我有一个数据集,其中包含一个人离开网络时的日期。一个人可以多次离开网络,因为他们可能在离开网络后再次加入网络。以下代码复制该场景。如何在R中构建高效循环查找
library(data.table)
Leaving_Date<- data.table(Id= c(1,2,3,4,3,5),Date = as.Date(
c("2017-01-01","2017-02-03","2017-01-01","2017-03-10","2017-02-09","2017-02-05")))
(IDS在此表中重复多次作为一个人可以离开网络多次给定他们又加入了它)
> Leaving_Date
Id Date
1: 1 2017-01-01
2: 2 2017-02-03
3: 3 2017-01-01
4: 4 2017-03-10
5: 3 2017-02-09
6: 5 2017-02-05
我有另外一个数据集给的日期,当特定的人之后这可以在他们离开网络之前或之后进行。以下代码复制该场景。
FOLLOWUPs <- data.table(Id = c(1,2,3,2,2,3,3,4,1,5),
Date =as.Date(c("2016-10-01","2017-02-04",
"2017-01-17","2017-02-23", "2017-03-03",
"2017-02-10","2017-02-11","2017-01-01",
"2017-01-15","2017-01-01")))
> FOLLOWUPs
Id Date
1: 1 2016-10-01
2: 2 2017-02-04
3: 3 2017-01-17
4: 2 2017-02-23
5: 2 2017-03-03
6: 3 2017-02-10
7: 3 2017-02-11
8: 4 2017-01-01
9: 1 2017-01-15
10: 5 2017-01-01
现在我想查找在Leaving_Date每种情况下,发现当他们进行随访日期和创建三列(SevenDay,FourteenDay,ThirtyDay),表明后续的时间段中的0(柜面,如果有任何)和1秒。我使用下面的代码:
SEVENDAY_FOLLOWUP <- vector()
FOURTEEN_FOLLOWUP <- vector()
THIRTYDAY_FOLLOWUP <- vector()
for(i in 1:nrow(Leaving_Date)){
sub_data <- FOLLOWUPs[Id== Leaving_Date[i,1]]
if(nrow(sub_data[Date > Leaving_Date[i,Date] &
Date < (Leaving_Date[i,Date]+7)])== 0){
SEVENDAY_FOLLOWUP <- rbind(SEVENDAY_FOLLOWUP,0)
}
else{
SEVENDAY_FOLLOWUP <- rbind(SEVENDAY_FOLLOWUP,1)
}
if(nrow(sub_data[Date > Leaving_Date[i,Date] &
Date < (Leaving_Date[i,Date]+14)])== 0){
FOURTEEN_FOLLOWUP <- rbind(FOURTEEN_FOLLOWUP,0)
}
else{
FOURTEEN_FOLLOWUP <- rbind(FOURTEEN_FOLLOWUP,1)
}
if(nrow(sub_data[Date > Leaving_Date[i,Date] &
Date < (Leaving_Date[i,Date]+30)])== 0){
THIRTYDAY_FOLLOWUP <- rbind(THIRTYDAY_FOLLOWUP,0)
}
else{
THIRTYDAY_FOLLOWUP <- rbind(THIRTYDAY_FOLLOWUP,1)
}
}
Leaving_Date$SEVENDAY <- as.vector(SEVENDAY_FOLLOWUP)
Leaving_Date$FOURTEENDAY <- as.vector(FOURTEEN_FOLLOWUP)
Leaving_Date$THIRTYDAY <- as.vector(THIRTYDAY_FOLLOWUP)
最终数据
> Leaving_Date
Id Date SEVENDAY FOURTEENDAY THIRTYDAY
1: 1 2017-01-01 0 0 1
2: 2 2017-02-03 1 1 1
3: 3 2017-01-01 0 0 1
4: 4 2017-03-10 0 0 0
5: 3 2017-02-09 1 1 1
6: 5 2017-02-05 0 0 0
此代码是非常低效的,因为我要运行它100K的观察,它需要大量的时间。有没有任何有效的方法来做到这一点。
您可能想要阅读[R Inferno](http://www.burns-stat.com/pages/Tutor/R_inferno.pdf)的第二个圆圈 – shayaa
@Frank我编辑了它 –