如何按时间间隔连接2个数据表并按因子变量汇总重叠和不重叠的时间段

How to join 2 data tables by time interval and summarize overlapping and non-overlapping time periods by factor variable

我有 2 个数据表,每个表都列出了观察工作的时间段和工作类型 (A、B、C)。 我想知道重叠和非重叠工作时间的持续时间。

我尝试用 data.table 和重叠来做到这一点,但无法弄清楚如何包括所有非重叠周期。

这是我的示例数据。我首先创建了 2 个包含努力时期的数据表。我的数据集将包括单个观察者努力工作的时间段。

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 foverlaps 来查找工作重叠的地方。我已经设置了 nomatch=NA,但这只是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)

我在这里继续展示我是如何尝试确定所有重叠和非重叠轮班时间的持续时间的。但我认为我也没有正确理解这一部分。

# 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 "EffortType"

的每个分组的观察工作量时间

最后得到这样的结果(数字只是示例,因为即使在 excel 中我也没有弄清楚如何正确计算)

EffortType  Duration(in minutes)
A           10
B           20
C           12
AA          8
BB          6
CC          1
AC          160
AB          200
BC          150

不是完整的答案(见最后一段)..但我认为这会让你得到你想要的。

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")

结果

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

有时同一事件会在同一分钟内发生 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

不确定你想如何总结...

如果您能将它们视为一分钟,那么:

> 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

我认为可以用几分钟来计算持续时间(尽管这可能以更聪明的方式完成)