英文:
Grouping Customer Sessions by Customer and Time until Next Transaction
问题
需要按照距离下一次交易的时间对客户的购物会话进行分组。一个示例数据框如下：
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_))
我想按照 customer_name 进行分组，并使每位客户的第一笔交易的 cust_session 值从1开始。对于下一条观察值，如果 time_until_next 小于等于30，则将 cust_session 的值保持与前一个观察值相同。如果 time_until_next 大于30，则将前一个 cust_session 的值加1。
最后，如果 time_until_next 是 NA，则将其设置为前一个 cust_session 的值。
处理后的成功数据框如下：
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))
希望这对你有帮助。
英文:
I need to bucket customer shopping sessions by time until next transaction. An example data frame is:
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_))
I would like to group by customer_name and have the first transaction per customer start at 1 for the value cust_session. For the next observation I'd like to do an if/then where if time_until_next is <= 30 then keep the same session number for cust_session as the previous observation. If time_until_next is > 30 then take the previous cust_session and add 1 to it.
Lastly, if time_until_next is NA then have it equal the previous cust_session.
A successful data frame after processing would look like this:
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))
答案1
得分: 1
library(dplyr)cust_transactions_before %>%group_by(customer_name) %>%mutate(cust_session = cumsum(lag(time_until_next, default = 31) > 30))
英文:
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 12 a 19 23 a 5 24 a 27 25 a 49 26 a 3 37 a 10 38 a 20 39 a 13 310 a NA 311 b 25 112 b 17 113 b 8 114 b 33 115 b 25 216 b 31 217 b 19 318 b 5 319 b 27 320 b NA 3
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