如何计算R中组内日期的时间间隔？答案

【问题标题】：How to calculate time intervals of dates within groups in R?如何计算R中组内日期的时间间隔？
【发布时间】：2020-06-10 21:03:22
【问题描述】：

我有 200 万笔交易的数据，其中包括用户 ID、发票编号、发票日期和购买的物品。我想了解购买之间的平均时间间隔（按客户）。

sample.data <- data.frame(userID = c("ID1", "ID2", "ID2", "ID2", "ID3","ID3","ID4"),
invoiceNr = c("INV01", "INV02","INV03", "INV04","INV05", "INV06", "INV07"),
invoiceDate = lubridate::ymd("2008-06-29", "2008-10-10", "2008-10-10","2008-06-12","2008-12-11","2008-03-15","2008-07-14"),
items = c(LETTERS[1:7]))

我尝试使用填充新列的 for 循环生成间隔。

sample.data$intervals <- NA

for (i in 2:nrow(sample.data) {
  # if ID matches ID in previous row, calculate difference between purchase dates
  ifelse(sample.data$userID[i] == sample.data$userID[i-1], 
         sample.data$intervals[i] <- as.numeric(difftime(sample.data$invoiceDate[i], sample.data$invoiceDate[i-1], units = "days")),
  # if the previous ID is different, then do not calculate the time difference, but mark as NA (this is the first purchase in customers history)
        sample.data$intervals[i] <- NA)
}

在此处形成表格，我将能够汇总数据并通过 userID 计算总体平均值或平均值。

然而，对于这么大的数据集，for 循环需要很长时间。有更快/更好的方法吗？

【问题讨论】：

标签： r difftime

【解决方案1】：

你可以使用dplyr:

library(dplyr)

sample.data %>% 
  group_by(userID) %>% 
  arrange(invoiceDate) %>% 
  mutate(timediff = c(NA, diff(invoiceDate))) %>% 
  summarise(mean_time_diff = mean(timediff, na.rm = TRUE))
#> # A tibble: 4 x 2
#>   userID mean_time_diff
#>   <chr>           <dbl>
#> 1 ID1               NaN
#> 2 ID2                60
#> 3 ID3               271
#> 4 ID4               NaN

显然，如果用户只进行了一次购买，则购买之间的平均时间为NA

【讨论】：

【解决方案2】：

data.table 解决方案：

library(data.table)
sample.data[order(userID,invoiceDate),
            .(lastVisit=difftime(invoiceDate,lag(invoiceDate,),unit = "days"),
              nbVisit = .N),
            by=userID]

   userID lastVisit nbVisit
1:    ID1   NA days       1
2:    ID2   NA days       3
3:    ID2  120 days       3
4:    ID2    0 days       3
5:    ID3   NA days       2
6:    ID3  271 days       2
7:    ID4   NA days       1

您也可以为每个客户平均：

sample.data[order(userID,invoiceDate),
            .(lastVisit=difftime(invoiceDate,lag(invoiceDate,),unit = "days"),
              nbVisit = .N),
            by=userID][,.(avg=mean(lastVisit,na.rm=T)),by=.(userID,nbVisit)]
   userID nbVisit      avg
1:    ID1       1 NaN days
2:    ID2       3  60 days
3:    ID3       2 271 days
4:    ID4       1 NaN days

【讨论】：