如何知道在上一个订单的 delivery/receiving 之前下了下一个订单的客户？在 R
How to know customers who placed next order before delivery/receiving of earlier order? In R
我有一个包含两个日期的大型数据库。例如。取超市数据(http://www.tableau.com/sites/default/files/training/global_superstore.zip)'Orders'Sheet.
一个日期是订单日期，另一个是 shipping/delivery 日期（假设它是交货日期）。我想知道下一个订单的客户的所有订单的详细信息，而没有等待他们之前任何一个订单的 shipping/delivery。
例如ID 为 'ZC-21910' 的客户于 2014 年 6 月 12 日下了 ID 为 CA-2014-133928 的订单，该订单于 2014 年 6 月 18 日发货。但是，同一客户在 2014 年 6 月 13 日下了 ID 为 'IT-2014-3511710' 的下一个订单，即2014 年 6 月 18 日之前（较早订单之一的发货日期）。
最好在单独的 dataframe/table.
中过滤掉所有此类订单（订单 ID）
如何在 R 中实现？
示例数据集
> dput(df)
structure(list(customer_id = c("A", "A", "A", "B", "B", "C",
"C"), order_id = structure(1:7, .Label = c("1", "2", "3", "4",
"5", "6", "7"), class = "factor"), order_date = structure(c(17897,
17901, 17912, 17901, 17902, 17903, 17905), class = "Date"), ship_date = structure(c(17926,
17906, 17914, 17904, 17904, 17904, 17906), class = "Date")), row.names = c(NA,
-7L), class = c("tbl_df", "tbl", "data.frame"))
以下是我将如何在 R 中构建此工作流，请注意：在 Tableau 中复制该功能将非常困难。
# Install pacakges if they are not already installed: necessary_packages => vector
necessary_packages <- c("readxl")
# Create a vector containing the names of any packages needing installation:
# new_pacakges => vector
new_packages <- necessary_packages[!(necessary_packages %in%
installed.packages()[, "Package"])]
# If the vector has more than 0 values, install the new pacakges
# (and their) associated dependencies:
if(length(new_packages) > 0){install.packages(new_packages, dependencies = TRUE)}
# Initialise the packages in the session:
lapply(necessary_packages, require, character.only = TRUE)
# Store a scalar of the link to the data: durl => character scalar
durl <- "http://www.tableau.com/sites/default/files/training/global_superstore.zip"
# Store the path to the temporary directory: tmpdir_path => character scalar
tmpdir_path <- tempdir()
# Store a character scalar denoting the link to the zipped directory
# that is to be created: zip_path => character scalar
zip_path <- paste0(tmpdir_path, "/tableau.zip")
# Store a character scalar denoting the link to the unzipped directory
# that is to be created: unzip_path => character scalar
unzip_path <- paste0(tmpdir_path, "/global_superstore")
# Download the zip file: global_superstore.zip => stdout (zip_path)
download.file(durl, zip_path)
# Unzip the file into the unzip directory: tableau.zip => stdout (global_superstore)
unzip(zipfile = zip_path, exdir = unzip_path)
# Read in the excel file: df => data.frame
df <- read_xls(normalizePath(list.files(unzip_path, full.names = TRUE)))
