Question

下面是我正在使用的表的示例。

df = data.frame(Test_ID = c('a1','a1','a1','a1','a1','a1','a1','a2','a2','a2','a2','a2','a2'), 
            Event_ID = c('Failure_x', 'Failure_x', 'Failure_y', 'Failure_y', 'Failure_x',
                         'Failure_x', 'Failure_y', 'Failure_x', 'Failure_y', 'Failure_y',
                         'Failure_x','Failure_x', 'Failure_y'),
            Fail_Date = c('2018-10-10 17:52:20', '2018-10-11 17:02:16', '2018-10-14 12:52:20',
                          '2018-11-11 16:18:34', '2018-11-12 17:03:06', '2018-11-25 10:50:10',
                          '2018-12-01 10:28:50', '2018-09-12 19:02:08', '2018-09-20 11:32:25',
                          '2018-10-13 14:43:30', '2018-10-15 14:22:28', '2018-10-30 21:55:45',
                          '2018-11-17 11:53:35'))

我只想在Failure_x之后出现Failure_y的地方减去故障日期（按Test_ID）。从Event_ID Failure_x的Fail_Date中减去Event_ID Failure_y的Fail_Date。在一个组中，我可以有多个Failure_y。从第一个Failure_y实例之后发生的Failure_x中减去第二个Failure_y。

我尝试使用dplyr创建TIME_BETWEEN_FAILURES列。

library(lubridate)
df$Fail_Date = as.POSIXct(as.character(as.factor(df$Fail_Date)),format="%Y-%m-%d %H:%M:%S")
df = df %>% group_by(Test_ID) %>% 
mutate(TIME_BETWEEN_FAILURES = ifelse(Event_ID == "Failure_y" & lag(Event_ID) == "Failure_x", 
                                    difftime(Fail_Date, first(Fail_Date),units = "hours"),''))`

我只能使用dplyr中的first（）为第一个实例正确创建Time_BETWEEN_FAILURES。那就是我目前停留的地方。在这方面的任何帮助将不胜感激。

This is result from the code snippet above.

分析所需的输出。
This is ideal response needed for my analysis.

谢谢。干杯。

Answer 1

df %>% 
  group_by(gr = rev(cumsum(rev(Event_ID)=="Failure_y")), Test_ID) %>%
  mutate(time_between_failures = ifelse(n() > 1 & Event_ID=="Failure_y", difftime(Fail_Date[n()], Fail_Date[1L], units = "hours"), NA)) 

# A tibble: 13 x 5
# Groups:   gr, Test_ID [6]
   Test_ID Event_ID  Fail_Date              gr time_between_failures
   <fct>   <fct>     <dttm>              <int>                 <dbl>
 1 a1      Failure_x 2018-10-10 17:52:20     6                   NA 
 2 a1      Failure_x 2018-10-11 17:02:16     6                   NA 
 3 a1      Failure_y 2018-10-14 12:52:20     6                   91 
 4 a1      Failure_y 2018-11-11 16:18:34     5                   NA 
 5 a1      Failure_x 2018-11-12 17:03:06     4                   NA 
 6 a1      Failure_x 2018-11-25 10:50:10     4                   NA 
 7 a1      Failure_y 2018-12-01 10:28:50     4                  449.
 8 a2      Failure_x 2018-09-12 19:02:08     3                   NA 
 9 a2      Failure_y 2018-09-20 11:32:25     3                  185.
10 a2      Failure_y 2018-10-13 14:43:30     2                   NA 
11 a2      Failure_x 2018-10-15 14:22:28     1                   NA 
12 a2      Failure_x 2018-10-30 21:55:45     1                   NA 
13 a2      Failure_y 2018-11-17 11:53:35     1                  790.

在r中使用dplyr根据条件减去日期

1 个答案: