我有2个数据表，每个数据表都列出了观察工作的周期以及工作的类型（A，B，C）。 我想知道重叠和不重叠工作的持续时间。
我尝试使用data.table和foverlaps进行此操作，但无法弄清楚如何包括所有非重叠时间段。
这是我的示例数据。我首先创建了2个包含工作时间的数据表。我的数据集将包括一个观察者正在努力的时间段。
library(data.table)
library(lubridate)
# times have been edited so not fixed to minute intervals - to make more realistic
set.seed(13)
EffortType = sample(c("A","B","C"), 100, replace = TRUE)
On = sample(seq(as.POSIXct('2016/01/01 01:00:00'), as.POSIXct('2016/01/03 01:00:00'), by = "1 sec"), 100, replace=F)
Off = On + minutes(sample(1:60, 100, replace=T))
Effort1 = data.table(EffortType, On, Off)
EffortType2 = sample(c("A","B","C"), 100, replace = TRUE)
On2 = sample(seq(as.POSIXct('2016/01/01 12:00:00'), as.POSIXct('2016/01/03 12:00:00'), by = "1 sec"), 100, replace=F)
Off2 = On2 + minutes(sample(1:60, 100, replace=T))
Effort2 = data.table(EffortType2, On2, Off2)
#prep for using foverlaps
setkey(Effort1, On, Off)
setkey(Effort2, On2, Off2)
然后，我使用保险杠来查找工作重叠的地方。我已经设置了nomatch = NA，但这给了我正确的外部连接。我想要完整的外部连接。所以我想知道更合适的功能是什么。
matches = foverlaps(Effort1,Effort2,type="any",nomatch=NA)
我在这里继续说明如何尝试确定所有重叠和不重叠的移位时间的持续时间。但是我也不认为这部分内容正确。
# find start and end of intersection of all shifts
matches$start = pmax(matches$On, matches$On2, na.rm=T)
matches$end = pmin(matches$Off, matches$Off2, na.rm=T)
# create intervals and find durations
matches$int = interval(matches$start, matches$end)
matches$dur = as.duration(matches$int)
然后我想总结每个“ EffortType”分组的观察努力时间
最后得到这样的结果（数字只是示例，因为即使在excel中，我也无法弄清楚如何正确地计算出这一点）
EffortType Duration(in minutes)
A 10
B 20
C 12
AA 8
BB 6
CC 1
AC 160
AB 200
BC 150
答案 0 :(得分：2)
不是全部答案（请参阅最后一段）..但我认为这将为您提供想要的东西。
library( data.table )
library( lubridate )
set.seed(13)
EffortType = sample(c("A","B","C"), 100, replace = TRUE)
On = sample(seq(as.POSIXct('2016/01/01 01:00:00'), as.POSIXct('2016/01/03 01:00:00'), by = "15 mins"), 100, replace=T)
Off = On + minutes(sample(1:60, 100, replace=T))
Effort1 = data.table(EffortType, On, Off)
EffortType2 = sample(c("A","B","C"), 100, replace = TRUE)
On = sample(seq(as.POSIXct('2016/01/01 12:00:00'), as.POSIXct('2016/01/03 12:00:00'), by = "15 mins"), 100, replace=T)
Off = On + minutes(sample(1:60, 100, replace=T))
Effort2 = data.table(EffortType2, On, Off)
#create DT of minutes, spanning your entire period.
dt.minutes <- data.table( On = seq(as.POSIXct('2016/01/01 01:00:00'), as.POSIXct('2016/01/03 12:00:00'), by = "1 mins"),
Off = seq(as.POSIXct('2016/01/01 01:00:00'), as.POSIXct('2016/01/03 12:00:00'), by = "1 mins") + 60 )
#prep for using foverlaps
setkey(Effort1, On, Off)
setkey(Effort2, On, Off)
#overlap join both efforts on the dt.minutes. note the use of "within" an "nomatch" to throw away minutes without events.
m1 <- foverlaps(dt.minutes, Effort1 ,type="within",nomatch=0L)
m2 <- foverlaps(dt.minutes, Effort2 ,type="within",nomatch=0L)
#bind together
result <- rbindlist(list(m1,m2))[, `:=`(On=i.On, Off = i.Off)][, `:=`(i.On = NULL, i.Off = NULL)]
#cast the result
result.cast <- dcast( result, On + Off ~ EffortType, value.var = "EffortType")
产生
head( result.cast, 10)
# On Off A B C
# 1: 2016-01-01 01:00:00 2016-01-01 01:01:00 1 0 1
# 2: 2016-01-01 01:01:00 2016-01-01 01:02:00 1 0 1
# 3: 2016-01-01 01:02:00 2016-01-01 01:03:00 1 0 1
# 4: 2016-01-01 01:03:00 2016-01-01 01:04:00 1 0 1
# 5: 2016-01-01 01:04:00 2016-01-01 01:05:00 1 0 1
# 6: 2016-01-01 01:05:00 2016-01-01 01:06:00 1 0 1
# 7: 2016-01-01 01:06:00 2016-01-01 01:07:00 1 0 1
# 8: 2016-01-01 01:07:00 2016-01-01 01:08:00 1 0 1
# 9: 2016-01-01 01:08:00 2016-01-01 01:09:00 1 0 1
# 10: 2016-01-01 01:09:00 2016-01-01 01:10:00 1 0 1
有时某个事件在同一分钟内发生2-3次，例如
# On Off A B C
#53: 2016-01-02 14:36:00 2016-01-02 14:37:00 2 2 3
不确定要如何求和...
如果您可以将它们视为一分钟，则：
> sum( result.cast[A>0 & B==0, C==0, ] )
[1] 476
> sum( result.cast[A==0 & B>0, C==0, ] )
[1] 386
> sum( result.cast[A==0 & B==0, C>0, ] )
[1] 504
> sum( result.cast[A>0 & B>0, C==0, ] )
[1] 371
> sum( result.cast[A==0 & B>0, C>0, ] )
[1] 341
> sum( result.cast[A>0 & B==0, C>0, ] )
[1] 472
> sum( result.cast[A>0 & B>0, C>0, ] )
[1] 265
可以在数分钟内获得持续时间的技巧（尽管这可以用更聪明的方式完成）