计算并输出客户第一次下单的日期

Question

数据：

DB <- data.frame(orderID = c(1,2,3,4,4,5,6,6,7,8),    
                 orderDate = c("1.1.12","1.1.12","1.1.12","13.1.12","13.1.12","12.1.12","10.1.12","10.1.12","21.1.12","24.1.12"),
                 itemID = c(2,3,2,5,12,4,2,3,1,5),   
                 customerID = c(1, 2, 3, 1, 1, 3, 2, 2, 1, 1),
                 itemPrice = c(9.99, 14.99, 9.99, 19.99, 29.99, 4.99, 9.99, 14.99, 49.99, 19.99))

预期结果：

DB <- data.frame(orderID = c(1,2,3,4,4,5,6,6,7,8),    
                 orderDate = c("1.1.12","2.1.12","3.1.12","13.1.12","13.1.12","12.1.12","10.1.12","10.1.12","21.1.12","24.1.12"),
                 itemID = c(2,3,2,5,12,4,2,3,1,5),   
                 customerID = c(1, 2, 3, 1, 1, 3, 2, 2, 1, 1),
                 itemPrice = c(9.99, 14.99, 9.99, 19.99, 29.99, 4.99, 9.99, 14.99, 49.99, 19.99),
                 DateOfFirstOrderofCustomer = c("1.1.12", "2.1.12", "3.1.12", "1.1.12", "1.1.12", "3.1.12", "2.1.12", "2.1.12", "1.1.12", "1.1.12"))

为了理解：

orderID 是连续的。同一天从同一customerID 订购的产品将获得相同的orderID。当同一客户在另一天订购产品时，他/她是新的orderID。

我想为包含客户第一个订单日期的每一行/条目添加一个额外的列（例如，客户 1 (customerID1) 在1.1.12 下了他的第一个订单，因此在所有订单中都输入了此日期来自该客户）。我们该怎么做？

原始数据大约有 500k 行：所以请给出一个只需要很少性能的解决方案。

Answer 1

仅使用基本 R 函数：

# convert the date column to date-format
DB$orderDate <- as.Date(DB$orderDate, format('%d.%m.%y'))

# get the first date for each customer
DB$DateFirstOrder <- with(DB, ave(orderDate, customerID, FUN = min))

那么结果是（使用 Mike Spencer 的数据）：

> DB
   orderID  orderDate itemID customerID itemPrice DateFirstOrder
1        1 2012-01-01      2          1      9.99     2012-01-01
2        2 2012-01-04      3          2     14.99     2012-01-04
3        3 2012-01-06      2          3      9.99     2012-01-06
4        4 2012-01-13      5          1     19.99     2012-01-01
5        4 2012-01-13     12          1     29.99     2012-01-01
6        5 2012-01-12      4          3      4.99     2012-01-06
7        6 2012-01-10      2          2      9.99     2012-01-04
8        6 2012-01-10      3          2     14.99     2012-01-04
9        7 2012-01-21      1          1     49.99     2012-01-01
10       8 2012-01-24      5          1     19.99     2012-01-01

---

对于最快的解决方案，我会推荐 data.table 包。要使用此包获得所需的结果，您需要执行以下操作：

library(data.table)
setDT(DB)[, orderDate := as.Date(orderDate, format('%d.%m.%y'))
          ][, DateFirstOrder := min(orderDate), by = customerID]

Answer 2

我更改了您的输入数据，因为您提供的所有数据都具有相同的每位客户首次购买的日期，因此无法判断代码是否有效。该示例使用 dplyr，您也可以使用 tapply，但您需要重新格式化命名向量。

# Dummy data
DB <- data.frame(orderID = c(1,2,3,4,4,5,6,6,7,8),    
                 orderDate = c("1.1.12","4.1.12","6.1.12","13.1.12","13.1.12","12.1.12","10.1.12","10.1.12","21.1.12","24.1.12"),
                 itemID = c(2,3,2,5,12,4,2,3,1,5),   
                 customerID = c(1, 2, 3, 1, 1, 3, 2, 2, 1, 1),
                 itemPrice = c(9.99, 14.99, 9.99, 19.99, 29.99, 4.99, 9.99, 14.99, 49.99, 19.99))

# -------------------------------------------

# Change dates to a readable format
DB$orderDate <- as.Date(DB$orderDate, format="%d.%m.%y")

# -------------------------------------------

library(dplyr)

DB <- DB %>% 
   group_by(customerID) %>% 
   mutate(DateOfFirstOrderofCustomer=min(orderDate))

Answer 3

我正在使用plyr 包。休息一切都是一样的。

DB <- data.frame(orderID = c(1,2,3,4,4,5,6,6,7,8),    
             orderDate = c("1.1.12","4.1.12","6.1.12","13.1.12","13.1.12","12.1.12","10.1.12","10.1.12","21.1.12","24.1.12"),
             itemID = c(2,3,2,5,12,4,2,3,1,5),   
             customerID = c(1, 2, 3, 1, 1, 3, 2, 2, 1, 1),
             itemPrice = c(9.99, 14.99, 9.99, 19.99, 29.99, 4.99, 9.99, 14.99, 49.99, 19.99))


install.packages("plyr")
library(plyr)

DB$orderDate <- as.Date(DB$orderDate, format="%d.%m.%y")
DB = ddply(DB, .(customerID), mutate, DateOfFirstOrderofCustomer = min(orderDate))