-02 01:00:00 |                                                               13 |
| A  |     1 | 2021-01-02 02:00:00 |                                                                1 |
| A  |     1 | 2021-01-03 03:00:00 |                                                                1 |
| A  |     2 | 2021-01-05 16:00:00 |                                                                0 |
| A  |     2 | 2021-01-06 13:00:00 |                                                               21 |
| A  |     2 | 2021-01-07 06:00:00 |                                                               17 |
| A  |     2 | 2021-01-08 08:00:00 |                                                                2 |
| A  |     2 | 2021-01-08 10:00:00 |                                                                2 |
| A  |     2 | 2021-01-08 11:00:00 |                                                                1 |
| A  |     3 | 2021-01-09 09:00:00 |                                                                0 |
| A  |     3 | 2021-01-10 19:00:00 |                                                               36 |
| A  |     3 | 2021-01-11 20:00:00 |                                                                1 |
| A  |     3 | 2021-01-12 22:00:00 |                                                                1 |
| B  |     1 | 2021-02-01 23:00:00 |                                                                0 |
| B  |     1 | 2021-02-02 15:00:00 |                                                               16 |
| B  |     1 | 2021-02-03 06:00:00 |                                                               15 |
| B  |     1 | 2021-02-04 08:00:00 |                                                               26 |
| B  |     2 | 2021-02-05 18:00:00 |                                                                0 |
| B  |     2 | 2021-02-05 19:00:00 |                                                                1 |
| B  |     2 | 2021-02-06 22:00:00 |                                                               27 |
| B  |     2 | 2021-02-06 23:00:00 |                                                                1 |
| B  |     2 | 2021-02-07 04:00:00 |                                                                5 |
| B  |     2 | 2021-02-08 02:00:00 |                                                               22 |
| B  |     3 | 2021-02-09 01:00:00 |                                                                0 |
| B  |     3 | 2021-02-10 03:00:00 |                                                               26 |
| B  |     3 | 2021-02-11 13:00:00 |                                                               11 |
| B  |     3 | 2021-02-12 14:00:00 |                                                                1 |
+----+-------+---------------------+------------------------------------------------------------------+

最后我想计算平均时间，这将是每组（ID，SUBID）的时间差总和除以每组的总计数，如下所示：

+----+-------+------------------------------------------------------------+
| ID | SUBID | Average  time (count per group/ total time diff of group ) |
+----+-------+------------------------------------------------------------+
| A  |     1 | 15/4                                                       |
| A  |     2 | 43/6                                                       |
| A  |     3 | 38/4                                                       |
| B  |     1 | 57/4                                                       |
| B  |     2 | 56/6                                                       |
| B  |     3 | 38/4                                                       |
+----+-------+------------------------------------------------------------+

我对 R 很陌生，我遇到了 lubridate 来帮助我格式化日期，我可以使用下面的代码来获取时间差异

df%>%
        group_by(ID, SUBID) %>%
        mutate(time_diff = difftime(date, lag(date), unit = 'min'))

但是，我在获取工作日时间的差异以及根据最后一张表获取平均时间时遇到了麻烦

Answer 1

欢迎！使用dplyr和lubridate：

使用的数据：

library(tidyverse)
library(lubridate)
df <- data.frame(ID = c("A","A","A","A"),
                 SUBID = c(1,1,2,2),
                 Date = lubridate::as_datetime(c("2021-01-01 12:00:00","2021-01-02 1:00:00","2021-01-01 2:00:00","2021-01-01 13:00:00")))

  ID SUBID                Date
1  A     1 2021-01-01 12:00:00
2  A     1 2021-01-02 01:00:00
3  A     2 2021-01-01 02:00:00
4  A     2 2021-01-01 13:00:00

代码：

df %>% 
  group_by(ID, SUBID) %>% 
  mutate(diff = Date - lag(Date)) %>% 
  mutate(diff = ifelse(is.na(diff), 0, diff)) %>% 
  summarise(Average = sum(diff)/n())

输出：

  ID    SUBID Average
  <chr> <dbl>   <dbl>
1 A         1     6.5
2 A         2     5.5

编辑：如何处理week_ends

对于周末，更简单的解决方案是将日期更改为下周一：

df %>% 
  mutate(week_day = wday(Date,label = TRUE, abbr = FALSE)) %>%
  mutate(Date = ifelse(week_day == "samedi", Date + days(2),
                       ifelse(week_day == "dimanche", Date + days(1), Date))) %>%
  mutate(Date = as_datetime(Date))

这将创建week_day具有当天名称的列。如果这一天是“samedi”（星期六）或“dimanche”（星期日），它会将 Date 增加 2 或 1 天，因此它变成了星期一。然后，您只需要重新排序日期（df %>% arrange(ID, SUBID, Date))并重新运行第一个代码。

由于我的本地语言是法语，因此您必须将samediand更改dimanche为saturdayandsunday

对于假期，您可以通过创建一个表示假期的时间间隔变量来应用相同的系统，测试每个日期是否在此时间间隔内，如果是，则将日期更改为此时间间隔的最后一天。