Question

我有一个数据集

dt <- data.table(Customer = c("a", "a", "c"), months = c(24, 12, 37), Date = c("2019-02-23","2019-03-31","2019-10-01"), Cost = c("100","200","370"))

我希望按年细分成本，并按行号回头客

months_to_year <- function(months){
  if(months%%12==0) y <- rep(12, months %/%  12) else y <- c(rep(12, months %/%  12), months %% 12)
  return(y)
}

dt$years<- dt$months/12
dt$Cost <- as.numeric(dt$Cost)
dt<- dt %>% mutate(Date = as.Date(Date), rn = row_number()) %>% 
  slice(rep(rn, ceiling(months/12)))%>%
  group_by(Customer, rn) %>%
  mutate(months1 = months_to_year(first(months)),
         Date = seq(first(Date), by="1 year", length.out=n()),
         Cost = Cost/months * months1)

我得到以下输出

  Customer months Date        Cost years    rn months1
  <chr>     <dbl> <date>     <dbl> <dbl> <int>   <dbl>
1 a            24 2019-02-23    50  2        1      12
2 a            24 2020-02-23    50  2        1      12
3 a            12 2019-03-31   200  1        2      12
4 c            37 2019-10-01   120  3.08     3      12
5 c            37 2020-10-01   120  3.08     3      12
6 c            37 2021-10-01   120  3.08     3      12
7 c            37 2022-10-01    10  3.08     3       1

现在我希望进一步按月细分

dt %>% mutate(Date = as.Date(Date), rn1 = row_number()) %>% 
  slice(rep(rn1, months1))%>%
  group_by(Customer, rn1) %>%
  mutate(New.Date = seq(first(Date), by="1 month", length.out=n()))

但是，第3行中的客户“ a”被索引为rn1 = 1且新的开始日期比客户“ a”的前一个rn = 1索引增加了1个月。请参阅“新日期”列的第12和25行...我希望在第25行中获取新的日期，以开始2019-03-31。

[![dt output][1]][1]

我将非常感谢您的帮助。

谢谢。

Answer 1

这可能有效。我对示例数据进行了一些编辑，因此请获得正确类型的列。

library(data.table)

dt <- data.table(Customer = c("a", "a", "c"), months_num = c(24, 12, 37), Date = c("2019-02-23","2019-03-31","2019-10-01"), Cost = c(100,200,370))
#set dates
dt[, Date := as.POSIXct( Date, format = "%Y-%m-%d" ) ]
dt[, EndDate := Date %m+% months( months_num )][]

str(dt)
# Classes ‘data.table’ and 'data.frame':    3 obs. of  5 variables:
# $ Customer  : chr  "a" "a" "c"
# $ months_num: num  24 12 37
# $ Date      : POSIXct, format: "2019-02-23" "2019-03-31" "2019-10-01"
# $ Cost      : num  100 200 370
# $ EndDate   : POSIXct, format: "2021-02-23" "2020-03-31" "2022-11-01"

代码

实际上是单线的，但是由于在SO上的易读性而溢出了多条线。

#monthly
dt[ , .( Customer = Customer, 
         month = seq(Date, 
                     by = "month", 
                     length.out = months_num ),
         Cost = Cost / months_num ), 
    by = .(id = 1:nrow(dt) )][]

通过按行对客户按年然后按月细分成本

1 个答案: