我有一個數據集,其中包含一個人離開網絡時的日期。一個人可以多次離開網絡,因爲他們可能在離開網絡後再次加入網絡。以下代碼複製該場景。如何在R中構建高效循環查找
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在此表中重複多次作爲一個人可以離開網絡多次給定他們又加入了它)
> 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
我有另外一個數據集給的日期,當特定的人之後這可以在他們離開網絡之前或之後進行。以下代碼複製該場景。
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
現在我想查找在Leaving_Date每種情況下,發現當他們進行隨訪日期和創建三列(SevenDay,FourteenDay,ThirtyDay),表明後續的時間段中的0(櫃面,如果有任何)和1秒。我使用下面的代碼:
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)
最終數據
> 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
此代碼是非常低效的,因爲我要運行它100K的觀察,它需要大量的時間。有沒有任何有效的方法來做到這一點。
您可能想要閱讀[R Inferno](http://www.burns-stat.com/pages/Tutor/R_inferno.pdf)的第二個圓圈 – shayaa
@Frank我編輯了它 –