根据日期在组内拆分重叠行

Question

我正在尝试根据现有行的重叠时间段创建新行。例如，我想把这个：

Customer_Product <- data.table(Customer=c("A01","A01","A01", "A02", "A02", "A02", "A03", "A03", "A03"), 
                Product=c("Prod1","Prod2","Prod3","Prod1","Prod2","Prod3","Prod1","Prod2","Prod3"), 
                Start_Date=c("1/1/2015", "3/1/2015", "4/1/2015", "1/1/2015", "3/1/2015", "4/1/2015", "1/1/2015", "3/1/2015", "4/1/2015"),
                End_Date=c("2/1/2015","5/1/2015","5/1/2015","2/1/2015","5/1/2015","6/1/2015","2/1/2015","6/1/2015","5/1/2015"))

   Customer Product Start_Date End_Date
1:      A01   Prod1   1/1/2015 2/1/2015
2:      A01   Prod2   3/1/2015 5/1/2015
3:      A01   Prod3   4/1/2015 5/1/2015
4:      A02   Prod1   1/1/2015 2/1/2015
5:      A02   Prod2   3/1/2015 5/1/2015
6:      A02   Prod3   4/1/2015 6/1/2015
7:      A03   Prod1   1/1/2015 2/1/2015
8:      A03   Prod2   3/1/2015 6/1/2015
9:      A03   Prod3   4/1/2015 5/1/2015

变成这样：

Customer_Product_Combo <- data.table(Customer=c("A01","A01","A01", "A02", "A02", "A02", "A02","A03", "A03","A03","A03"),
                Product_or_Combination=c("Prod1","Prod2","Prod2/Prod3","Prod1","Prod2","Prod2/Prod3","Prod3","Prod1","Prod2","Prod2/Prod3","Prod2"),
                Start_Date=c("1/1/2015","3/1/2015","4/1/2015","1/1/2015","3/1/2015","4/1/2015","5/1/2015","1/1/2015","3/1/2015","4/1/2015","5/1/2015"),
                End_Date=c("2/1/2015","4/1/2015","5/1/2015","2/1/2015","4/1/2015","5/1/2015","6/1/2015","2/1/2015","4/1/2015","5/1/2015","6/1/2015"))

    Customer Product_or_Combination Start_Date End_Date
 1:      A01                  Prod1   1/1/2015 2/1/2015
 2:      A01                  Prod2   3/1/2015 4/1/2015
 3:      A01            Prod2/Prod3   4/1/2015 5/1/2015
 4:      A02                  Prod1   1/1/2015 2/1/2015
 5:      A02                  Prod2   3/1/2015 4/1/2015
 6:      A02            Prod2/Prod3   4/1/2015 5/1/2015
 7:      A02                  Prod3   5/1/2015 6/1/2015
 8:      A03                  Prod1   1/1/2015 2/1/2015
 9:      A03                  Prod2   3/1/2015 4/1/2015
10:      A03            Prod2/Prod3   4/1/2015 5/1/2015
11:      A03                  Prod2   5/1/2015 6/1/2015

我一直在研究 IRanges，因为看起来 disjoin() 可能是一个可能的解决方案，但我看不到 inherit/merge "Prod" 数据的任何方法。

我也一直在尝试在 dplyr 中使用 lead/lag 勾勒出一些东西，然后是 gather/merge 循环，但同样值得注意的是，我可以有超过 2 "Prod"重叠，然后逻辑就变得混乱了。

有没有合理的方法来做到这一点？非常感谢任何帮助！

Answer 1

我正在使用您发布的数据（作为 data.frame）

Customer_Product <- data.frame(Customer=c("A01","A01","A01", "A02", "A02", "A02", "A03", "A03", "A03"), 
                               Product=c("Prod1","Prod2","Prod3","Prod1","Prod2","Prod3","Prod1","Prod2","Prod3"), 
                               Start_Date=c("1/1/2015", "3/1/2015", "4/1/2015", "1/1/2015", "3/1/2015", "4/1/2015", "1/1/2015", "3/1/2015", "4/1/2015"),
                               End_Date=c("2/1/2015","5/1/2015","5/1/2015","2/1/2015","5/1/2015","6/1/2015","2/1/2015","6/1/2015","5/1/2015"))

这是一个可能的解决方案：

library(tidyverse)
library(data.table)
library(lubridate)

Customer_Product %>%
  mutate_at(vars(matches("Date")), dmy) %>%                          # update to date columns (if needed)
  mutate(day = map2(Start_Date, End_Date, ~seq(.x, .y, "day"))) %>%  # create sequence of days between start and end
  unnest() %>%                                                       # unnest data
  group_by(Customer, day) %>%                                        # for each customer and day
  summarise(Product = paste0(Product, collapse = "/")) %>%           # find corresponding products
  group_by(Customer, Product, id = rleid(Product)) %>%               # for each customer, product combination and position of product combination
  summarise(Start_Date = min(day),                                   # get start date
            End_Date = max(day)) %>%                                 # get end date
  ungroup() %>%                                                      # ungroup
  select(-id) %>%                                                    # remove id column
  arrange(Customer, Start_Date)                                      # order rows (if needed)


# # A tibble: 11 x 4
#   Customer Product     Start_Date End_Date  
#   <fct>    <chr>       <date>     <date>    
# 1 A01      Prod1       2015-01-01 2015-01-02
# 2 A01      Prod2       2015-01-03 2015-01-03
# 3 A01      Prod2/Prod3 2015-01-04 2015-01-05
# 4 A02      Prod1       2015-01-01 2015-01-02
# 5 A02      Prod2       2015-01-03 2015-01-03
# 6 A02      Prod2/Prod3 2015-01-04 2015-01-05
# 7 A02      Prod3       2015-01-06 2015-01-06
# 8 A03      Prod1       2015-01-01 2015-01-02
# 9 A03      Prod2       2015-01-03 2015-01-03
#10 A03      Prod2/Prod3 2015-01-04 2015-01-05
#11 A03      Prod2       2015-01-06 2015-01-06

请注意，此解决方案不允许输出中的日期范围重叠 table。

例如，如果您在 4/1/2015 - 5/1/2015 期间获得 Prod2/Prod3，则在 5/1/2015 - 6/1/2015 期间不会获得 Prod2，但是 6/1/2015 - 6/1/2015，因为 [=18] =] 包含在 Prod2/Prod3.

中

根据日期在组内拆分重叠行

Splitting overlapping rows, within groups, based on dates

r

iranges

dplyr

data.table