计算从纬度/经度矢量行进的总里程

Question

我有一个数据框,其中包含有关驱动程序及其遵循的路径的数据.我想弄清楚旅行的总里程数.我正在使用该geosphere软件包,但无法弄清楚应用它的正确方法并以英里为单位获得答案.

> head(df1)
  id       routeDateTime driverId      lat       lon
1  1 2012-11-12 02:08:41      123 76.57169 -110.8070
2  2 2012-11-12 02:09:41      123 76.44325 -110.7525
3  3 2012-11-12 02:10:41      123 76.90897 -110.8613
4  4 2012-11-12 03:18:41      123 76.11152 -110.2037
5  5 2012-11-12 03:19:41      123 76.29013 -110.3838
6  6 2012-11-12 03:20:41      123 76.15544 -110.4506

到目前为止我已经尝试过了

spDists(cbind(df1$lon,df1$lat))

和其他几个功能似乎无法得到一个合理的答案.

有什么建议？

> dput(df1)
structure(list(id = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 
13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 
29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40), routeDateTime = c("2012-11-12 02:08:41", 
"2012-11-12 02:09:41", "2012-11-12 02:10:41", "2012-11-12 03:18:41", 
"2012-11-12 03:19:41", "2012-11-12 03:20:41", "2012-11-12 03:21:41", 
"2012-11-12 12:08:41", "2012-11-12 12:09:41", "2012-11-12 12:10:41", 
"2012-11-12 02:08:41", "2012-11-12 02:09:41", "2012-11-12 02:10:41", 
"2012-11-12 03:18:41", "2012-11-12 03:19:41", "2012-11-12 03:20:41", 
"2012-11-12 03:21:41", "2012-11-12 12:08:41", "2012-11-12 12:09:41", 
"2012-11-12 12:10:41", "2012-11-12 02:08:41", "2012-11-12 02:09:41", 
"2012-11-12 02:10:41", "2012-11-12 03:18:41", "2012-11-12 03:19:41", 
"2012-11-12 03:20:41", "2012-11-12 03:21:41", "2012-11-12 12:08:41", 
"2012-11-12 12:09:41", "2012-11-12 12:10:41", "2012-11-12 02:08:41", 
"2012-11-12 02:09:41", "2012-11-12 02:10:41", "2012-11-12 03:18:41", 
"2012-11-12 03:19:41", "2012-11-12 03:20:41", "2012-11-12 03:21:41", 
"2012-11-12 12:08:41", "2012-11-12 12:09:41", "2012-11-12 12:10:41"
), driverId = c(123, 123, 123, 123, 123, 123, 123, 123, 123, 
123, 456, 456, 456, 456, 456, 456, 456, 456, 456, 456, 789, 789, 
789, 789, 789, 789, 789, 789, 789, 789, 246, 246, 246, 246, 246, 
246, 246, 246, 246, 246), lat = c(76.5716897079255, 76.4432530414779, 
76.9089707506355, 76.1115217276383, 76.2901271982118, 76.155437662499, 
76.4115052509587, 76.8397977722343, 76.3357809444424, 76.032417796785, 
76.5716897079255, 76.4432530414779, 76.9089707506355, 76.1115217276383, 
76.2901271982118, 76.155437662499, 76.4115052509587, 76.8397977722343, 
76.3357809444424, 76.032417796785, 76.5716897079255, 76.4432530414779, 
76.9089707506355, 76.1115217276383, 76.2901271982118, 76.155437662499, 
76.4115052509587, 76.8397977722343, 76.3357809444424, 76.032417796785, 
76.5716897079255, 76.4432530414779, 76.9089707506355, 76.1115217276383, 
76.2901271982118, 76.155437662499, 76.4115052509587, 76.8397977722343, 
76.3357809444424, 76.032417796785), lon = c(-110.80701574916, 
-110.75247172825, -110.861284852726, -110.203674311982, -110.383751512505, 
-110.450569844106, -110.22185564111, -110.556956546381, -110.24483308522, 
-110.217355202651, -110.80701574916, -110.75247172825, -110.861284852726, 
-110.203674311982, -110.383751512505, -110.450569844106, -110.22185564111, 
-110.556956546381, -110.24483308522, -110.217355202651, -110.80701574916, 
-110.75247172825, -110.861284852726, -110.203674311982, -110.383751512505, 
-110.450569844106, -110.22185564111, -110.556956546381, -110.24483308522, 
-110.217355202651, -110.80701574916, -110.75247172825, -110.861284852726, 
-110.203674311982, -110.383751512505, -110.450569844106, -110.22185564111, 
-110.556956546381, -110.24483308522, -110.217355202651)), .Names = c("id", 
"routeDateTime", "driverId", "lat", "lon"), row.names = c(NA, 
-40L), class = "data.frame")

Answer 1

这个怎么样？

## Setup
library(geosphere)
metersPerMile <- 1609.34
pts <- df1[c("lon", "lat")]

## Pass in two derived data.frames that are lagged by one point
segDists <- distVincentyEllipsoid(p1 = pts[-nrow(df),], 
                                  p2 = pts[-1,])
sum(segDists)/metersPerMile
# [1] 1013.919

(要使用的更快的距离计算算法中的一个,只是替换distCosine,distVincentySphere或distHaversine用于distVincentyEllipsoid在呼叫的上方.)