data <- data.frame(
record_id = c(1,1,1,2,3,4,4,5,6,7,8,8,9,10,10,10),
visit_date = c("2020-09-24", "2020-12-05", "2021-03-01", "2021-10-03", "2021-10-01", "2021-10-05", "NA", "2021-08-25", "2021-09-19", "2021-10-01", "2021-09-27", "2021-09-07",
"2021-10-03", "2021-10-08", "2022-03-22", "2022-07-12"),
repeat_instance = c(0,1,2,0,0,0,1,0,0,0,0,1,0,0,1,2),
Time_Since_Appointment = c("NA", "72d 1H 0M 0S", "86d 0H 0M 0S", "NA", "NA", "NA", "NA",
"NA", "NA", "NA", "NA", "1076d 0H 0M 0S", "NA", "NA", "165d 0H 0M 0S", "112d 0H 0M 0S")
)
data1 <- data %>%
group_by(record_id) %>%
mutate(Time_Since_Appointment = visit_date - lag(visit_date))
我有兴趣计算一个新的变量，该变量将使用最新的repeat_instance来捕捉自今天以来的天数或周数，每个记录都不同，或者如果没有repeat_intance，则计算自基线就诊日期以来的时间。
非常感谢，如果需要提供更多信息，请告诉我。
首先，我会取每个ID的最大(最近(日期。接下来，我将使用difftime((函数来查找最近一次访问和今天访问之间的差异。在这种情况下，函数today((调用今天的日期，但可以用硬编码的日期代替。difftime((函数允许您选择想要的单位。在这种情况下，我计算了几天和几周的差异。
使用data.table包：
data <- data.table(
record_id = c(1,1,1,2,3,4,4,5,6,7,8,8,9,10,10,10),
visit_date = c("2020-09-24", "2020-12-05", "2021-03-01", "2021-10-03",
"2021-10-01", "2021-10-05", "NA", "2021-08-25",
"2021-09-19", "2021-10-01", "2021-09-27", "2021-09-07",
"2021-10-03", "2021-10-08", "2022-03-22", "2022-07-12"))
data <- data[,.(last_visit = max(as.Date(visit_date,"%Y-%m-%d"))),by=.(record_id)]
data <- data[,days_since_last_visit := difftime(today(),last_visit,units="days")]
data <- data[,weeks_since_last_visit := difftime(today(),last_visit,units="weeks")]
data
record_id last_visit days_since_last_visit weeks_since_last_visit
1: 1 2021-03-01 546 days 78.000000 weeks
2: 2 2021-10-03 330 days 47.142857 weeks
3: 3 2021-10-01 332 days 47.428571 weeks
4: 4 <NA> NA days NA weeks
5: 5 2021-08-25 369 days 52.714286 weeks
6: 6 2021-09-19 344 days 49.142857 weeks
7: 7 2021-10-01 332 days 47.428571 weeks
8: 8 2021-09-27 336 days 48.000000 weeks
9: 9 2021-10-03 330 days 47.142857 weeks
10: 10 2022-07-12 48 days 6.857143 weeks