如何根据R中的产品列表汇总客户购买的产品数量答案

【问题标题】：How to summarize number of products purchased by a customer based on a product list in R如何根据R中的产品列表汇总客户购买的产品数量
【发布时间】：2021-03-07 05:51:02
【问题描述】：

我在 R 中有两个数据框，一个由产品 SKU（产品 ID）列表组成，另一个是包含订单号、客户电子邮件、购买日期和产品 ID（product_sku）的购买日志和购买的数量。

purchases_dataframe：

order_number | email                | product_sku | quantity | purchase_date
1000         |customer1@sample.com  | RT-100      | 2        | 2020-01-01
1000         |customer1@sample.com  | CT-300      | 1        | 2020-01-01
1000         |customer1@sample.com  | Phone-100   | 1        | 2020-01-01
2000         |customer2@sample.com  | Phone-200   | 1        | 2020-04-20
2000         |customer2@sample.com  | OM-200      | 1        | 2020-04-20
3000         |customer3@sample.com  | CT-300      | 3        | 2020-03-15
4000         |customer1@sample.com  | OM-200      | 5        | 2020-07-07
5000         |customer4@sample.com  | Phone-200   | 3        | 2020-08-19
6000         |customer3@sample.com  | Phone-100   | 1        | 2020-09-22
6000         |customer3@sample.com  | RT-100      | 1        | 2020-09-22

tv_list：

 SKU
    RT-100
    CT-300
    OM-200
    LL-400
    ...

我想计算客户在其一生中购买的电视总数，而忽略所有其他产品（例如手机）。数据框 tv_list 应该可以帮助我识别哪些 SKU 是电视，哪些不是，因为我有各种不同的电视 SKU，以上只是一个较小的示例。理想情况下，生成的数据框如下所示：

email                | number_purchased_tv
customer1@sample.com | 8
customer2@sample.com | 1
customer3@sample.com | 4
customer4@sample.com | 0

为了重现性和为了更容易理解我的示例，这里是上面两个 sample_tables 的代码：

purchase_dataframe <- data.frame(order_number = c(1000,1000,1000, 2000,2000, 3000, 4000, 5000, 6000, 6000),
                      email = c("customer1@sample.com","customer1@sample.com", "customer1@sample.com","customer2@sample.com",
                                "customer2@sample.com","customer3@sample.com","customer1@sample.com","customer4@sample.com",
                                "customer3@sample.com","customer3@sample.com"),
                      product_sku = c("RT-100", "CT-300", "Phone-100", "Phone-200", "OM-200", "CT-300", "OM-200", "Phone-200", "Phone-100", "RT-100"),
                      quantity = c(2,1,1,1,1,3,5,3,1,1),
                      purchase_date = c("2020-01-01","2020-01-01","2020-01-01","2020-04-20","2020-04-20","2020-03-15","2020-07-07","2020-08-19","2020-09-22","2020-09-22"))

tv_list <- data.frame(SKU = c("RT-100", "OM-200", "CT-300", "LL-400", "ZV-700"))

非常感谢！

【问题讨论】：

标签： r dplyr aggregate tidyr summarize

【解决方案1】：

下面使用dplyr执行您的要求

library(dplyr)
library(data.table)
purchase_dataframe %>% dplyr::group_by(email) %>% dplyr::summarise(sumtv = sum(quantity[product_sku %in% unique(tv_list$SKU)]))
# A tibble: 4 x 2
email                sumtv
<chr>                <dbl>
  1 customer1@sample.com     8
2 customer2@sample.com     1
3 customer3@sample.com     4
4 customer4@sample.com     0

编辑请在上面找到关于sumtv 数字的更正和下面的data.table 解决方案

library(dplyr)
library(data.table)
purchase_datatable <- purchase_dataframe
purchase_datatable %>% setDT
> purchase_datatable[,sumtv := sum(quantity[product_sku %in% unique(tv_list$SKU)]), by="email"][
  +   ,.(email, sumtv)] %>% unique
email sumtv
1: customer1@sample.com     8
2: customer2@sample.com     1
3: customer3@sample.com     4
4: customer4@sample.com     0

microbenchmarking 为data.table 解决方案带来了近 50% 的优势，IMO 是一个非常值得通过这些vignettes 学习的优秀软件包

library(microbenchmark)
microbenchmark(purchase_datatable[,sumtv := sum(quantity[product_sku %in% unique(tv_list$SKU)]), by="email"][
  ,.(email, sumtv)] %>% unique, purchase_dataframe %>% dplyr::group_by(email) %>% dplyr::summarise(sumtv = sum(quantity[product_sku %in% unique(tv_list$SKU)]))
)
min      lq     mean  median     uq    max neval
1.268 1.42700 1.823445 1.80300 2.0887 2.8332   100
2.715 2.98025 3.250287 3.20355 3.3509 8.8255   100

