删除不需要的观察，并计算时间差答案

【问题标题】：Removing unwanted Observation, and calculating Time Difference删除不需要的观察，并计算时间差
【发布时间】：2018-11-30 09:10:26
【问题描述】：

我有一个数据框，当审核一个 id 时，它可以在特定时间通过或失败。如何计算，将状态从 Failed 更改为 Passed 状态 w.r.t id 所需的所有时间的总和。如果一个 id 已经失败，失败，Passed 状态。我必须添加从第一次失败到第二次失败通过状态更改状态所需的时间

time <- c("08-10-2018 08:36", "12-10-2018 07:53", "23-10-2018 23:09", "30-10-2018 18:24","07-11-2018 18:13","10-11-2018 05:47","19-11-2018 21:26","26-11-2018 14:04","16-10-2018 03:19","07-11-2018 19:00","09-11-2018 23:25","20-11-2018 19:24", "22-11-2018 01:12","28-11-2018 03:46","04-10-2018 15:05","15-10-2018 15:32","20-10-2018 06:21","26-10-2018 04:51","02-11-2018 00:28","09-11-2018 22:43","15-11-2018 22:39","21-11-2018 04:10","26-11-2018 13:29")
id <-c("A1","A1","A1","A1","A1","A1","A1","A1","A2","A2","A2","A2","A2","A2","A3","A3","A3","A3","A3","A3","A3","A3","A3")
status <- c("FAILED","PASSED","FAILED","PASSED","FAILED","PASSED","PASSED","PASSED","PASSED","FAILED","PASSED","FAILED","PASSED","PASSED","PASSED","FAILED","PASSED","PASSED","PASSED","FAILED","PASSED","PASSED","FAILED")

df <- data.frame(id, time, status)

要求的格式：

ids <- c("A1","A2", "A3")
diff_time <- c(13.25, 3.46, 10.61)
df2 <- data.frame(ids,diff_time)

提前致谢

【问题讨论】：

但是你有几个从 Failed -> Passed for each Id 的转换。为什么每个只有 1 个值？
您需要详细说明。这不是编码服务，您需要帮助我们来帮助您。提供更多详细信息，例如如何计算 diff_time
我有一个数据框，当一个 id 被审计时，它可以在那个特定的时间通过或失败。如何计算，将状态从 Failed 更改为 Passed 状态 w.r.t id 所需的所有时间的总和。如果一个 id 已经失败，失败，Passed 状态。我必须添加从第一次失败到通过和第二次失败通过状态更改状态所花费的时间。

标签： r data.table tidyverse reshape2

【解决方案1】：

如果我理解正确，OP 想要测量每个FAILED 事件与下一个后续PASSED 事件之间的时间差（对于每个id）。最后，需要对每个id 汇总测量的差异。

这可以通过data.table 包中提供的向后滚动连接来解决。

首先，我们必须将FAILED 和PASSED 事件分开。然后，使用 右连接 以便为每个 FAILED 事件查找后续的 PASSED 事件。这两个子集在id 和time 上连接，其中时间向后滚动（NOCB = 下一个观察值向后移动）。

library(data.table)
# coerce df to data.table, coerce time to POSIXct
setDT(df)[, time := lubridate::dmy_hm(time)]
# create subset PASSED
dfp <- df[status == "PASSED"][, timep := time]
# create subset FAILED
dff <- df[status == "FAILED"][, timef := time]
# backward rolling join
dfp[dff, on = .(id, time),  roll = -Inf]

   id                time status               timep i.status               timef
1: A1 2018-10-08 08:36:00 PASSED 2018-10-12 07:53:00   FAILED 2018-10-08 08:36:00
2: A1 2018-10-23 23:09:00 PASSED 2018-10-30 18:24:00   FAILED 2018-10-23 23:09:00
3: A1 2018-11-07 18:13:00 PASSED 2018-11-10 05:47:00   FAILED 2018-11-07 18:13:00
4: A2 2018-11-07 19:00:00 PASSED 2018-11-09 23:25:00   FAILED 2018-11-07 19:00:00
5: A2 2018-11-20 19:24:00 PASSED 2018-11-22 01:12:00   FAILED 2018-11-20 19:24:00
6: A3 2018-10-15 15:32:00 PASSED 2018-10-20 06:21:00   FAILED 2018-10-15 15:32:00
7: A3 2018-11-09 22:43:00 PASSED 2018-11-15 22:39:00   FAILED 2018-11-09 22:43:00
8: A3 2018-11-26 13:29:00   <NA>                <NA>   FAILED 2018-11-26 13:29:00

# rolling join and aggregate by id
dfp[dff, on = .(id, time),  roll = -Inf][, .(diff_time = sum(timep - timef, na.rm = TRUE)), by = id]

   id      diff_time
1: A1 13.254167 days
2: A2  3.425694 days
3: A3 10.614583 days

【讨论】：