# Regex the vector names to fit with R convention: names(df) => character vector
names(df) <- gsub("\W+", "_", tolower(trimws(names(df), "both")))
# Allocate some memory by creating an empty list the same size as the number of
# customers: df_list => list
df_list <- vector("list", length(unique(df$customer_id)))
# Split the data.frame into the list by the customer_id: df_list => lis
df_list <- with(df, split(df, customer_id))
# Sort the data (by date) and test whether or not each customer waited for their
# order before ordering again: orders_prior_to_delivery => data.frame
orders_prior_to_delivery <- data.frame(do.call("rbind", Map(function(x){
# Order the data.frame: y => data.frame
y <- x[order(x$order_date),]
# Return only the observations where the customer didn't wait:
# data.frame => GlobalEnv()
with(y, y[c(FALSE,
apply(data.frame(sapply(order_date[-1], `<`, ship_date[-nrow(y)])), 2, any)),])
},
df_list)), row.names = NULL, stringsAsFactors = FALSE)
# Unique customers and orders that were ordered prior to shipping the
# previous order: cust_orders_prior_to_delivery => data.frame
cust_orders_prior_to_delivery <-
unique(orders_prior_to_delivery[,c("order_id", "customer_id")])
编辑：我之前的回答没有正确处理订单日期 == 发货日期的情况。
我假设您已经将数据加载到名为 df 的对象中。您可以使用@hello_friend 代码的第一部分来获取它。
library(tidyverse)
df %>%
distinct(`Customer ID`, `Order ID`, `Order Date`, `Ship Date`) %>%
arrange(`Customer ID`, `Order Date`, `Ship Date`) %>%
mutate(sort_key = row_number()) %>%
pivot_longer(c(`Order Date`, `Ship Date`), names_to = "Activity", names_pattern = "(.*) Date", values_to = "Date") %>%
mutate(Activity = factor(Activity, ordered = TRUE, levels = c("Order", "Ship")),
Open = if_else(Activity == "Order", 1, -1)) %>%
group_by(`Customer ID`) %>%
arrange(Date, sort_key, Activity, .by_group = TRUE) %>%
mutate(Open = cumsum(Open)) %>%
ungroup %>%
filter(Open > 1, Activity == "Order") %>%
select(`Customer ID`, `Order ID`)
首先，只接受不同的订单和客户 ID，否则来自同一订单的多个项目会混淆并导致不正确的结果。然后，旋转数据，使每个订单成为两行，每行代表一个不同的 activity：订购或发货。我们创建了 运行 个未结订单总数。你要找什么时候变成两个或更多。
我使用 Activity 的有序因子来确保我总是在关闭订单之前打开订单。当订单日期和发货日期相同时，这很重要。
我使用一个特殊的 sort_key 栏来确保我在打开新订单之前关闭旧订单，以防客户在同一天订购其他东西。您可能需要相反的逻辑。
所有这些都假设给定的客户 ID 和订单 ID 在数据中只出现一次，这在您的数据集中实际上是不正确的，如您所见：
df %>% group_by(`Customer ID`, `Order ID`) %>% filter(n_distinct(`Ship Date`)> 1) %>% select(1:9)
我有一个包含两个日期的大型数据库。例如。取超市数据(http://www.tableau.com/sites/default/files/training/global_superstore.zip)'Orders'Sheet.
一个日期是订单日期，另一个是 shipping/delivery 日期（假设它是交货日期）。我想知道下一个订单的客户的所有订单的详细信息，而没有等待他们之前任何一个订单的 shipping/delivery。
例如ID 为 'ZC-21910' 的客户于 2014 年 6 月 12 日下了 ID 为 CA-2014-133928 的订单，该订单于 2014 年 6 月 18 日发货。但是，同一客户在 2014 年 6 月 13 日下了 ID 为 'IT-2014-3511710' 的下一个订单，即2014 年 6 月 18 日之前（较早订单之一的发货日期）。
最好在单独的 dataframe/table.
