我在R中有一个数据帧，如下所示：
df <- data.frame("Type" = c("Item A","Item B"), "Frequency" = c("Quarterly","Other"), "Date" = as.Date(c("2021-02-05","2021-05-05")),"endDate" = as.Date("2021-12-12"), stringsAsFactors = F)
我正在尝试生成Date和endDate之间的日期序列作为每一行。我正在使用下面的代码来生成序列
df <- df %>%
dplyr::mutate(id = 1:nrow(df),deliveryDate = ifelse(
df$Frequency == "Quarterly", list(seq(as.Date(df$Date), as.Date(df$endDate), by = "3 month")),
ifelse(df$Frequency == "Monthly", list(seq(as.Date(df$Date), as.Date(df$endDate), by = "month")),
ifelse(df$Frequency %in% c("Other"),list(seq(as.Date(df$Date), as.Date(df$Date), by = "month")),df$Date)))) %>%
tidyr::unnest(deliveryDate) %>%
dplyr::group_by(Type) %>%
dplyr::mutate(deliveryNumber = row_number()) %>%
dplyr::select(deliveryNumber,Type, Frequency, deliveryDate) %>%
为了更具描述性，将根据类型的频率生成日期序列。因此，为了处理这种情况，我使用了dplyr::mutate()。
但我得到了一个错误如下：
Error: Problem with `mutate()` input `deliveryDate`.
x 'from' must be of length 1
ℹ Input `deliveryDate` is `ifelse(...)`.
有人能在R帮我解决这个问题吗？提前感谢！！！
您应该考虑一个命名向量：
图书馆
vec<-c(Quarterly = "3 months", Other = "month")
df %>%
rowwise() %>%
mutate(deliveryDate = list(seq(Date,endDate, by = vec[Frequency]))) %>%
unnest(deliveryDate)
# A tibble: 12 x 5
Type Frequency Date endDate deliveryDate
<chr> <chr> <date> <date> <date>
1 Item A Quarterly 2021-02-05 2021-12-12 2021-02-05
2 Item A Quarterly 2021-02-05 2021-12-12 2021-05-05
3 Item A Quarterly 2021-02-05 2021-12-12 2021-08-05
4 Item A Quarterly 2021-02-05 2021-12-12 2021-11-05
5 Item B Other 2021-05-05 2021-12-12 2021-05-05
6 Item B Other 2021-05-05 2021-12-12 2021-06-05
7 Item B Other 2021-05-05 2021-12-12 2021-07-05
8 Item B Other 2021-05-05 2021-12-12 2021-08-05
9 Item B Other 2021-05-05 2021-12-12 2021-09-05
10 Item B Other 2021-05-05 2021-12-12 2021-10-05
11 Item B Other 2021-05-05 2021-12-12 2021-11-05
12 Item B Other 2021-05-05 2021-12-12 2021-12-05
使用complete
df %>% group_by(Type) %>% mutate(DeliveryDate = Date,
Frequency = case_when(Frequency %in% "Quarterly"~ "quarter",
Frequency %in% "Monthly" ~ "month",
Frequency %in% "Weekly" ~ "week",
TRUE ~ "month")) %>%
complete(DeliveryDate = seq.Date(Date, endDate, by = Frequency)) %>%
fill(Frequency, Date, endDate)
# A tibble: 12 x 5
# Groups: Type [2]
Type DeliveryDate Frequency Date endDate
<chr> <date> <chr> <date> <date>
1 Item A 2021-02-05 quarter 2021-02-05 2021-12-12
2 Item A 2021-05-05 quarter 2021-02-05 2021-12-12
3 Item A 2021-08-05 quarter 2021-02-05 2021-12-12
4 Item A 2021-11-05 quarter 2021-02-05 2021-12-12
5 Item B 2021-05-05 month 2021-05-05 2021-12-12
6 Item B 2021-06-05 month 2021-05-05 2021-12-12
7 Item B 2021-07-05 month 2021-05-05 2021-12-12
8 Item B 2021-08-05 month 2021-05-05 2021-12-12
9 Item B 2021-09-05 month 2021-05-05 2021-12-12
10 Item B 2021-10-05 month 2021-05-05 2021-12-12
11 Item B 2021-11-05 month 2021-05-05 2021-12-12
12 Item B 2021-12-05 month 2021-05-05 2021-12-12
这里有一种方法。不清楚你想要什么"；其他"；与"；月"；，所以我把它设置为"；周"；。
请注意，使用mutate()时不需要引用数据帧，因为函数中调用的所有内容都设置为数据帧的环境。另外，请考虑使用case_when()，而不是使用嵌套的ifelse()调用。
library(tidyverse)
df %>%
mutate(Frequency2 = case_when(Frequency == "Quarterly" ~ "3 month",
Frequency == "Month" ~ "month",
TRUE ~ "week")) %>%
group_by(Type, Frequency2) %>%
nest() %>%
mutate(middates = map2(data, Frequency2, ~ seq.Date(min(.x$Date), max(.x$endDate), by = .y))) %>%
unnest(c(data, middates)) %>%
ungroup()
# A tibble: 36 x 6
Type Frequency Frequency2 Date endDate middates
<chr> <chr> <chr> <date> <date> <date>
1 Item A Quarterly 3 month 2021-02-05 2021-12-12 2021-02-05
2 Item A Quarterly 3 month 2021-02-05 2021-12-12 2021-05-05
3 Item A Quarterly 3 month 2021-02-05 2021-12-12 2021-08-05
4 Item A Quarterly 3 month 2021-02-05 2021-12-12 2021-11-05
5 Item B Other week 2021-05-05 2021-12-12 2021-05-05
6 Item B Other week 2021-05-05 2021-12-12 2021-05-12
7 Item B Other week 2021-05-05 2021-12-12 2021-05-19
8 Item B Other week 2021-05-05 2021-12-12 2021-05-26
9 Item B Other week 2021-05-05 2021-12-12 2021-06-02
10 Item B Other week 2021-05-05 2021-12-12 2021-06-09
# ... with 26 more rows