R中的条件计算

Question

我在根据条件计算事件之间的时间时遇到了一些问题。我想确定向客户退款与他们上次购买之间的时间。也就是说，退款时间减去他们最后一次按 ID 购买的时间。有多个用户按 ID 分组，每个用户都有多个事件（购买或退款）由 TIMESTAMP 索引。 table 的相关行如下所示：

View(df1)
TIMESTAMP   ID  Order_Type
2017-05-04  55  Purchase
2017-05-12  55  Purchase
2017-05-18  55  Purchase
2017-06-16  55  Refund 
2017-05-06  36  Purchase
2017-05-14  36  Purchase
2017-05-22  36  Purchase
2017-06-14  36  Purchase
2017-06-28  36  Refund 
2017-07-10  36  Purchase

与 table 一样，在某些情况下，客户收到了退款，但随后又进行了购买。我只想计算从上一个购买日期到退款。我在想我可以使用类似于聚合函数的东西。

输出为：

View(df2)
TIMESTAMP   ID  Days_Since_Last_Purchase
2017-06-16  55     29
2017-06-28  36     14

感谢您的任何意见。

Answer 1

这是基于 R 的有效解决方案：

df$TIMESTAMP <- as.Date.character(df$TIMESTAMP, format = "%Y-%m-%d")

inds <- which(df$Order_Type == "Refund")
df2  <- df[inds, ]

df2$Days_Since <- unlist(Map(`-`, df$TIMESTAMP[inds], df$TIMESTAMP[inds-1]))
#    TIMESTAMP  ID Order_Type Days_Since_Last_Purchase
#    2017-06-16 55     Refund                       29
#    2017-06-28 36     Refund                       14

在大多数（所有？）情况下，您也可以选择 mapply 而不是 Map：

df2$Days_Since <- mapply(difftime, df$TIMESTAMP[inds], df$TIMESTAMP[inds-1])

注意：这种方法的一个好处是它只使用基础 R。但是，正如 Moody_Mudskipper 在评论中指出的那样，这种解决方案仅适用于数据按时间顺序排列，每个退款记录之前都有其相应的购买记录。在大多数实际情况下，这很重要！

Answer 2

一个解决方案使用 dplyr 和 tidyr。

library(dplyr)
library(tidyr)

dt2 <- dt %>%
  mutate(RowID = 1:n()) %>%
  mutate(TIMESTAMP = as.Date(TIMESTAMP)) %>%
  spread(Order_Type, TIMESTAMP) %>%
  fill(Refund, .direction = "up") %>%
  mutate(Days_Since_Last_Purchase = Refund - Purchase) %>%
  filter(Days_Since_Last_Purchase > 0) %>%
  arrange(ID, Refund, Days_Since_Last_Purchase) %>%
  group_by(ID, Refund) %>%
  slice(1) %>%
  select(TIMESTAMP = Refund, ID, Days_Since_Last_Purchase)
dt2
# A tibble: 2 x 3
# Groups:   ID, TIMESTAMP [2]
   TIMESTAMP    ID Days_Since_Last_Purchase
      <date> <int>                   <time>
1 2017-06-28    36                  14 days
2 2017-06-16    55                  29 days

数据

dt <- read.table(text = "TIMESTAMP   ID  Order_Type
2017-05-04  55  Purchase
                 2017-05-12  55  Purchase
                 2017-05-18  55  Purchase
                 2017-06-16  55  Refund 
                 2017-05-06  36  Purchase
                 2017-05-14  36  Purchase
                 2017-05-22  36  Purchase
                 2017-06-14  36  Purchase
                 2017-06-28  36  Refund 
                 2017-07-10  36  Purchase",
                 header = TRUE, stringsAsFactors = FALSE)

Answer 3

另一个dplyr/tidyr解决方案

library(dplyr)
library(lubridate)
library(tidyr)
df %>%
  mutate(TIMESTAMP = as_date(TIMESTAMP)) %>%
  arrange(ID,TIMESTAMP) %>%
  group_by(ID) %>%
  mutate(refund_group = lag(cumsum(Order_Type == "Refund"),1,0)) %>% # as table is sorted, and we're inside a given group, every instance of "Refund" marks the end of a refund_group
  group_by(ID,refund_group,Order_Type) %>%
  do({tail(.,1)}) %>% # we keep the last instance of Purchase & Refund for each refund_group
  ungroup %>%
  spread(Order_Type,TIMESTAMP) %>%
  mutate(Days_Since_Last_Purchase = Refund - Purchase)   %>% # that's basically the final table, but we strip it further to get exactly the expected output
  select(TIMESTAMP = Refund,ID,Days_Since_Last_Purchase) %>%
  filter(!is.na(Days_Since_Last_Purchase))

结果

# A tibble: 2 x 3
   TIMESTAMP    ID Days_Since_Last_Purchase
      <date> <int>                   <time>
1 2017-06-28    36                  14 days
2 2017-06-16    55                  29 days

Answer 4

这是一个 for 循环，用于显示新 column 中相同 table 中的日期差异：

for(i in 2:nrow(df1)){
  df1$Days_Since_Last_Purchase[1] <- ""
  if((df1$Order_Type[i] == "Refund" & df1$Order_Type[i-1] == "Purchase") & (df1$ID[i] == df1$ID[i-1])){
      df1$Days_Since_Last_Purchase[i] <- difftime(df1$Timestamp[i],df1$Timestamp[i-1], units = c("days"))
  } else{
    df1$Days_Since_Last_Purchase[i] <- ""
  }
}

> df1
    Timestamp ID Order_Type Days_Since_Last_Purchase
1  2017-05-04 55   Purchase                         
2  2017-05-12 55   Purchase                         
3  2017-05-18 55   Purchase                         
4  2017-06-16 55     Refund                       29
5  2017-05-06 36   Purchase                         
6  2017-05-14 36   Purchase                         
7  2017-05-22 36   Purchase                         
8  2017-06-14 36   Purchase                         
9  2017-06-28 36     Refund                       14
10 2017-07-10 36   Purchase

R中的条件计算

Conditional calculation in R

r

intervals