几天前，我打开了此相关线程：Time-interval overlap match by group
但是，现在我要处理一个事实，我需要重叠多个时间间隔列，并在出现这种情况时返回标志= 1的第一个row_number值。
例如，我有以下df：
id flag row_number time_1 time_2 result
1 1 1 2001-04-01 UTC--2001-05-01 UTC 1960-01-01 UTC--1962-01-01 UTC NA
1 1 2 2007-08-01 UTC--2007-12-01 UTC 1980-01-01 UTC--1982-01-01 UTC NA
1 1 3 2010-03-01 UTC--2011-03-01 UTC 1949-01-01 UTC--1951-01-01 UTC NA
1 0 4 2001-04-15 UTC--2001-04-20 UTC 1981-01-01 UTC--1983-01-01 UTC NA
1 0 5 2001-04-17 UTC--2001-05-15 UTC 1959-01-01 UTC--1961-01-01 UTC 1
1 0 6 2007-09-01 UTC--2007-12-01 UTC 1980-01-01 UTC--1983-01-01 UTC 2
1 0 7 2011-01-01 UTC--2011-03-05 UTC 1994-01-01 UTC--1996-01-01 UTC NA
1 0 8 2018-01-01 UTC--2017-12-01 UTC 1949-01-01 UTC--1951-01-01 UTC NA
使用以下代码创建：
library(dplyr)
library(purrr)
library(lubridate)
df <- data.frame(id=c(1, 1, 1, 1, 1, 1, 1, 1),
flag=c(1, 1, 1, 0, 0, 0, 0, 0),
row_number=c(1,2,3,4,5,6,7,8),
time_1=c(interval(ymd(20010401), ymd(20010501)),
interval(ymd(20070801), ymd(20071201)),
interval(ymd(20100301), ymd(20110301)),
interval(ymd(20010415), ymd(20010420)),
interval(ymd(20010417), ymd(20010515)),
interval(ymd(20070801), ymd(20071201)),
interval(ymd(20110101), ymd(20110305)),
interval(ymd(20180101), ymd(20171201))),
time_2=c(interval(ymd(19600101), ymd(19620101)),
interval(ymd(19800101), ymd(19820101)),
interval(ymd(19490101), ymd(19510101)),
interval(ymd(19810101), ymd(19830101)),
interval(ymd(19590101), ymd(19610101)),
interval(ymd(19800101), ymd(19820101)),
interval(ymd(19940101), ymd(19960101)),
interval(ymd(19490101), ymd(19510101))),
result = c(NA, NA, NA, NA, 1, 2, NA, NA))
这是，我需要找到标记为0的行的time_1和time_2与标记为1的行的所有time_1和time_2变量重叠的地方。
结果应该是具有row_number值且其标记为0的行与标记1的行的time_1和time_2间隔重叠的行之间第一次匹配的值的列。为此，我尝试了lubridate包中的int_overlap（）函数。
使用此代码，我可以利用map_int（）函数确定标志= 0的一行与标志[=] 1的any行之间是否存在time_1重叠
library(tidyverse)
library(lubridate)
df %>%
group_by(id) %>%
mutate(value = ifelse(flag == 0, map_int(time_1, ~ any(int_overlaps(.x, time_1[flag == 1]))), NA))
一个相关的问题，可能会有所帮助：R Find overlap among time periods
编辑：我想获得一个用row_number变量标识的列，该列是具有时间1和时间2与值0行重叠的值的第一标志1行。
id flag row_number time_1 time_2 result
1 1 1 2001-04-01 UTC--2001-05-01 UTC 1960-01-01 UTC--1962-01-01 UTC NA
1 0 5 2001-04-17 UTC--2001-05-15 UTC 1959-01-01 UTC--1961-01-01 UTC 1
例如row_number 1和5满足条件。结果是一个整数列，指示row_number 5（标志0行）具有time_1和time_2与row_number 1（标志1）重叠。
希望这可以澄清。
我很确定我不完全了解您的要求。在您的数据中，time_1和time_2相距很远，并且从未相交。那是对的吗？
也许这会使球滚动。这是您想要的吗？
df %>%
mutate(test = case_when(
int_overlaps(time_1,time_2) & flag == 1 ~ T,
int_overlaps(time_1,time_2) & flag == 0 ~ F,
T ~ NA
))
这里是通过两次执行重叠联接使用data.table的选项：
setkey(setDT(df), id, time_1_start, time_1_end)
ol1 <- foverlaps(df, df, nomatch=0L)[
row_number!=i.row_number & i.flag==0L & flag==1L,
.(id, irn=i.row_number, rn=row_number, flag=i.flag,
time_2_start=i.time_2_start, time_2_end=i.time_2_end)]
setkey(df, id, time_2_start, time_2_end)
setkey(ol1, id, time_2_start, time_2_end)
olaps <- foverlaps(ol1, df)[row_number!=irn & row_number==rn & i.flag==0L & flag==1L,
.(id, irn, xrn=row_number)]
df[olaps, on=.(id, row_number=irn), res := xrn]
setorder(df, row_number)
df
输出：
id flag row_number time_1_start time_1_end time_2_start time_2_end res
1: 1 1 1 2001-04-01 2001-05-01 1960-01-01 1962-01-01 NA
2: 1 1 2 2007-08-01 2007-12-01 1980-01-01 1982-01-01 NA
3: 1 1 3 2010-03-01 2011-03-01 1949-01-01 1951-01-01 NA
4: 1 0 4 2001-04-15 2001-04-20 1981-01-01 1983-01-01 NA
5: 1 0 5 2001-04-17 2001-05-15 1959-01-01 1961-01-01 1
6: 1 0 6 2007-08-01 2007-12-01 1980-01-01 1982-01-01 2
7: 1 0 7 2011-01-01 2011-03-05 1994-01-01 1996-01-01 NA
8: 1 0 8 2017-12-01 2018-01-01 1949-01-01 1951-01-01 NA
数据：
library(data.table)
dtfun <- function(x) as.IDate(x, format="%Y%m%d")
df <- data.frame(id=c(1, 1, 1, 1, 1, 1, 1, 1),
flag=c(1, 1, 1, 0, 0, 0, 0, 0),
row_number=c(1,2,3,4,5,6,7,8),
time_1_start=dtfun(c("20010401","20070801","20100301","20010415",
"20010417","20070801","20110101","20171201")),
time_1_end=dtfun(c("20010501","20071201","20110301","20010420","
20010515","20071201","20110305","20180101")),
time_2_start=dtfun(c("19600101","19800101","19490101","19810101",
"19590101","19800101","19940101","19490101")),
time_2_end=dtfun(c("19620101","19820101","19510101","19830101",
"19610101","19820101","19960101","19510101")))