使用 data.table 逐行更新相互依赖的两列

Question

我想创建一个数据表，其中包含巴士站之间的出发和到达时间。这是我的data.table 的格式。 （下面的可重现数据集）

    trip_id stop_sequence arrival_time departure_time travel_time
 1:       a             1     07:00:00       07:00:00    00:00:00
 2:       a             2     00:00:00       00:00:00    00:02:41
 3:       a             3     00:00:00       00:00:00    00:01:36
 4:       a             4     00:00:00       00:00:00    00:02:39
 5:       a             5     00:00:00       00:00:00    00:02:28
 6:       b             1     07:00:00       07:00:00    00:00:00
 7:       b             2     00:00:00       00:00:00    00:00:00
 8:       b             3     00:00:00       00:00:00    00:01:36
 9:       b             4     00:00:00       00:00:00    00:00:37
10:       b             5     00:00:00       00:00:00    00:03:00

这是它应该如何工作的。这个想法是车辆按照停止顺序行驶。例如，在行程a 中，车辆从停止1 行驶到停止2 需要00:02:41。给定乘客在每个停靠站进出车辆的固定时间为 40 秒，巴士将从停靠站2 出发"07:03:21"

这里的问题是，这是两列之间的逐行迭代过程。直觉上，我会选择for set loop in data.table，但我无法理解这一点。帮忙？

可重现的数据集：

library(data.table)
library(chron)

dt <- structure(list(trip_id = c("a", "a", "a", "a", "a", "b", "b", 
      "b", "b", "b"), stop_sequence = c(1L, 2L, 3L, 4L, 5L, 1L, 2L, 
      3L, 4L, 5L), arrival_time = structure(c(0.291666666666667, 0, 
      0, 0, 0, 0.291666666666667, 0, 0, 0, 0), format = "h:m:s", class = "times"), 
      departure_time = structure(c(0.291666666666667, 0, 0, 0, 
      0, 0.291666666666667, 0, 0, 0, 0), format = "h:m:s", class = "times"), 
      travel_time = structure(c(0, 0.00186598685444013, 0.00110857958406301, 
      0.00183749407361369, 0.00171664297781446, 0, 0.000522388450578203, 
      0.00111473367541453, 0.000427755975518318, 0.00207918951573377
      ), format = "h:m:s", class = "times")), .Names = c("trip_id", 
      "stop_sequence", "arrival_time", "departure_time", "travel_time"
      ), class = c("data.table", "data.frame"), row.names = c(NA, -10L
      ))

预期输出：前四行

   trip_id stop_sequence arrival_time departure_time travel_time
1:       a             1     07:00:00       07:00:00    00:00:00
2:       a             2     07:02:41       07:03:21    00:02:41
3:       a             3     07:04:57       07:05:37    00:01:36
4:       a             4     07:08:16       07:08:56    00:02:39

Answer 1

我认为不用循环也可以做到这一点。我认为您可以在不循环的情况下计算departure_time，然后一旦有了，arrival_time 就是departure_time - 40 seconds：

dt2 <- copy(dt)
dt2[,c("arrival_time", "departure_time") := .(cumsum(arrival_time + ifelse(travel_time==0, 0, travel_time + times("00:00:40"))) - ifelse(travel_time == 0 , 0, times("00:00:40")),
                                              cumsum(arrival_time + ifelse(travel_time==0, 0, travel_time + times("00:00:40")))),
    by = trip_id]

dt2

 #   trip_id stop_sequence arrival_time departure_time travel_time
 #1:       a             1     07:00:00       07:00:00    00:00:00
 #2:       a             2     07:02:41       07:03:21    00:02:41
 #3:       a             3     07:04:57       07:05:37    00:01:36
 #4:       a             4     07:08:16       07:08:56    00:02:39
 #5:       a             5     07:11:24       07:12:04    00:02:28
 #6:       b             1     07:00:00       07:00:00    00:00:00
 #7:       b             2     07:00:45       07:01:25    00:00:45
 #8:       b             3     07:03:01       07:03:41    00:01:36
 #9:       b             4     07:04:18       07:04:58    00:00:37
#10:       b             5     07:07:58       07:08:38    00:03:00

或者，因此您不必为departure_time 重复很长的cumsum 来获得arrival_time，您可以这样做：

dt2[,departure_time := cumsum(arrival_time + ifelse(travel_time==0, 0, travel_time + times("00:00:40"))), by = trip_id]
dt2[, arrival_time := departure_time - ifelse(travel_time == 0 , 0, times("00:00:40"))]

---

@eddi 发布的第三个选项：

dt[, departure_time := arrival_time[1] + cumsum(travel_time) + (0:(.N-1))*times('00:00:40'), by = trip_id]
dt[, arrival_time := c(arrival_time[1], tail(departure_time, -1) - times('00:00:40')), by = trip_id]