r - 算法效率——时差循环

Question

我有一个名为 vistsPerDay 的数据集，它看起来像这样，但有 405,890 行和 10,406 个唯一的 CUST_ID:

> CUST_ID   Date
> 1         2013-09-19
> 1         2013-10-03
> 1         2013-10-08
> 1         2013-10-12
> 1         2013-10-20
> 1         2013-10-25
> 1         2013-11-01
> 1         2013-11-02
> 1         2013-11-08
> 1         2013-11-15
> 1         2013-11-23
> 1         2013-12-02
> 1         2013-12-04
> 1         2013-12-09
> 2         2013-09-16
> 2         2013-09-17
> 2         2013-09-18

我想做的是创建一个新变量，它是访问日期之间的滞后差异。这是我目前使用的代码:

visitsPerDay <- visitsPerDay[order(visitsPerDay$CUST_ID), ]
cust_id <- 0
for (i in 1:nrow(visitsPerDay)) {
  if (visitsPerDay$CUST_ID[i] != cust_id) {
    cust_id <- visitsPerDay$CUST_ID[i]
    visitsPerDay$MTBV <- NA
  } else {
    visitsPerDay$MBTV <- as.numeric(visitsPerDay$Date[i] - visitsPerDay$Date[i-1])
  }
}

我觉得这肯定不是最有效的方法。有没有人有更好的方法来处理它？谢谢!

Answer 1

这是data.table 解决方案。这可能会更快并且更具可读性:

dt = data.table(visitsPerDay)

dt[, MBTV := c(NA, diff(as.Date(Date))), by = CUST_ID]
dt
#    CUST_ID       Date    MBTV
# 1:       1 2013-09-19 NA days
# 2:       1 2013-10-03 14 days
# 3:       1 2013-10-08  5 days
# 4:       1 2013-10-12  4 days
# 5:       1 2013-10-20  8 days
# 6:       1 2013-10-25  5 days
# 7:       1 2013-11-01  7 days
# 8:       1 2013-11-02  1 days
# 9:       1 2013-11-08  6 days
#10:       1 2013-11-15  7 days
#11:       1 2013-11-23  8 days
#12:       1 2013-12-02  9 days
#13:       1 2013-12-04  2 days
#14:       1 2013-12-09  5 days
#15:       2 2013-09-16 NA days
#16:       2 2013-09-17  1 days
#17:       2 2013-09-18  1 days