按客户和时间对客户会话进行分组，直到下一次交易

Question

我需要按时间对客户购物会话进行分类，直到下一次交易为止。一个示例数据框是：

library(tidyverse)

cust_transactions_before <- 
  tibble(
    customer_name = c("a", "a", "a", "a", "a", "a", "a", "a", "a", "a", "b", "b", "b", "b", "b", "b", "b", "b", "b", "b"),
    time_until_next =c(41, 19, 5, 27, 49, 3, 10, 20, 13, NA_integer_, 25, 17, 8, 33, 25, 31, 19, 5, 27, NA_integer_))

我想按 customer_name 分组，并让每个客户的第一笔交易从 1 开始，值 cust_session。对于下一次观察，我想做一个 if/then，如果 time_until_next 是 <= 30，那么保持与之前观察相同的 cust_session 会话号。如果 time_until_next 大于 30，则取前一个 cust_session 并向其加 1。

最后，如果 time_until_next 是 NA，那么让它等于之前的 cust_session。

处理后成功的数据框如下所示：

cust_transactions_after <- 
  tibble(
    customer_name = c("a", "a", "a", "a", "a", "a", "a", "a", "a", "a", "b", "b", "b", "b", "b", "b", "b", "b", "b", "b"),
    time_until_next =c(41, 19, 5, 27, 49, 3, 10, 20, 13, NA_integer_, 25, 17, 8, 33, 25, 31, 19, 5, 27, NA_integer_), 
    cust_session = c(1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 1, 1, 1, 1, 2, 2, 3, 3, 3, 3))

Answer 1

library(dplyr)
cust_transactions_before %>% 
  group_by(customer_name) %>% 
  mutate(cust_session = cumsum(lag(time_until_next, default = 31) > 30))

   customer_name time_until_next cust_session
   <chr>                   <dbl>        <int>
 1 a                          41            1
 2 a                          19            2
 3 a                           5            2
 4 a                          27            2
 5 a                          49            2
 6 a                           3            3
 7 a                          10            3
 8 a                          20            3
 9 a                          13            3
10 a                          NA            3
11 b                          25            1
12 b                          17            1
13 b                           8            1
14 b                          33            1
15 b                          25            2
16 b                          31            2
17 b                          19            3
18 b                           5            3
19 b                          27            3
20 b                          NA            3

Answer 2

在 dplyr 1.1.0 中，我们可以使用：

cust_transactions_before |>
  mutate(cust_session = consecutive_id(time_until_next > 30), .by = customer_name)