r - 通过ID计算时差

Question

我有这样的数据:

Incident.ID.. = c(rep("INCFI0000029582",4), rep("INCFI0000029587",4))
date = c("2014-09-25 08:39:45", "2014-09-25 08:39:48", "2014-09-25 08:40:44", "2014-10-10 23:04:00", "2014-09-25 08:33:32", "2014-09-25 08:34:41", "2014-09-25 08:35:24", "2014-10-10 23:04:00")
df = data.frame(Incident.ID..,date, stringsAsFactors = FALSE)

df

   Incident.ID..                date
1  INCFI0000029582 2014-09-25 08:39:45
2  INCFI0000029582 2014-09-25 08:39:48
3  INCFI0000029582 2014-09-25 08:40:44
4  INCFI0000029582 2014-10-10 23:04:00
5  INCFI0000029587 2014-09-25 08:33:32
6  INCFI0000029587 2014-09-25 08:34:41
7  INCFI0000029587 2014-09-25 08:35:24
8  INCFI0000029587 2014-10-10 23:04:00

我使用这个函数来计算以秒为单位的时间差:

padded.diff = function(x) c(0L, diff(x)) 

df2=within(df, {
  date        = strptime(date, format="%Y-%m-%d %H:%M:%S")
  date.diff   = padded.diff(as.numeric(date)) 
})

df2

Incident.ID..      date                date.diff
1  INCFI0000029582 2014-09-25 08:39:45         0
2  INCFI0000029582 2014-09-25 08:39:48         3
3  INCFI0000029582 2014-09-25 08:40:44        56
4  INCFI0000029582 2014-10-10 23:04:00   1347796
5  INCFI0000029587 2014-09-25 08:33:32  -1348228
6  INCFI0000029587 2014-09-25 08:34:41        69
7  INCFI0000029587 2014-09-25 08:35:24        43
8  INCFI0000029587 2014-10-10 23:04:00   1348116

但是我如何计算差异，以便它从每个“事件.ID ..”的零开始？:

 Incident.ID..                date date.diff
1  INCFI0000029582 2014-09-25 08:39:45         0
2  INCFI0000029582 2014-09-25 08:39:48         3
3  INCFI0000029582 2014-09-25 08:40:44        56
4  INCFI0000029582 2014-10-10 23:04:00   1347796
5  INCFI0000029587 2014-09-25 08:33:32         0
6  INCFI0000029587 2014-09-25 08:34:41        69
7  INCFI0000029587 2014-09-25 08:35:24        43
8  INCFI0000029587 2014-10-10 23:04:00   1348116

Answer 1

使用基础 R，您可以简单地将其包装在 ave 中

ave(as.numeric(as.POSIXct(date)), Incident.ID.., FUN = padded.diff) 

或使用 data.table (根据@akruns 评论)

library(data.table) 
setDT(df)[, date.diff := padded.diff(as.POSIXct(date)), by = Incident.ID..]