英文:
Assign Missing End Dates based on another date in group
问题
我有一个类似的数据集，其中日期结束列有一些NA值，我想要用适当的日期填充它们。我的意图是查看组中的下一个日期开始变量，并用前一个日期结束填充日期，日期减去一天，如果没有可接受的日期填充，则为日期开始添加一年以用作日期结束。
我想请问如何解决这个问题？
英文:
I have a dataset similar to this
id <- c("A1", "A1", "A1", "A1", "B2","B2","C1", "C2")date_start <- as.Date(c("2022-01-01", "2022-01-06", "2022-01-10", "2022-05-23","2023-01-02", "2022-07-19","2022-12-07", "2023-03-23"))date_end <- as.Date(c("2022-01-05", NA, "2022-04-13", NA, NA, "2023-10-06", "2023-01-18", "2023-08-15"))df <- data.frame(id, date_start, date_end)tibble(df)
You can see that the date_end column has NA values which I would like to fill in with an appropriate date. The intention is to look at the next date_start variable in the group and fill in the previous date_end with a date less a day and if there is no acceptable date to fill add a year to the date_start for the date_end instead.
I can not come up with a solution, and was hoping for some insight.
答案1
得分: 2
df %>%
按id分组 %>%
变异(date_end=coalesce(date_end,lead(date_start-1),date_start+years(1)))
A tibble: 8 × 3
分组: id [4]
id date_start date_end
1 A1 2022-01-01 2022-01-05
2 A1 2022-01-06 2022-01-09
3 A1 2022-01-10 2022-04-13
4 A1 2022-05-23 2023-05-23
5 B2 2023-01-02 2022-07-18
6 B2 2022-07-19 2023-10-06
7 C1 2022-12-07 2023-01-18
8 C2 2023-03-23 2023-08-15
如果您有更新的tidyverse包：
df %>%
变异(date_end =
coalesce(date_end, lead(date_start-1), date_start + years(1)),
.by = id)
英文:
df %>%group_by(id) %>%mutate(date_end=coalesce(date_end,lead(date_start-1),date_start+years(1)))# A tibble: 8 × 3# Groups: id [4]id date_start date_end<chr> <date> <date>1 A1 2022-01-01 2022-01-052 A1 2022-01-06 2022-01-093 A1 2022-01-10 2022-04-134 A1 2022-05-23 2023-05-235 B2 2023-01-02 2022-07-186 B2 2022-07-19 2023-10-067 C1 2022-12-07 2023-01-188 C2 2023-03-23 2023-08-15
if you have the newer tidyverse package:
df %>%mutate(date_end =coalesce(date_end, lead(date_start-1), date_start + years(1)),.by = id)
答案2
得分: 1
这是一个包含if_else语句的解决方案：
library(dplyr) #> 1.1.0df %>%mutate(date_end = if_else(is.na(date_end),lead(date_start - days(1), default = last(date_start) + years(1)),date_end), .by=id)id date_start date_end1 A1 2022-01-01 2022-01-052 A1 2022-01-06 2022-01-093 A1 2022-01-10 2022-04-134 A1 2022-05-23 2023-05-235 B2 2023-01-02 2022-07-186 B2 2022-07-19 2023-10-067 C1 2022-12-07 2023-01-188 C2 2023-03-23 2023-08-15
希望这对你有帮助。
英文:
Here is a solution with an if_else statmentent:
library(dplyr) #> 1.1.0df %>%mutate(date_end = if_else(is.na(date_end),lead(date_start - days(1), default = last(date_start) + years(1)),date_end), .by=id)id date_start date_end1 A1 2022-01-01 2022-01-052 A1 2022-01-06 2022-01-093 A1 2022-01-10 2022-04-134 A1 2022-05-23 2023-05-235 B2 2023-01-02 2022-07-186 B2 2022-07-19 2023-10-067 C1 2022-12-07 2023-01-188 C2 2023-03-23 2023-08-15
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