R 统计data.table中事件发生后的天数

我们可以使用

zoo

中的

na.locf

创建一个新的日期列，其中

count>0

的最后一天的日期向下填充，并取该列的

lag

，因为如果给定日期本身具有

count>0

，我们不希望返回0天。相反，我们要从上一个非零日期开始计算天数。最后，我们找到

date

和

date2

之间的

difftime

：

c(NA, 1, 2, 4, 1, 1, 1, 1, 1, 2, 5, 6, 9, 10, 2, 3, NA, NA, 2, 2, 3, 1, 2, 1, 1, 2, 5, 1, 4, 1, 2, 3)

结果：

library(data.table)
library(zoo)

dt[,count2 := difftime(date, na.locf(lag(ifelse(count > 0, date, NA)), na.rm = FALSE)), by = id]

library(data.table)
library(zoo)

dt[,count2 := difftime(date, na.locf(lag(ifelse(count > 0, date, NA)), na.rm = FALSE)), by = id]

      id       date count  count2
 1: 0151 2011-08-09     7 NA days
 2: 0151 2011-08-10     1  1 days
 3: 0151 2011-08-12     0  2 days
 4: 0151 2011-08-14     4  4 days
 5: 0151 2011-08-15     1  1 days
 6: 0151 2011-08-16     4  1 days
 7: 0151 2011-08-17     2  1 days
 8: 0151 2011-08-18     1  1 days
 9: 0151 2011-08-19     0  1 days
10: 0151 2011-08-20     0  2 days
11: 0151 2011-08-23     0  5 days
12: 0151 2011-08-24     0  6 days
13: 0151 2011-08-27     0  9 days
14: 0151 2011-08-28     1 10 days
15: 0151 2011-08-30     0  2 days
16: 0151 2011-08-31     1  3 days
17: 4615 2011-08-09     0 NA days
18: 4615 2011-08-10     1 NA days
19: 4615 2011-08-12     3  2 days
20: 4615 2011-08-14     0  2 days
21: 4615 2011-08-15     1  3 days
22: 4615 2011-08-16     0  1 days
23: 4615 2011-08-17     5  2 days
24: 4615 2011-08-18     1  1 days
25: 4615 2011-08-19     0  1 days
26: 4615 2011-08-20     0  2 days
27: 4615 2011-08-23     5  5 days
28: 4615 2011-08-24     0  1 days
29: 4615 2011-08-27     1  4 days
30: 4615 2011-08-28     2  1 days
31: 4615 2011-08-30     0  2 days
32: 4615 2011-08-31     1  3 days
      id       date count  count2