使用R和Google Map API获取两点之间的驾驶距离（lat，lon）

本文介绍了使用R和Google Map API获取两点之间的驾驶距离（lat，lon）的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 我试图获得两个点之间的行驶距离，并给出经纬度。 我可以手动将它们放到Google地图中并获得驾驶距离，但我想以编程方式完成所有这些。 我猜JavaScript是要去的语言。但是，我不懂JavaScript，而且我对R的使用相当熟悉。我宁愿在R中执行它，因为我正在做R中的所有数据分析。 I我正在寻找路上的距离而不是乌鸦飞的距离。经过几个小时的尝试之后，我在R中写了下面的函数（这和 this one 帮助）。你有没有更好的方法来获得这个函数内的距离，或者有什么更简单的方法？ library（XML） latlon2ft< - 函数（来源，目标） { xml.url< - paste0（'http://maps.googleapis.com/maps/api/distancematrix/ xml？origins ='，origin，'& destinations ='，destination，'& mode = driving& sensor = false'） xmlfile< - xmlTreeParse（xml.url） xmltop = xmlRoot（xmlfile） distance< - xmltop [['row']] [[1]] [5] [1] [['distance']] [['value']] [[ 1]] distance ft return（ft ）} latlon2ft（origin = '37 .193489，-121.07395'，destination = '37 .151616，-121.046586'） RESULT = 17224.41 解决方案 RCurl 或同等价格。 图书馆（XML）库（bithop）库（RCurl） latlon2ft< - 函数（来源，目标）{ xml.url< - paste0（'http：/ /maps.googleapis.com/maps/api/distancematrix/xml?origins=',origin,'&destinations=',destination,'&mode=driving&sensor=false'） xmlfile< - xmlParse（getURL（xml.url）） dist distance< - as。数字（小（km，，dist）） ft 返回（英尺）} latlon2ft（origin = '37 .193489，-121.07395'，destination = '37 .151616，-121.046586'） 结果： [1] 17224.41 I am trying to get the driving distance between two points with lat/lon given. I can manually put them into google map and get the driving distance but I want to do all this programatically. I guess JavaScript is the language to go. But, I do not know JavaScript and I am fairly familiar using R. I would prefer to do it in R since I am doing all the data analysis in R. I am looking for distance along the road not crow-fly distance. After few hours of trying, I wrote the following function in R (This and this one helped). Do you have any better way to get the distance either within this function or anything very very simpler?library(XML)latlon2ft <- function(origin,destination){xml.url <- paste0('http://maps.googleapis.com/maps/api/distancematrix/xml?origins=',origin,'&destinations=',destination,'&mode=driving&sensor=false')xmlfile <- xmlTreeParse(xml.url)xmltop = xmlRoot(xmlfile)distance <- xmltop[['row']][[1]][5][1][['distance']][['value']][[1]]distance <- as.numeric(unclass(distance)[['value']])ft <- distance*3.28084 # FROM METER TO FEETreturn(ft)}latlon2ft(origin='37.193489,-121.07395',destination='37.151616,-121.046586')RESULT = 17224.41 解决方案 You need RCurl or an equivalent here.library(XML)library(bitops)library(RCurl)latlon2ft <- function(origin,destination){ xml.url <- paste0('http://maps.googleapis.com/maps/api/distancematrix/xml?origins=',origin,'&destinations=',destination,'&mode=driving&sensor=false') xmlfile <- xmlParse(getURL(xml.url)) dist <- xmlValue(xmlChildren(xpathApply(xmlfile,"//distance")[[1]])$value) distance <- as.numeric(sub(" km","",dist)) ft <- distance*3.28084 # FROM METER TO FEET return(ft)}latlon2ft(origin='37.193489,-121.07395',destination='37.151616,-121.046586')Result:[1] 17224.41 这篇关于使用R和Google Map API获取两点之间的驾驶距离（lat，lon）的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！