中过滤掉所有此类订单（订单 ID）如何在 R 中实现？
示例数据集
> dput(df)
structure(list(customer_id = c("A", "A", "A", "B", "B", "C",
"C"), order_id = structure(1:7, .Label = c("1", "2", "3", "4",
"5", "6", "7"), class = "factor"), order_date = structure(c(17897,
17901, 17912, 17901, 17902, 17903, 17905), class = "Date"), ship_date = structure(c(17926,
17906, 17914, 17904, 17904, 17904, 17906), class = "Date")), row.names = c(NA,
-7L), class = c("tbl_df", "tbl", "data.frame"))
以下是我将如何在 R 中构建此工作流，请注意：在 Tableau 中复制该功能将非常困难。
# Install pacakges if they are not already installed: necessary_packages => vector
necessary_packages <- c("readxl")
# Create a vector containing the names of any packages needing installation:
# new_pacakges => vector
new_packages <- necessary_packages[!(necessary_packages %in%
installed.packages()[, "Package"])]
# If the vector has more than 0 values, install the new pacakges
# (and their) associated dependencies:
if(length(new_packages) > 0){install.packages(new_packages, dependencies = TRUE)}
# Initialise the packages in the session:
lapply(necessary_packages, require, character.only = TRUE)
# Store a scalar of the link to the data: durl => character scalar
durl <- "http://www.tableau.com/sites/default/files/training/global_superstore.zip"
# Store the path to the temporary directory: tmpdir_path => character scalar
tmpdir_path <- tempdir()
# Store a character scalar denoting the link to the zipped directory
# that is to be created: zip_path => character scalar
zip_path <- paste0(tmpdir_path, "/tableau.zip")
# Store a character scalar denoting the link to the unzipped directory
# that is to be created: unzip_path => character scalar
unzip_path <- paste0(tmpdir_path, "/global_superstore")
# Download the zip file: global_superstore.zip => stdout (zip_path)
download.file(durl, zip_path)
# Unzip the file into the unzip directory: tableau.zip => stdout (global_superstore)
unzip(zipfile = zip_path, exdir = unzip_path)
# Read in the excel file: df => data.frame
df <- read_xls(normalizePath(list.files(unzip_path, full.names = TRUE)))
# Regex the vector names to fit with R convention: names(df) => character vector
names(df) <- gsub("\W+", "_", tolower(trimws(names(df), "both")))
# Allocate some memory by creating an empty list the same size as the number of
# customers: df_list => list
df_list <- vector("list", length(unique(df$customer_id)))
# Split the data.frame into the list by the customer_id: df_list => lis
df_list <- with(df, split(df, customer_id))
# Sort the data (by date) and test whether or not each customer waited for their
# order before ordering again: orders_prior_to_delivery => data.frame
orders_prior_to_delivery <- data.frame(do.call("rbind", Map(function(x){
# Order the data.frame: y => data.frame
y <- x[order(x$order_date),]
# Return only the observations where the customer didn't wait:
# data.frame => GlobalEnv()
with(y, y[c(FALSE,
apply(data.frame(sapply(order_date[-1], `<`, ship_date[-nrow(y)])), 2, any)),])
},
df_list)), row.names = NULL, stringsAsFactors = FALSE)
# Unique customers and orders that were ordered prior to shipping the
# previous order: cust_orders_prior_to_delivery => data.frame
cust_orders_prior_to_delivery <-
unique(orders_prior_to_delivery[,c("order_id", "customer_id")])
编辑：我之前的回答没有正确处理订单日期 == 发货日期的情况。
我假设您已经将数据加载到名为 df 的对象中。您可以使用@hello_friend 代码的第一部分来获取它。
library(tidyverse)
df %>%
distinct(`Customer ID`, `Order ID`, `Order Date`, `Ship Date`) %>%
arrange(`Customer ID`, `Order Date`, `Ship Date`) %>%
mutate(sort_key = row_number()) %>%
pivot_longer(c(`Order Date`, `Ship Date`), names_to = "Activity", names_pattern = "(.*) Date", values_to = "Date") %>%
mutate(Activity = factor(Activity, ordered = TRUE, levels = c("Order", "Ship")),
Open = if_else(Activity == "Order", 1, -1)) %>%
group_by(`Customer ID`) %>%
arrange(Date, sort_key, Activity, .by_group = TRUE) %>%
mutate(Open = cumsum(Open)) %>%
ungroup %>%
filter(Open > 1, Activity == "Order") %>%
select(`Customer ID`, `Order ID`)
首先，只接受不同的订单和客户 ID，否则来自同一订单的多个项目会混淆并导致不正确的结果。然后，旋转数据，使每个订单成为两行，每行代表一个不同的 activity：订购或发货。我们创建了 运行 个未结订单总数。你要找什么时候变成两个或更多。
我使用 Activity 的有序因子来确保我总是在关闭订单之前打开订单。当订单日期和发货日期相同时，这很重要。
我使用一个特殊的 sort_key 栏来确保我在打开新订单之前关闭旧订单，以防客户在同一天订购其他东西。您可能需要相反的逻辑。
所有这些都假设给定的客户 ID 和订单 ID 在数据中只出现一次，这在您的数据集中实际上是不正确的，如您所见：
df %>% group_by(`Customer ID`, `Order ID`) %>% filter(n_distinct(`Ship Date`)> 1) %>% select(1:9)