所以我一直在努力解决这个问题，但是我不知道该怎么做。
这是一个例子：
ID Hosp. date Discharge date
1 2006-02-02 2006-02-04
1 2006-02-04 2006-02-18
1 2006-02-22 2006-03-24
1 2008-08-09 2008-09-14
2 2004-01-03 2004-01-08
2 2004-01-13 2004-01-15
2 2004-06-08 2004-06-28
我想要的是一种通过ID组合行的方法，前提是该异议日期与医院相同。日期（或+ -7天）在下一行中。所以看起来像这样：
ID Hosp. date Discharge date
1 2006-02-02 2006-03-24
1 2008-08-09 2008-09-14
2 2004-01-03 2004-01-15
2 2004-06-08 2004-06-28
最佳答案
使用data.table -package：
# load the package
library(data.table)
# convert to a 'data.table'
setDT(d)
# make sure you have the correct order
setorder(d, ID, Hosp.date)
# summarise
d[, grp := cumsum(Hosp.date > (shift(Discharge.date, fill = Discharge.date[1]) + 7))
, by = ID
][, .(Hosp.date = min(Hosp.date), Discharge.date = max(Discharge.date))
, by = .(ID,grp)]
你得到：
ID grp Hosp.date Discharge.date
1: 1 0 2006-02-02 2006-03-24
2: 1 1 2008-08-09 2008-09-14
3: 2 0 2004-01-03 2004-01-15
4: 2 1 2004-06-08 2004-06-28
dplyr的逻辑相同：library(dplyr)
d %>%
arrange(ID, Hosp.date) %>%
group_by(ID) %>%
mutate(grp = cumsum(Hosp.date > (lag(Discharge.date, default = Discharge.date[1]) + 7))) %>%
group_by(grp, add = TRUE) %>%
summarise(Hosp.date = min(Hosp.date), Discharge.date = max(Discharge.date))
使用的数据：
d <- structure(list(ID = c(1L, 1L, 1L, 1L, 2L, 2L, 2L),
Hosp.date = structure(c(13181, 13183, 13201, 14100, 12420, 12430, 12577), class = "Date"),
Discharge.date = structure(c(13183, 13197, 13231, 14136, 12425, 12432, 12597), class = "Date")),
.Names = c("ID", "Hosp.date", "Discharge.date"), class = "data.frame", row.names = c(NA, -7L))