R找到時間段之間的重疊

Question

經過大量的思考和谷歌搜索後，我找不到解決我的問題，我希望你能幫助我。

我有一個ID列可以重複2次以上的大數據框，一個開始和結束日期列將構成一個時間段。我想知道，按照ID進行分組，如果該ID的任何時間段與另一個ID重疊，並且如果是，則通過例如創建新列來標記該ID，說明該ID是否具有重疊。

下面是一個例子的數據幀已經具有期望的新列：

structure(list(ID= c(34L, 34L, 80L, 80L, 81L, 81L, 81L, 94L, 
94L), Start = structure(c(1072911600, 1262300400, 1157061600, 
1277935200, 1157061600, 1277935200, 1157061600, 1075590000, 1285891200 
), class = c("POSIXct", "POSIXt"), tzone = ""), End = structure(c(1262214000, 
1409436000, 1251669600, 1404079200, 1251669600, 1404079200, 1251669600, 
1264892400, 1475193600), class = c("POSIXct", "POSIXt"), tzone = ""), 
    Overlap = c(FALSE, FALSE, FALSE, FALSE, TRUE, TRUE, TRUE, 
    FALSE, FALSE)), .Names = c("ID", "Start", "End", "Overlap" 
), row.names = c(NA, -9L), class = "data.frame") 


ID    Start     End Overlap 
34 2004-01-01 00:00:00 2009-12-31 00:00:00 FALSE 
34 2010-01-01 00:00:00 2014-08-31 00:00:00 FALSE 
80 2006-09-01 00:00:00 2009-08-31 00:00:00 FALSE 
80 2010-07-01 00:00:00 2014-06-30 00:00:00 FALSE 
81 2006-09-01 00:00:00 2009-08-31 00:00:00 TRUE 
81 2010-07-01 00:00:00 2014-06-30 00:00:00 TRUE 
81 2006-09-01 00:00:00 2009-08-31 00:00:00 TRUE 
94 2004-02-01 00:00:00 2010-01-31 00:00:00 FALSE 
94 2010-10-01 02:00:00 2016-09-30 02:00:00 FALSE 

在這種情況下，對於ID「81」有兩個時間週期之間的重疊，所以我想標誌的所有行ID = 81爲TRUE，這意味着在該ID的至少兩行中發現重疊。這只是一個理想的解決方案，但總的來說，我想要做的就是根據ID進行分組時找出重疊部分，因此標記它的方式可以靈活，以便簡化事情。

在此先感謝您的幫助。

Answer 1

我認爲這是你正在尋找的代碼？讓我知道。

data<- structure(list(ID= c(34L, 34L, 80L, 80L, 81L, 81L, 81L, 94L, 
          94L), Start = structure(c(1072911600, 1262300400, 1157061600, 
                 1277935200, 1157061600, 1277935200, 1157061600, 1075590000, 1285891200 
          ), class = c("POSIXct", "POSIXt"), tzone = ""), End = structure(c(1262214000, 
                           1409436000, 1251669600, 1404079200, 1251669600, 1404079200, 1251669600, 
                           1264892400, 1475193600), class = c("POSIXct", "POSIXt"), tzone = ""), 
         Overlap = c(FALSE, FALSE, FALSE, FALSE, TRUE, TRUE, TRUE, 
            FALSE, FALSE)), .Names = c("ID", "Start", "End", "Overlap" 
           ), row.names = c(NA, -9L), class = "data.frame") 

library("dplyr") 
library("lubridate") 

overlaps<- function(intervals){ 
     for(i in 1:(length(intervals)-1)){ 
       for(j in (i+1):length(intervals)){ 
         if(int_overlaps(intervals[i],intervals[j])){ 
           return(TRUE) 
         } 
       } 
     } 
     return(FALSE) 
} 

data %>% 
     mutate(Interval=interval(Start,End))%>% 
     group_by(ID) %>% 
     do({ 
       df<-. 
       ovl<- overlaps(df$Interval) 
       return(data.frame(ID=df$ID[1], ovl)) 
     }) 

而且，我希望有人想出了一個更好的解決方案，以我的overlaps功能..

Answer 2

另一種選擇 - 假設df包含您的數據幀，則：

library(data.table) 
dt <- data.table(df, key=c("Start", "End"))[, `:=`(Overlap=NULL, row=1:nrow(df))] 
overlapping <- unique(foverlaps(dt, dt)[ID==i.ID & row!=i.row, ID]) 
dt[, `:=`(Overlap=FALSE, row=NULL)][ID %in% overlapping, Overlap:=TRUE][order(ID, Start)] 
# ID    Start     End Overlap 
# 1: 34 2004-01-01 00:00:00 2009-12-31 00:00:00 FALSE 
# 2: 34 2010-01-01 00:00:00 2014-08-31 00:00:00 FALSE 
# 3: 80 2006-09-01 00:00:00 2009-08-31 00:00:00 FALSE 
# 4: 80 2010-07-01 00:00:00 2014-06-30 00:00:00 FALSE 
# 5: 81 2006-09-01 00:00:00 2009-08-31 00:00:00 TRUE 
# 6: 81 2006-09-01 00:00:00 2009-08-31 00:00:00 TRUE 
# 7: 81 2010-07-01 00:00:00 2014-06-30 00:00:00 TRUE 
# 8: 94 2004-02-01 00:00:00 2010-01-31 00:00:00 FALSE 
# 9: 94 2010-10-01 02:00:00 2016-09-30 02:00:00 FALSE