R处理药物处方数据：区分同ID同药处方段并计算随访总费用

基础信息 #

• 分析场景为带日期维度的药物处方数据分析，示例数据结构代码如下：

structure(list(id = c(4L, 4L, 5L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 
6L), claim = c(1L, 2L, 1L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L), start_date = structure(c(12267, 
12298, 12626, 12818, 12846, 12877, 12907, 12938, 13091, 13121, 
13152), class = "Date"), drug = c("a", "a", "a", "b", "b", "a", 
"a", "a", "a", "a", "b"), total.price = c(100L, 100L, 100L, 100L, 
100L, 100L, 100L, 100L, 100L, 100L, 100L), dose = c(1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), IBD = c("CD", "CD", "CD", "CD", 
"CD", "CD", "CD", "CD", "CD", "CD", "CD"), naivety = c(1, 1, 
1, 1, 1, 1, 1, 1, 1, 1, 1), diff_drug = c(0, 0, 0, 0, 0, 1, 0, 
0, 0, 0, 1)), class = c("grouped_df", "tbl_df", "tbl", "data.frame"
), row.names = c(NA, -11L), groups = structure(list(id = 4:6, 
    .rows = structure(list(1:2, 3L, 4:11), ptype = integer(0), class = c("vctrs_list_of", 
    "vctrs_vctr", "list"))), row.names = c(NA, -3L), class = c("tbl_df", 
"tbl", "data.frame"), .drop = TRUE))

• 初始分析目标：计算每个id对应各段药物处方的开始日期与结束日期，判断逻辑为：
  • 若变量discontinuation==1，该受试者对应处方的结束日期取停药日期
  • 若discontinuation == 0，结束日期取处方记录的最晚start_date（即max(start_date)）

已尝试的实现代码 #

bio_naive <- bio_naive %>% arrange(id,start_date) %>%  mutate(Diff = lead(start_date) - (start_date))

bio_naive <- bio_naive %>% group_by(id, drug) %>% mutate(discontinuation = ifelse(Diff > 90, '1', '0')) 

bio_naive$discontinuation[is.na(bio_naive $discontinuation)] <- 0

bio_naive %>%
  group_by(id,drug) %>% 
  summarise(discont=max(discontinuation), start = min(start_date, na.rm = TRUE), dc_final_date = if_else(any(discontinuation == 1), start_date[match(1, discontinuation)], max(start_date)))

现存问题 #

上述代码运行后未达到预期效果：id为6、drug为b的两段独立药物处方被错误合并统计，无法正确拆分两段独立的drug b处方记录。

待解决问题 #

1. 如何区分id=6对应的两段drug b处方，实现同id下同种药物不同独立处方段的正确拆分？
2. 如何编写代码，按id、药物、处方段分组，计算随访期（从处方开始日期到dc_final_date）内的total.price总和？

预期结果参考 #

内容的提问来源于stack exchange，提问作者Moon