如何使用数据表在 R 中获取自上次日期(滞后)以来的天数时差

发布于04月05日

person_id  diag_date concept_id  event diff_prev_event
1:         1 2012-01-15    4265600 comorb         NA secs
2:         1 2012-01-15     201820 comorb          0 secs
3:         1 2012-03-15    4265600 comorb    5184000 secs
4:         2 2012-03-15     201820 comorb         NA secs
5:         2 2012-06-22     201820 comorb    8553600 secs
6:         2 2012-06-22    4265600 comorb          0 secs

我试图计算每个人自上次活动以来的天数.我遇到了两个问题.

时差以秒为单位显示.我需要休息几天.(5184000秒=30天)
如果两天是相同的日期，那么第二天显示0，此时它应该查看不同的日期.第5行和第6行的日期相同，因此它们的日期差相同.

这是我试过的代码:

dt[order(diag_date),diff_prev_event := difftime(diag_date, lag( diag_date)), by = c("person_id") ]

推荐答案

指定units

library(数据.table)
dt[order(diag_date),diff_prev_event := difftime(diag_date, 
     lag( diag_date), units = 'days'), by = c("person_id") ]

然后，我们按"person_id"和"diag_date"分组，如果有多行，则将值更改为max

dt[, diff_prev_event := if(.N > 1) max(diff_prev_event, 
    na.rm = TRUE) else diff_prev_event, .(person_id, diag_date)]
> dt
   person_id  diag_date concept_id  event diff_prev_event
       <int>     <Date>      <int> <char>      <difftime>
1:         1 2012-01-15    4265600 comorb          0 days
2:         1 2012-01-15     201820 comorb          0 days
3:         1 2012-03-15    4265600 comorb         60 days
4:         2 2012-03-15     201820 comorb         NA days
5:         2 2012-06-22     201820 comorb         99 days
6:         2 2012-06-22    4265600 comorb         99 days

-输出

数据

dt <- structure(list(person_id = c(1L, 1L, 1L, 2L, 2L, 2L), diag_date = structure(c(15354, 
15354, 15414, 15414, 15513, 15513), class = "Date"), concept_id = c(4265600L, 
201820L, 4265600L, 201820L, 201820L, 4265600L), event = c("comorb", 
"comorb", "comorb", "comorb", "comorb", "comorb")), row.names = c(NA, 
-6L), class = c("数据.table", "数据.frame"))