我有一张这样的桌子：
| Activation Month | Disabled Month | Month.Fee | Custr
| 21/4/2018 | N/A | 10 | A
| 21/3/2018 | 21/6/2018 | 20 | B
我想根据有效范围来转换此表，在“激活”和“禁用”列中，以便每月有1个条目（假设今天是2018年11月30日），例如：
Month | Enrolled | Activation Month | Disabled Month | Month.Fee | Cust.
1 | N | 21/4/2018 | N/A | 10 | A
2 | N | 21/4/2018 | N/A | 10 | A
3 | N | 21/4/2018 | N/A | 10 | A
4 | Y | 21/4/2018 | N/A | 10 | A
5 | Y | 21/4/2018 | N/A | 10 | A
6 | Y | 21/4/2018 | N/A | 10 | A
7 | Y | 21/4/2018 | N/A | 10 | A
8 | Y | 21/4/2018 | N/A | 10 | A
9 | Y | 21/4/2018 | N/A | 10 | A
10 | Y | 21/4/2018 | N/A | 10 | A
11 | Y | 21/4/2018 | N/A | 10 | A
12 | Y | 21/4/2018 | N/A | 10 | A
1 | N | 21/3/2018 | 21/6/2018 | 10 | B
2 | N | 21/3/2018 | 21/6/2018 | 10 | B
3 | Y | 21/3/2018 | 21/6/2018 | 10 | B
4 | Y | 21/3/2018 | 21/6/2018 | 10 | B
5 | Y | 21/3/2018 | 21/6/2018 | 10 | B
6 | Y | 21/3/2018 | 21/6/2018 | 10 | B
7 | N | 21/3/2018 | 21/6/2018 | 10 | B
8 | N | 21/3/2018 | 21/6/2018 | 10 | B
9 | N | 21/3/2018 | 21/6/2018 | 10 | B
10 | N | 21/3/2018 | 21/6/2018 | 10 | B
11 | N | 21/3/2018 | 21/6/2018 | 10 | B
12 | N | 21/3/2018 | 21/6/2018 | 10 | B
有没有办法无循环地管理它？
答案 0 :(得分：0)
一个选项是
library(tidyverse)
df1 %>%
mutate(Enrolled = map2(Activation.Month, Disabled.Month, ~ {
x1 <- month(dmy(.x))
x2 <- month(dmy(.y))
ind <- if(is.na(x2)) x1:12 else x1:x2
1:12 %in% ind})) %>%
unnest %>%
mutate(Month = rep(1:12, length.out = n()))
# Activation.Month Disabled.Month Month.Fee Custr Enrolled Month
#1 21/4/2018 <NA> 10 A FALSE 1
#2 21/4/2018 <NA> 10 A FALSE 2
#3 21/4/2018 <NA> 10 A FALSE 3
#4 21/4/2018 <NA> 10 A TRUE 4
#5 21/4/2018 <NA> 10 A TRUE 5
#6 21/4/2018 <NA> 10 A TRUE 6
#7 21/4/2018 <NA> 10 A TRUE 7
#8 21/4/2018 <NA> 10 A TRUE 8
#9 21/4/2018 <NA> 10 A TRUE 9
#10 21/4/2018 <NA> 10 A TRUE 10
#11 21/4/2018 <NA> 10 A TRUE 11
#12 21/4/2018 <NA> 10 A TRUE 12
#13 21/3/2018 21/6/2018 20 B FALSE 1
#14 21/3/2018 21/6/2018 20 B FALSE 2
#15 21/3/2018 21/6/2018 20 B TRUE 3
#16 21/3/2018 21/6/2018 20 B TRUE 4
#17 21/3/2018 21/6/2018 20 B TRUE 5
#18 21/3/2018 21/6/2018 20 B TRUE 6
#19 21/3/2018 21/6/2018 20 B FALSE 7
#20 21/3/2018 21/6/2018 20 B FALSE 8
#21 21/3/2018 21/6/2018 20 B FALSE 9
#22 21/3/2018 21/6/2018 20 B FALSE 10
#23 21/3/2018 21/6/2018 20 B FALSE 11
#24 21/3/2018 21/6/2018 20 B FALSE 12
注意：在这里，我们将TRUE/FALSE用作“已注册”而不是“是/否”，因为使用逻辑列进行子集变得更容易
df1 <- structure(list(Activation.Month = c("21/4/2018", "21/3/2018"),
Disabled.Month = c(NA, "21/6/2018"), Month.Fee = c(10L, 20L
), Custr = c("A", "B")), class = "data.frame", row.names = c(NA,
-2L))