[英]R calculate time within time interval between start End time
我有一個數據框如下
timedf <- data.frame(spaceNum=c(1,1,1,2,2,2), starttime= c("2015-09-03 00:00",
"2015-09-04 23:18", "2015-09-05 05:59", "2015-09-03 06:19",
"2015-09-06 09:03", "2015-09-06 09:10"), endtime = c("2015-09-04 20:05", "2015-09-05 05:52",
"2015-09-05 06:15", "2015-09-05 16:36",
"2015-09-06 09:06", "2015-09-06 20:42"))
timedf$staytime <- round(as.numeric(difftime(as.POSIXct(timedf$endtime), as.POSIXct(timedf$starttime),units="hours")), digits=2)
timedf
spaceNum starttime endtime staytime
1 1 2015-09-03 00:00 2015-09-04 20:05 44.08
2 1 2015-09-04 23:18 2015-09-05 05:52 6.57
3 1 2015-09-05 05:59 2015-09-05 06:15 0.27
4 2 2015-09-03 06:19 2015-09-05 16:36 58.28
5 2 2015-09-06 09:03 2015-09-06 09:06 0.05
6 2 2015-09-06 09:10 2015-09-06 20:42 11.53
我想分別計算0To7,7To19和19To24的時間段,即我希望得到低於結果
spaceNum starttime endtime staytime 0To7 7To19 19To24
1 1 2015-09-03 00:00 2015-09-04 20:05 44.08 14 24 6.08
2 1 2015-09-04 23:18 2015-09-05 05:52 6.57 5.87 0 0.7
3 1 2015-09-05 05:59 2015-09-05 06:15 0.27 0.27 0 0
4 2 2015-09-03 06:19 2015-09-05 16:36 58.28 14.683 33.6 10
5 2 2015-09-06 09:03 2015-09-06 09:06 0.05 0 0.05 0
6 2 2015-09-06 09:10 2015-09-06 20:42 11.53 0 9.83 1.7
請評論我怎么能這樣做,我正在尋找具有強大時間功能的lubridate
,但我仍然無法得到答案。
如果您使用data.tables,IMO整個過程效率更高:
# your data as provided in the question
timedf <- data.frame(spaceNum=c(1,1,1,2,2,2),
starttime= c("2015-09-03 00:00","2015-09-04 23:18", "2015-09-05 05:59", "2015-09-03 06:19","2015-09-06 09:03", "2015-09-06 09:10"),
endtime = c("2015-09-04 20:05", "2015-09-05 05:52", "2015-09-05 06:15", "2015-09-05 16:36","2015-09-06 09:06", "2015-09-06 20:42"))
library(data.table)
# convert starttime and endtime to POSIXct - note that you need tz="GMT"
setDT(timedf)[,c("starttime","endtime"):=lapply(.SD, as.POSIXct, tz="GMT"), .SDcols=2:3]
# add staytime column
timedf[,staytime:=round(as.numeric(difftime(endtime,starttime,units="hours")),digits=2)]
# this does the main work
brks <- c(0,7*3600,19*3600,24*3600)
lbls <- c("0To7","7To19","19To24")
sec.per.day <- 24*60*60
get.cuts <- function(x,y) as.list(table(cut((x:y)%%sec.per.day, breaks=brks, labels=lbls))/3600)
timedf[,c(lbls):=get.cuts(starttime,endtime), by=1:nrow(timedf)]
timedf
# spaceNum starttime endtime staytime 0To7 7To19 19To24
# 1: 1 2015-09-03 00:00:00 2015-09-04 20:05:00 44.08 14.0000000 24.00000000 6.083056
# 2: 1 2015-09-04 23:18:00 2015-09-05 05:52:00 6.57 5.8666667 0.00000000 0.700000
# 3: 1 2015-09-05 05:59:00 2015-09-05 06:15:00 0.27 0.2669444 0.00000000 0.000000
# 4: 2 2015-09-03 06:19:00 2015-09-05 16:36:00 58.28 14.6836111 33.60000000 9.999444
# 5: 2 2015-09-06 09:03:00 2015-09-06 09:06:00 0.05 0.0000000 0.05027778 0.000000
# 6: 2 2015-09-06 09:10:00 2015-09-06 20:42:00 11.53 0.0000000 9.83361111 1.700000
這種方法利用了POSIXct變量只是自1970-01-01以來的秒數這一事實。 get.cuts(...)
函數從starttime:endtime
生成一個整數向量,然后將每個元素模數為24 * 60 * 60,它將一天中的秒數除去並留下余數。 然后我們使用cut(...)
根據范圍00:00:00 - 07:00:00,07:00:01 - 19:00:00和19:00:01將每個元素分配到一個bin - 24:00:00,我們使用table(...)
來計算每個bin中的秒數。 然后我們通過除以3600轉換為小時並將結果作為列表返回(這是data.table所必需的)。
最后,我們為get.cuts(...)
中的每一行調用get.cuts(...)
。
如您所見,代碼比說明更短......
