如何按日期范围和 ID 合并两个表？

Question

我有一个个人特征表，例如：

person <- data.frame(group.id = c("N","N","P"), person.id = c("A", "B", "C"), strt = c(as.Date(x = "2002-07-01"), as.Date(x = "2003-08-01"), as.Date(x = "2004-06-23")), end = c(as.Date(x = "2003-08-01"), as.Date(x = "2004-09-01"), as.Date(x = "2006-07-01")), c = 1:3, d = 3:5)

以及一组组特征的组表，例如：

group <- data.frame(group.id = c("N", "N", "N", "O", "O", "O", "P", "P", "P"), report.date = c(as.Date(x = "2002-07-01"), as.Date(x = "2002-08-01"), as.Date(x = "2002-09-01")), a = c(1:3), b = c(4:6))

我想按 group.id 和适用的日期范围合并它们，例如：

group2 <- data.frame(group, person.id = c("A", "A", "A", NA, NA, NA, NA, NA, NA), strt = c(as.Date(x = "2002-07-01"), as.Date(x = "2002-07-01"), as.Date(x = "2002-07-01"), NA, NA, NA, NA, NA, NA), end = c(as.Date(x = "2003-08-01"), as.Date(x = "2003-08-01"), as.Date(x = "2003-08-01"), NA, NA, NA, NA, NA, NA), c = c(1, 1, 1, NA, NA, NA, NA, NA, NA), d = c(3, 3, 3, NA, NA, NA, NA, NA, NA))

  group.id report.date a b person.id       strt        end  c  d
1        N  2002-07-01 1 4         A 2002-07-01 2003-08-01  1  3
2        N  2002-08-01 2 5         A 2002-07-01 2003-08-01  1  3
3        N  2002-09-01 3 6         A 2002-07-01 2003-08-01  1  3
4        O  2002-07-01 1 4      <NA>       <NA>       <NA> NA NA
5        O  2002-08-01 2 5      <NA>       <NA>       <NA> NA NA
6        O  2002-09-01 3 6      <NA>       <NA>       <NA> NA NA
7        P  2002-07-01 1 4      <NA>       <NA>       <NA> NA NA
8        P  2002-08-01 2 5      <NA>       <NA>       <NA> NA NA
9        P  2002-09-01 3 6      <NA>       <NA>       <NA> NA NA

有没有人建议如何在 R 中做到这一点？

Answer 1

person <- data.frame(group_id = c("N","N","P"), person_id = c("A", "B", "C"), strt = c(as.Date(x = "2002-07-01"), as.Date(x = "2003-08-01"), as.Date(x = "2004-06-23")), end = c(as.Date(x = "2003-08-01"), as.Date(x = "2004-09-01"), as.Date(x = "2006-07-01")), c = 1:3, d = 3:5)

group <- data.frame(group_id = c("N", "N", "N", "O", "O", "O", "P", "P", "P"), report_date = c(as.Date(x = "2002-07-01"), as.Date(x = "2002-08-01"), as.Date(x = "2002-09-01")), a = c(1:3), b = c(4:6))

group2 <- data.frame(group, person_id = c("A", "A", "A", NA, NA, NA, NA, NA, NA), strt = c(as.Date(x = "2002-07-01"), as.Date(x = "2002-07-01"), as.Date(x = "2002-07-01"), NA, NA, NA, NA, NA, NA), end = c(as.Date(x = "2003-08-01"), as.Date(x = "2003-08-01"), as.Date(x = "2003-08-01"), NA, NA, NA, NA, NA, NA), c = c(1, 1, 1, NA, NA, NA, NA, NA, NA), d = c(3, 3, 3, NA, NA, NA, NA, NA, NA))


library(sqldf)


sqldf("select a.*, b.* from 'group' a left join person b on a.group_id = b.group_id and (a.report_date >= b.strt and a.report_date <= b.end)")

  group_id report_date a b group_id person_id       strt        end  c  d
1        N  2002-07-01 1 4        N         A 2002-07-01 2003-08-01  1  3
2        N  2002-08-01 2 5        N         A 2002-07-01 2003-08-01  1  3
3        N  2002-09-01 3 6        N         A 2002-07-01 2003-08-01  1  3
4        O  2002-07-01 1 4     <NA>      <NA>       <NA>       <NA> NA NA
5        O  2002-08-01 2 5     <NA>      <NA>       <NA>       <NA> NA NA
6        O  2002-09-01 3 6     <NA>      <NA>       <NA>       <NA> NA NA
7        P  2002-07-01 1 4     <NA>      <NA>       <NA>       <NA> NA NA
8        P  2002-08-01 2 5     <NA>      <NA>       <NA>       <NA> NA NA
9        P  2002-09-01 3 6     <NA>      <NA>       <NA>       <NA> NA NA

请注意，group 是保留字，因此我必须将其放在单引号中才能将其用作表格。我还将列名中的.s 更改为_ 以避免出现问题，但您可以保留. 并引用所有列名。