下一个发生记录的索引

Question

我有一个自行车轨迹的样本数据集.我的目标是平均计算出访问B站之间的时间.

到目前为止,我已经能够简单地对数据集进行排序:

test[order(test$starttime, decreasing = FALSE),]

并找到的行索引start_station和end_station等于B.

 which(test$start_station == 'B')
 which(test$end_station == 'B')

接下来的部分是我遇到麻烦的地方.为了计算出失误的时候自行车是在站B之间,我们必须采取的时间difftime(),其中间start_station = "B"(自行车离开)和下一个出现的创纪录的 地方end_station= "B", 即使记录恰好是在同一行中(见第6行).

使用下面的数据集,我们知道自行车在B站之间7:30:00和16:00:00之外花费了510分钟,B站之间和之外30分钟,18:00:00以及18:30:00B站之间19:00:00和22:30:00之外210分钟,平均为250 minutes.

如何使用R在R中重现此输出difftime()？

> test
   bikeid start_station           starttime end_station             endtime
1       1             A 2017-09-25 01:00:00           B 2017-09-25 01:30:00
2       1             B 2017-09-25 07:30:00           C 2017-09-25 08:00:00
3       1             C 2017-09-25 10:00:00           A 2017-09-25 10:30:00
4       1             A 2017-09-25 13:00:00           C 2017-09-25 13:30:00
5       1             C 2017-09-25 15:30:00           B 2017-09-25 16:00:00
6       1             B 2017-09-25 18:00:00           B 2017-09-25 18:30:00
7       1             B 2017-09-25 19:00:00           A 2017-09-25 19:30:00
8       1             ? 2017-09-25 20:00:00           C 2017-09-25 20:30:00
9       1             C 2017-09-25 22:00:00           B 2017-09-25 22:30:00
10      1             B 2017-09-25 23:00:00           C 2017-09-25 23:30:00

以下是示例数据:

> dput(test)
structure(list(bikeid = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1), start_station = c("A", 
"B", "C", "A", "C", "B", "B", "?", "C", "B"), starttime = structure(c(1506315600, 
1506339000, 1506348000, 1506358800, 1506367800, 1506376800, 1506380400, 
1506384000, 1506391200, 1506394800), class = c("POSIXct", "POSIXt"
), tzone = ""), end_station = c("B", "C", "A", "C", "B", "B", 
"A", "C", "B", "C"), endtime = structure(c(1506317400, 1506340800, 
1506349800, 1506360600, 1506369600, 1506378600, 1506382200, 1506385800, 
1506393000, 1506396600), class = c("POSIXct", "POSIXt"), tzone = "")), .Names = c("bikeid", 
"start_station", "starttime", "end_station", "endtime"), row.names = c(NA, 
-10L), class = "data.frame")

Answer 1

这将按照出现的顺序计算差异，但不会将其附加到data.frame

lapply(df1$starttime[df1$start_station == "B"], function(x, et) difftime(et[x < et][1], x, units = "mins"), et = df1$endtime[df1$end_station == "B"])

[[1]]
Time difference of 510 mins

[[2]]
Time difference of 30 mins

[[3]]
Time difference of 210 mins

[[4]]
Time difference of NA mins

计算平均时间：

v1 <- sapply(df1$starttime[df1$start_station == "B"], function(x, et) difftime(et[x < et][1], x, units = "mins"), et = df1$endtime[df1$end_station == "B"])
mean(v1, na.rm = TRUE)

[1] 250