最后我為此編寫了一個函數,我知道函數是丑陋而不是優化,但這是我目前可以獲得的結果,如果有人能提供更簡單的解決方案,那將會很棒。
getp1TOp2 <- function(x,y,p1,p2) {
#Calcuate the numofdays
numdays <- as.numeric(difftime( update(as.POSIXct(y), hour=0, min=0, sec=0), update(as.POSIXct(x), hour=0, min=0, sec=0),units = "days"))
#Calculate the FirstdayStart, FirstdayEnd, LastdayStart and LastdayEnd
FirstdayStart <- update(as.POSIXct(x), hour=p1, min=0, sec=0)
FirstdayEnd <- update(as.POSIXct(x), hour=p2, min=0, sec=0)
LastdayStart <- update(as.POSIXct(y), hour=p1, min=0, sec=0)
LastdayEnd <- update(as.POSIXct(y), hour=p2, min=0, sec=0)
# For case Start and End in the same day , eg (2015-09-07 03:00, 2015-09-07 06:00)
if (numdays==0)
{
#For case StartTime >= FirstdayStart
if (as.POSIXct(x) >= FirstdayStart)
{
#For case EndTime < FirstdayEnd
if (as.POSIXct(y) <= FirstdayEnd)
{
#Example fall into this case (2015-09-07 08:00-2015-09-07 18:32), just return diff of start/end time
outval<- as.numeric(difftime(as.POSIXct(y), as.POSIXct(x), units="hours"))
}
else
{
#Example fall into this case (2015-09-07 08:00-2015-09-07 20:11 ), return diff of start/ FirstdayEnd time
outval<- as.numeric(difftime(FirstdayEnd, as.POSIXct(x), units="hours"))
}
}
else #For case StartTime < FirstdayStart
{
#For case EndTime < FirstdayEnd
if (as.POSIXct(y) <= FirstdayEnd)
{
#Example fall into this case (2015-09-07 06:00-2015-09-07 18:32), just return diff of FirstdayStart/end time
outval<- as.numeric(difftime(as.POSIXct(y), FirstdayStart, units="hours"))
}
else
{
#Example fall into this case (2015-09-07 06:00-2015-09-07 20:11 ), return diff of FirstdayStart/ FirstdayEnd time
outval<- as.numeric(difftime(FirstdayEnd, FirstdayStart, units="hours"))
}
}
}
else # For case Start and End not in the same day , eg (2015-09-07 03:00", 2015-09-07 06:00)
{
# For case starttime < FirstdayStart Calculate the Firstday period first, 2015-09-03 02:00 2015-09-04 16:00
if (as.POSIXct(x)< FirstdayStart )
{
#Example fall into this case (2015-09-07 06:00 2015-09-08 20:11 ), firstdayPeriod return diff of FirstdayStart/ FirstdayEnd time
firstdayPeriod <- as.numeric(difftime(FirstdayEnd, FirstdayStart, units="hours"))
}
else
{
if (as.POSIXct(x) <= FirstdayEnd)
{
#Example fall into this case (2015-09-07 08:00 2015-09-08 20:11 ), firstdayPeriod = of Startime/ FirstdayEnd time
firstdayPeriod <- as.numeric(difftime(FirstdayEnd, as.POSIXct(x), units="hours"))
}
else
{ #Example fall into this case (2015-09-07 20:00 2015-09-08 20:11 ), firstdayPeriod=0
firstdayPeriod <- c(0)
}
}
#Calculate the last day period
if (as.POSIXct(y) > LastdayEnd)
{
#Example fall into this case (2015-09-07 08:00 2015-09-08 21:00 ), lastdayPeriod = of LastdayEnd/ LastdayStart time
lastdayPeriod <- as.numeric(difftime(FirstdayEnd, FirstdayStart, units="hours"))
}
else
{
if (as.POSIXct(y) >= LastdayStart)
{
#Example fall into this case (2015-09-07 08:00 2015-09-08 18:00 ), lastdayPeriod = of LastdayEnd/ endtime time
lastdayPeriod <- as.numeric(difftime(as.POSIXct(y),LastdayStart, units="hours"))
}
else
{
lastdayPeriod <- c(0)
}
}
#Calculate the overrall time
outval <- lastdayPeriod +firstdayPeriod + (numdays-1)* as.numeric(difftime(FirstdayEnd,FirstdayStart, units="hours"))
}
outval <- round(outval, digits=2)
if (outval < 0)
{
outval <- c(0)
}
outval
}
我得到如下結果
timedf$P0To7 <- apply(timedf[,c(2,3)], 1, function(x) getp1TOp2(x[1], x[2], 0, 7))
timedf$P7To19 <- apply(timedf[,c(2,3)], 1, function(x) getp1TOp2(x[1], x[2], 7, 19))
timedf$P19To24 <- apply(timedf[,c(2,3)], 1, function(x) getp1TOp2(x[1], x[2], 19, 24))
> timedf
spaceNum starttime endtime staytime P0To7 P7To19 P19To24
1 1 2015-09-03 00:00 2015-09-04 20:05 44.08 14.00 24.00 6.08
2 1 2015-09-04 23:18 2015-09-05 05:52 6.57 5.87 0.00 0.70
3 1 2015-09-05 05:59 2015-09-05 06:15 0.27 0.27 0.00 0.00
4 2 2015-09-03 06:19 2015-09-05 16:36 58.28 14.68 33.60 10.00
5 2 2015-09-06 09:03 2015-09-06 09:06 0.05 0.00 0.05 0.00
6 2 2015-09-06 09:10 2015-09-06 20:42 11.53 0.00 9.83 1.70
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