计算经纬度向量的总里程数 [英] Calculate total miles traveled from vectors of lat / lon

问题描述

我有一个数据框，里面有关于驱动程序的数据和他们遵循的路线。我试图找出旅行总里程。我正在使用 geosphere 包，但无法找出正确的方法来应用它并获得以英里为单位的答案。

 > 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年11月12日2点10分41秒123 76.90897 -110.8613 
 4 4 2012年11月12日三点18分41秒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））
  

以及其他一些函数，但似乎无法得到合理的答案。

有什么建议吗？

 > 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 ，
2 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，
 76.4115052509587，76.8397977722 343，76.3357809444424，76.032417796785，
 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，
 -110.55 6956546381，-110.24483308522，-110.217355202651）），.Names = C（ ID，
 的routeDateTime， driverId， LAT， LON），row.names = C（NA，
 -40L），class =data.frame）
  

解决方案

这个怎么样？

  ##设置
 library（geosphere）
 metersPerMile < - > 1609.34 
 pts < -  df1 [c（lon，lat）] 
 
 ##传入两个滞后一个点的派生数据帧
 segDists <  -  distVincentyEllipsoid（P1 = PTS [-nrow（DF），]，
 P2 = PTS [-1，]）
总和（segDists）/ metersPerMile 
＃[1] 1013.919 $ （要使用更快的距离计算算法之一，只需替换 distCosine  b 
 
 
 $ b ， distVincentySphere 或 distHaversine 为 distVincentyEllipsoid 在上面的调用中。）
 
I have a data frame with data about a driver and the route they followed.  I'm trying to figure out the total mileage traveled.  I'm using the geosphere package but can't figure out the correct way to apply it and get an answer in miles.
> 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
so far I've tried 
spDists(cbind(df1$lon,df1$lat))
and several other functions but can't seem to get a reasonable answer. 

Any suggestions?
> 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")

解决方案
How about this?
## 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
(To use one of the faster distance calculation algorithms, just substitute distCosine, distVincentySphere, or distHaversine for distVincentyEllipsoid in the call above.)

                        
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