【讨论】：

嘿！感谢您的快速答复！不幸的是，这不会导致正确的数量总和 - 似乎只汇总了订单。对于客户1，如果我没记错的话，总和应该是8。
感谢您的快速修复！出于某种原因，data.table 解决方案给了我以下警告： In [.data.table(purchase_datatable, product_sku %in% unique(tv_list$SKU), : Invalid .internal.selfref 通过获取（浅）副本检测并修复data.table 以便 := 可以通过引用添加此新列。在较早的时候，此 data.table 已由 R 复制（或使用 structure() 或类似方法手动创建）。避免使用名称
抱歉，由于“警告信息”，字符已用完。出于某种原因，它也没有返回“唯一”表——但这可能与上面的警告消息有关。
您是否同时加载了dplyr 和data.table？如果不是你应该这样做，我会编辑答案
刚刚再次检查。如果我没记错的话，我认为 dplyr 解决方案有效。谢谢！但是，data.table 解决方案不会聚合客户（正如您在输出中看到的那样，客户 1 和客户 3 存在两次，具有 2 个不同的值）

【解决方案2】：

一个使用base R的选项：

#Match and index
purchase_dataframe$ProductIndex <- tv_list[match(purchase_dataframe$product_sku,tv_list$SKU),'SKU']
purchase_dataframe$Counter <- ifelse(is.na(purchase_dataframe$ProductIndex),0,purchase_dataframe$quantity)
#Aggregate
Res <- aggregate(Counter~email,data=purchase_dataframe,sum,na.rm=T)

输出：

                 email Counter
1 customer1@sample.com       8
2 customer2@sample.com       1
3 customer3@sample.com       4
4 customer4@sample.com       0

【讨论】：

【解决方案3】：

这是您提供的数据：

library('dplyr')
purchase_dataframe <- data.frame(order_number = c(1000,1000,1000, 2000,2000, 3000, 4000, 5000, 6000, 6000),
                      email = c("customer1@sample.com","customer1@sample.com", "customer1@sample.com","customer2@sample.com",
                                "customer2@sample.com","customer3@sample.com","customer1@sample.com","customer4@sample.com",
                                "customer3@sample.com","customer3@sample.com"),
                      product_sku = c("RT-100", "CT-300", "Phone-100", "Phone-200", "OM-200", "CT-300", "OM-200", "Phone-200", "Phone-100", "RT-100"),
                      quantity = c(2,1,1,1,1,3,5,3,1,1),
                      purchase_date = c("2020-01-01","2020-01-01","2020-01-01","2020-04-20","2020-04-20","2020-03-15","2020-07-07","2020-08-19","2020-09-22","2020-09-22"))

tv_list <- data.frame(SKU = c("RT-100", "OM-200", "CT-300", "LL-400", "ZV-700"))

这将为您提供摘要但省略任何电子邮件（尚未购买电视的客户）

total_tvs_by_cusomter <- purchase_dataframe %>%
  filter(product_sku %in% tv_list$SKU) %>%
  group_by(email) %>%
  mutate(quantity = as.numeric(quantity)) %>%
  summarise(number_purchased_tv = sum(quantity))

结果：

# A tibble: 3 x 2
  email                number_purchased_tv
  <chr>                              <dbl>
1 customer1@sample.com                   8
2 customer2@sample.com                   1
3 customer3@sample.com                   4

如果您想保留尚未购买电视的电子邮件/客户并将其添加为 0，以防万一

total_tvs_by_cusomter <- left_join(unique(purchase_dataframe %>%
            select(email)), total_tvs_by_cusomter)

total_tvs_by_cusomter[is.na(total_tvs_by_cusomter)] <- 0

结果：

                 email number_purchased_tv
1 customer1@sample.com                   8
2 customer2@sample.com                   1
3 customer3@sample.com                   4
4 customer4@sample.com                   0

【讨论】：

【解决方案4】：

tv_purchases <-
purchase_dataframe %>% 
  group_by(email) %>% 
  filter(product_sku %in% tv_list$SKU) %>%
  summarise(number_purchased_tv = sum(as.numeric(quantity)))

## join tv_purchases on distinct emails, to also have the 'customer4@sample.com     0' row

purchase_dataframe %>%
  distinct(email) %>%
  left_join(tv_purchases) %>% ## emails which are not in tv_purchases will have NAs
  mutate(number_purchased_tv = case_when(is.na(number_purchased_tv) ~ 0, ## NAs become zeros
                                         TRUE ~ number_purchased_tv) ## non-NAs stay as they are
         )

【讨论】：