这是一些数据的样本
Tag.ID TimeStep.coa Latitude.coa Longitude.coa
<chr> <dttm> <dbl> <dbl>
1 1657 2017-08-17 12:00:00 72.4 -81.1
2 1657 2017-08-17 18:00:00 72.3 -81.1
3 1658 2017-08-14 18:00:00 72.3 -81.2
4 1658 2017-08-15 00:00:00 72.3 -81.3
5 1659 2017-08-14 18:00:00 72.3 -81.1
6 1659 2017-08-15 00:00:00 72.3 -81.2
7 1660 2017-08-20 18:00:00 72.3 -81.1
8 1660 2017-08-21 00:00:00 72.3 -81.2
9 1660 2017-08-21 06:00:00 72.3 -81.2
10 1660 2017-08-21 12:00:00 72.3 -81.3
11 1661 2017-08-28 12:00:00 72.4 -81.1
12 1661 2017-08-28 18:00:00 72.3 -81.1
13 1661 2017-08-29 06:00:00 72.3 -81.2
14 1661 2017-08-29 12:00:00 72.3 -81.2
15 1661 2017-08-30 06:00:00 72.3 -81.2
16 1661 2017-08-30 18:00:00 72.3 -81.2
17 1661 2017-08-31 00:00:00 72.3 -81.2
18 1661 2017-08-31 06:00:00 72.3 -81.2
19 1661 2017-08-31 12:00:00 72.3 -81.2
20 1661 2017-08-31 18:00:00 72.4 -81.1
我正在寻找一种获取每个ID行驶距离的方法。我将在VTrack包中使用ComputeDistance函数（尽管可以使用其他函数）。该函数如下所示：
ComputeDistance( Lat1, Lat2, Lon1, Lon2)
这将计算纬度/经度坐标之间的直线距离。 我最终想要一个具有四列Tag.ID，Timestep1，Timestep2和distance的数据框。这是一个示例：
Tag.ID Timestep1 Timestep2 Distance
1657 2017-08-17 12:00:00 2017-08-17 18:00:00 ComputeDistance(72.4,72.3,-81.1,-81.1)
1658 2017-08-14 18:00:00 2017-08-15 00:00:00 ComputeDistance(72.3,72.3,-81.2,-81.3)
1659 2017-08-14 18:00:00 2017-08-15 00:00:00 ComputeDistance(72.3,72.3,-81.1,-81.2)
1660 2017-08-20 18:00:00 2017-08-21 00:00:00 ComputeDistance(72.3,72.3,-81.1,-81.2)
1660 2017-08-21 00:00:00 2017-08-21 06:00:00 ComputeDistance(72.3,72.3,=81.1,-81.2
依此类推
编辑： 这是我使用的代码（感谢AntoniosK）。 COASpeeds2与上面的示例df完全相同：
test <- COASpeeds2 %>%
group_by(Tag.ID) %>%
mutate(Timestep1 = TimeStep.coa,
Timestep2 = lead(TimeStep.coa),
Distance = ComputeDistance(Latitude.coa, lead(Latitude.coa),
Longitude.coa, lead(Longitude.coa))) %>%
ungroup() %>%
na.omit() %>%
select(Tag.ID, Timestep1, Timestep2, Distance)
这是我得到的df。
Tag.ID Timestep1 Timestep2 Distance
<fct> <dttm> <dttm> <dbl>
1 1657 2017-08-17 12:00:00 2017-08-17 18:00:00 2.76
2 1657 2017-08-17 18:00:00 2017-08-14 18:00:00 1.40
3 1658 2017-08-14 18:00:00 2017-08-15 00:00:00 6.51
4 1658 2017-08-15 00:00:00 2017-08-14 18:00:00 10.5
5 1659 2017-08-14 18:00:00 2017-08-15 00:00:00 7.51
6 1659 2017-08-15 00:00:00 2017-08-20 18:00:00 7.55
7 1660 2017-08-20 18:00:00 2017-08-21 00:00:00 3.69
8 1660 2017-08-21 00:00:00 2017-08-21 06:00:00 4.32
9 1660 2017-08-21 06:00:00 2017-08-21 12:00:00 3.26
10 1660 2017-08-21 12:00:00 2017-08-28 12:00:00 10.5
11 1661 2017-08-28 12:00:00 2017-08-28 18:00:00 1.60
12 1661 2017-08-28 18:00:00 2017-08-29 06:00:00 1.94
13 1661 2017-08-29 06:00:00 2017-08-29 12:00:00 5.22
14 1661 2017-08-29 12:00:00 2017-08-30 06:00:00 0.759
15 1661 2017-08-30 06:00:00 2017-08-30 18:00:00 1.94
16 1661 2017-08-30 18:00:00 2017-08-31 00:00:00 0.342
17 1661 2017-08-31 00:00:00 2017-08-31 06:00:00 0.281
18 1661 2017-08-31 06:00:00 2017-08-31 12:00:00 4.21
19 1661 2017-08-31 12:00:00 2017-08-31 18:00:00 8.77
答案 0 :(得分：1)
library(tidyverse)
library(VTrack)
# example data
dt = read.table(text = "
Tag.ID TimeStep.coa Latitude.coa Longitude.coa
1 1657 2017-08-17_12:00:00 72.4 -81.1
2 1657 2017-08-17_18:00:00 72.3 -81.1
3 1658 2017-08-14_18:00:00 72.3 -81.2
4 1658 2017-08-15_00:00:00 72.3 -81.3
5 1659 2017-08-14_18:00:00 72.3 -81.1
6 1659 2017-08-15_00:00:00 72.3 -81.2
7 1660 2017-08-20_18:00:00 72.3 -81.1
8 1660 2017-08-21_00:00:00 72.3 -81.2
