R查找时间段之间的重叠

Question

经过大量思考和谷歌搜索，我找不到解决问题的方法，希望您能帮助我。

我有一个大型数据框，其中包含一个可以重复超过 2 次的 ID 列，一个构成时间段的开始和结束日期列。我想按 ID 分组找出该 ID 的任何时间段是否与另一个时间段重叠，如果是，则通过创建一个新列来标记它，例如，说明该 ID 是否重叠。

这是一个已经包含所需新列的示例数据框：

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

在这种情况下，对于 ID“81”，两个时间段之间存在重叠，因此我想将 ID = 81 的所有行标记为 TRUE，这意味着在该 ID 的至少两行中发现了重叠.这只是一个理想的解决方案，但总的来说，我要做的就是在按 ID 分组时找出重叠，因此标记它的方式可以灵活，以防简化事情。

提前感谢您的帮助。

Answer 1

另一种选择 - 假设 df 包含您的数据框，那么：

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

Answer 2

我认为这是您正在寻找的代码？告诉我。

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))
       })

另外，我希望有人对我的overlaps 函数提出更优雅的解决方案..