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，我只能使用以下命令对数据集进行排序：

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

并找到start_station和end_station等于B的行索引。

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

下一部分是我遇到麻烦的地方。为了计算自行车在B站之间的时间，我们必须在difftime()（自行车离开）和下一个出现的记录之间取start_station = "B"。其中end_station= "B"，即使记录恰好在同一行（见第6行）。

使用下面的数据集，我们知道自行车在B站外的7:30:00和16:00:00之间，在车站外18:00:00和18:30:00之间的30分钟内花了510分钟B，在B站外19:00:00和22:30:00之间210分钟，平均为250 minutes.

如何使用difftime()在R中重现此输出？

> 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             А 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", "А", "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

这将根据发生的顺序计算差异，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

Answer 2

另一种可能性：

library(data.table)
d <- setDT(test)[ , {
  start = starttime[start_station == "B"]
  end = endtime[end_station == "B"]
  .(start = start, end = end, duration = difftime(end, start, units = "min"))
}
, by = .(trip = cumsum(start_station == "B"))]
d
#    trip               start                 end duration
# 1:    0                <NA> 2017-09-25 01:30:00  NA mins
# 2:    1 2017-09-25 07:30:00 2017-09-25 16:00:00 510 mins
# 3:    2 2017-09-25 18:00:00 2017-09-25 18:30:00  30 mins
# 4:    3 2017-09-25 19:00:00 2017-09-25 22:30:00 210 mins
# 5:    4 2017-09-25 23:00:00                <NA>  NA mins


d[ , mean(duration, na.rm = TRUE)]
# Time difference of 250 mins

# or
d[ , mean(as.integer(duration), na.rm = TRUE)]
# [1] 250

数据按计数器分组，每次自行车从“B”开始增加1（by = cumsum(start_station == "B")）。

下一个发生记录的索引

2 个答案: