R 定义具有重叠和串联时间间隔的展示周期/情节

我试图通过处方确定药物暴露的时期/发作。如果这些处方分开30天，则被视为一个新的暴露期/事件。处方可以在特定时间重叠，也可以是连续的。如果连续两次处方的间隔天数之和大于30天，则不视为新的发作

我有这样的数据：

id = c(rep(1,3), rep(2,6), rep(3,5))
start = as.Date(c("2017-05-10", "2017-07-28", "2017-11-23", "2017-01-27", "2017-10-02", "2018-05-14", "2018-05-25", "2018-11-26", "2018-12-28", "2016-01-01", "2016-03-02", "2016-03-20", "2016-04-25", "2016-06-29"))
end = as.Date(c("2017-07-27", "2018-01-28", "2018-03-03", "2017-04-27", "2018-05-13", "2018-11-14", "2018-11-25", "2018-12-27", "2019-06-28", "2016-02-15", "2016-03-05", "2016-03-24", "2016-04-29", "2016-11-01"))

DT = data.table(id, start, end)

DT
    id      start        end
 1:  1 2017-05-10 2017-07-27
 2:  1 2017-07-28 2018-01-28
 3:  1 2017-11-23 2018-03-03
 4:  2 2017-01-27 2017-04-27
 5:  2 2017-10-02 2018-05-13
 6:  2 2018-05-14 2018-11-14
 7:  2 2018-05-25 2018-11-25
 8:  2 2018-11-26 2018-12-27
 9:  2 2018-12-28 2019-06-28
10:  3 2016-01-01 2016-02-15
11:  3 2016-03-02 2016-03-05
12:  3 2016-03-20 2016-03-24
13:  3 2016-04-25 2016-04-29
14:  3 2016-06-29 2016-11-01

我计算了开始观察和最后观察的差异（最后几天）

这将显示何时发生重叠（负值）或不发生重叠（正值）。我认为在这里使用ifelse/fcase语句是个坏主意，我不喜欢这样做

我认为这项工作的良好输出应该是：

    id      start        end last_diffdays noexp_days period
 1:  1 2017-05-10 2017-07-27        0 days          0      1
 2:  1 2017-07-28 2018-01-28        1 days          1      1
 3:  1 2017-11-23 2018-03-03      -66 days          0      1
 4:  2 2017-01-27 2017-04-27        0 days          0      1
 5:  2 2017-10-02 2018-05-13      158 days        158      2
 6:  2 2018-05-14 2018-11-14        1 days          1      2
 7:  2 2018-05-25 2018-11-25     -173 days          0      2
 8:  2 2018-11-26 2018-12-27        1 days          1      2
 9:  2 2018-12-28 2019-06-28        1 days          1      2
10:  3 2016-01-01 2016-02-15        0 days          0      1
11:  3 2016-03-02 2016-03-05       16 days         16      1
12:  3 2016-03-20 2016-03-24       15 days         15      1
13:  3 2016-04-25 2016-04-29       32 days         32      2
14:  3 2016-06-29 2016-11-01       61 days         61      3

我手工计算了处方前未暴露的天数（noexp_天数）

我不知道我的路径是否正确，但我认为我需要计算noexp\u days变量，然后生成一个

cumsum（（noexp\u days）>30）+1

如果有一个更好的解决方案，我没有看到或任何其他可能性，我没有考虑过，我将感谢阅读他们

提前感谢您的帮助！：）

试试看：

library(data.table)

DT[, noexp_days := pmax(as.integer(last_diffdays), 0)]
DT[, period := cumsum(noexp_days > 30) + 1, id]
DT
#    id      start        end last_diffdays noexp_days period
# 1:  1 2017-05-10 2017-07-27        0 days          0      1
# 2:  1 2017-07-28 2018-01-28        1 days          1      1
# 3:  1 2017-11-23 2018-03-03      -66 days          0      1
# 4:  2 2017-01-27 2017-04-27        0 days          0      1
# 5:  2 2017-10-02 2018-05-13      158 days        158      2
# 6:  2 2018-05-14 2018-11-14        1 days          1      2
# 7:  2 2018-05-25 2018-11-25     -173 days          0      2
# 8:  2 2018-11-26 2018-12-27        1 days          1      2
# 9:  2 2018-12-28 2019-06-28        1 days          1      2
#10:  3 2016-01-01 2016-02-15        0 days          0      1
#11:  3 2016-03-02 2016-03-05       16 days         16      1
#12:  3 2016-03-20 2016-03-24       15 days         15      1
#13:  3 2016-04-25 2016-04-29       32 days         32      2
#14:  3 2016-06-29 2016-11-01       61 days         61      3

library(data.table)

DT[, noexp_days := pmax(as.integer(last_diffdays), 0)]
DT[, period := cumsum(noexp_days > 30) + 1, id]
DT
#    id      start        end last_diffdays noexp_days period
# 1:  1 2017-05-10 2017-07-27        0 days          0      1
# 2:  1 2017-07-28 2018-01-28        1 days          1      1
# 3:  1 2017-11-23 2018-03-03      -66 days          0      1
# 4:  2 2017-01-27 2017-04-27        0 days          0      1
# 5:  2 2017-10-02 2018-05-13      158 days        158      2
# 6:  2 2018-05-14 2018-11-14        1 days          1      2
# 7:  2 2018-05-25 2018-11-25     -173 days          0      2
# 8:  2 2018-11-26 2018-12-27        1 days          1      2
# 9:  2 2018-12-28 2019-06-28        1 days          1      2
#10:  3 2016-01-01 2016-02-15        0 days          0      1
#11:  3 2016-03-02 2016-03-05       16 days         16      1
#12:  3 2016-03-20 2016-03-24       15 days         15      1
#13:  3 2016-04-25 2016-04-29       32 days         32      2
#14:  3 2016-06-29 2016-11-01       61 days         61      3