9 1660 2017-08-21_06:00:00 72.3 -81.2
10 1660 2017-08-21_12:00:00 72.3 -81.3
", header=T)
dt %>%
group_by(Tag.ID) %>%
mutate(Timestep1 = TimeStep.coa,
Timestep2 = lead(TimeStep.coa),
Distance = ComputeDistance(Latitude.coa, lead(Latitude.coa),
Longitude.coa, lead(Longitude.coa))) %>%
ungroup() %>%
na.omit() %>%
select(Tag.ID, Timestep1, Timestep2, Distance)
因此，您得到以下信息：
# # A tibble: 6 x 4
# Tag.ID Timestep1 Timestep2 Distance
# <int> <fct> <fct> <dbl>
# 1 1657 2017-08-17_12:00:00 2017-08-17_18:00:00 11.1
# 2 1658 2017-08-14_18:00:00 2017-08-15_00:00:00 3.38
# 3 1659 2017-08-14_18:00:00 2017-08-15_00:00:00 3.38
# 4 1660 2017-08-20_18:00:00 2017-08-21_00:00:00 3.38
# 5 1660 2017-08-21_00:00:00 2017-08-21_06:00:00 0.0000949
# 6 1660 2017-08-21_06:00:00 2017-08-21_12:00:00 3.38
答案 1 :(得分：0)
假设起点和终点是有序的，并且有一对配对。
这是另一个选择：
#identify the start and end of each trip
df$leg<-rep(c("Start", "End"), nrow(df)/2)
#label each trip
df$trip <- rep(1:(nrow(df)/2), each=2)
#change the shape
library(tidyr)
output<-pivot_wider(df, id_cols = c(Tag.ID, trip),
names_from = leg,
values_from = c(TimeStep.coa, Latitude.coa, Longitude.coa))
#calcuate distance (use your package of choice)
library(geosphere)
output$distance<-distGeo(output[ ,c("Longitude.coa_Start", "Latitude.coa_Start")],
output[ ,c("Longitude.coa_End", "Latitude.coa_End")])
# #remove undesired columns
# output <- output[, -c(5, 6, 7, 8)]
output
> output[, -c(5, 6, 7, 8)]
# A tibble: 10 x 5
Tag.ID trip TimeStep.coa_Start TimeStep.coa_End distance
<int> <int> <fct> <fct> <dbl>
1 1657 1 2017-08-17 12:00:00 2017-08-17 18:00:00 11159.
2 1658 2 2017-08-14 18:00:00 2017-08-15 00:00:00 3395.
3 1659 3 2017-08-14 18:00:00 2017-08-15 00:00:00 3395.
4 1660 4 2017-08-20 18:00:00 2017-08-21 00:00:00 3395.
5 1660 5 2017-08-21 06:00:00 2017-08-21 12:00:00 3395.
6 1661 6 2017-08-28 12:00:00 2017-08-28 18:00:00 11159.
7 1661 7 2017-08-29 06:00:00 2017-08-29 12:00:00 0
8 1661 8 2017-08-30 06:00:00 2017-08-30 18:00:00 0
9 1661 9 2017-08-31 00:00:00 2017-08-31 06:00:00 0
10 1661 10 2017-08-31 12:00:00 2017-08-31 18:00:00 11661.
答案 2 :(得分：0)
您可以通过geosphere::distGeo方法使用by。
library(geosphere)
do.call(rbind.data.frame, by(dat, dat$Tag.ID, function(s) {
t.diff <- (s$TimeStep.coa[length(s$TimeStep.coa)] - s$TimeStep.coa[1])
d.diff <- sum(mapply(function(x, y)
distGeo(s[x, 3:4], s[y, 3:4]), x=1:(nrow(s)-1), y=2:nrow(s)))/1e3
`colnames<-`(cbind(t.diff, d.diff), c("hours", "km"))
}))
# hours km
# 1657 6.00 1.727882
# 1658 6.00 11.166785
# 1659 6.00 11.166726
# 1660 18.00 22.333511
# 1661 3.25 24.192753
数据：
dat <- structure(list(Tag.ID = c(1657L, 1657L, 1658L, 1658L, 1659L,
1659L, 1660L, 1660L, 1660L, 1660L, 1661L, 1661L, 1661L, 1661L,
1661L, 1661L, 1661L, 1661L, 1661L, 1661L), TimeStep.coa = structure(c(1502964000,
1502985600, 1502726400, 1502748000, 1502726400, 1502748000, 1503244800,
1503266400, 1503288000, 1503309600, 1503914400, 1503936000, 1503979200,
1504000800, 1504065600, 1504108800, 1504130400, 1504152000, 1504173600,
1504195200), class = c("POSIXct", "POSIXt"), tzone = ""), Latitude.coa = c(72.4,
72.3, 72.3, 72.3, 72.3, 72.3, 72.3, 72.3, 72.3, 72.3, 72.4, 72.3,
72.3, 72.3, 72.3, 72.3, 72.3, 72.3, 72.3, 72.4), Longitude.coa = c(-81.1,
-81.1, -81.2, -81.3, -81.1, -81.2, -81.1, -81.2, -81.2, -81.3,
-81.1, -81.1, -81.2, -81.2, -81.2, -81.2, -81.2, -81.2, -81.2,
-81.1)), row.names = c(NA, -20L), class = "data.frame")