Question

我试图通过给出lat / lon来获得两点之间的行驶距离。我可以手动将它们放入谷歌地图并获得驾驶距离，但我想以编程方式完成所有这些操作。

我猜JavaScript是最常用的语言。但是，我不知道JavaScript，我很熟悉使用R.我更喜欢在R中这样做，因为我在R中进行所有数据分析。

我正在寻找沿着道路的距离而不是飞行距离。经过几个小时的尝试，我在R中写了以下函数（This和this one帮助）。你有没有更好的方法来获得这个功能或任何非常简单的距离？

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 FEET
return(ft)
}

latlon2ft(origin='37.193489,-121.07395',destination='37.151616,-121.046586')

结果= 17224.41

Answer 1

您需要RCurl或同等资格。

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')

结果：

[1] 17224.41

Answer 2

我撰写了gmapsdistance包以做到这一点。它可以在CRAN上使用。您可以通过以下方式使用该功能：

results = gmapsdistance(origin = "38.1621328+24.0029257",
                        destination = "37.9908372+23.7383394",
                        mode = "walking") results
# $Time
# [1] 30025
# 
# $Distance
# [1] 39507
# 
# $Status
# [1] "OK"

您还可以包含原点和目的地的矢量，并获得生成的距离矩阵。它还支持方向，并有许多选项：

results = gmapsdistance(origin = c("Washington+DC", "New+York+NY", "Seattle+WA", "Miami+FL"), 
                        destination = c("Los+Angeles+CA", "Austin+TX", "Chicago+IL", "Philadelphia+PA"), 
                        mode = "bicycling", 
                        departure = 1514742000)
results
# $Time
#              or Time.Los+Angeles+CA Time.Austin+TX Time.Chicago+IL Time.Philadelphia+PA
# 1 Washington+DC              856621         535146          247765                54430
# 2   New+York+NY              917486         596011          308630                32215
# 3    Seattle+WA              374692         678959          674989               956702
# 4      Miami+FL              829039         416667          452035               411283
# 
# $Distance
#              or Distance.Los+Angeles+CA Distance.Austin+TX Distance.Chicago+IL Distance.Philadelphia+PA
# 1 Washington+DC                 4567470            2838519             1303067                   266508
# 2   New+York+NY                 4855086            3126136             1590684                   160917
# 3    Seattle+WA                 1982354            3562970             3588297                  5051951
# 4      Miami+FL                 4559205            2279966             2381610                  2169382
# 
# $Status
#              or status.Los+Angeles+CA status.Austin+TX status.Chicago+IL status.Philadelphia+PA
# 1 Washington+DC                    OK               OK                OK                     OK
# 2   New+York+NY                    OK               OK                OK                     OK
# 3    Seattle+WA                    OK               OK                OK                     OK
# 4      Miami+FL                    OK               OK                OK                     OK

Answer 3

我需要为一堆地址计算行驶距离，所以我为它编写了一个简短的函数并将其放入同样小的数据包中。您可以在我的GitHub仓库中找到它：https://github.com/JanMultmeier/GeoData/blob/master/GeoDataPackage/R/GetDist.R

这应该让它运行：

require(devtools)
install_github("JanMultmeier/GeoData/GeoDataPackage")
library(GeoData)
getDist(from="1 Infinity Loop, Cupertino, CA 95014", to="1600 Amphitheatre Pkwy, Mountain View, CA 94043",modus="driving",get="distance")

它应该返回14.8公里。

Barryhunter已经暗示谷歌的使用限制，将此API的使用与在Google地图上显示结果联系在一起。

希望这仍然可以帮助一些偶然发现这篇文章的人（像我一样）......

Answer 4

我已使用Google Maps API编写googleway包来执行此操作

特别是，google_directions()功能会为您提供行车距离，路线，路线，腿，步等。 google_distance()函数将为您提供所有起点/目的地的距离矩阵

您需要Google API密钥才能使用其API

library(googleway)

## your valid API key
key <- "your_api_key_here"

directions <- google_directions(origin = c(37.193489,-121.07395),
                                destination = c(37.151616,-121.046586),
                                key = key, 
                                simplify = T)

directions$routes$legs
# [[1]]
# distance.text distance.value duration.text duration.value duration_in_traffic.text duration_in_traffic.value                 end_address
# 1        5.2 km           5250        3 mins            161                   3 mins                       156 I-5, Gustine, CA 95322, USA
# end_location.lat end_location.lng               start_address start_location.lat start_location.lng
# 1         37.15162        -121.0466 I-5, Gustine, CA 95322, USA           37.19349           -121.074
# steps
# 1 5.2 km, 5250, 3 mins, 161, 37.1516163, -121.0465852, Head <b>southeast</b> on <b>I-5 S</b>, ij_bFfg~aVpBgA`DkB~FeDbIwEpEgCtaAsj@nAs@lDqBxIaF~FgDlHcEjC{AdFuCrBkAhC{A|A{@|A}@bAk@rBkArBkA|A{@`DiB|A}@vDwBdAm@dAm@rBkA|A{@zA{@~J{FpC_B~A}@tBkAjHeEvGuDlMmHtBkAVO, 37.1934864, -121.0739565, DRIVING
#   traffic_speed_entry via_waypoint
#   1                NULL         NULL



google_distance(origins = list(c(37.193489,-121.07395)),
                destinations = list(c(37.151616,-121.046586)),
                key = key, 
                simplify = T,
                units = "imperial")

# $destination_addresses
# [1] "I-5, Gustine, CA 95322, USA"
# 
# $origin_addresses
# [1] "I-5, Gustine, CA 95322, USA"
# 
# $rows
# elements
# 1 3.3 mi, 5250, 3 mins, 161, 3 mins, 157, OK
# 
# $status
# [1] "OK"

鉴于google_directions()函数返回折线（搜索路线时您在Google地图上获得的线），我们可以在Google地图上绘制它

key <- 'your_map_api_key'

df_route <- decode_pl(directions$routes$overview_polyline$points)

google_map(data = df_route, key = key, height = 800, search_box = T) %>%
    add_markers()
## or you can use `add_polyline()` to view the entire line

Answer 5

在撰写本文时，Renjin（基于Java的R解释器）没有很多软件包来帮助解决这个问题。这是一个不依赖于额外包的实现。

# Computes the distance between two locations in meters. This uses an online
# map API and therefore an Internet connection is required for an accurate
# result. If no connection is found, this will use the Haversine formula
# to provide a rough estimate for the distance.
#
# @param src The source latitude and longitude.
# @param dst The destination latitude and longitude.
# @param mode Driving, cycling, walking, etc.
distance <- function( lat1, lon1, lat2, lon2, mode = 'driving' ) {
  lat1 = as.numeric( lat1 )
  lon1 = as.numeric( lon1 )
  lat2 = as.numeric( lat2 )
  lon2 = as.numeric( lon2 )

  # Create the URL to use to get the distance data.
  url = paste0(
    'https://maps.googleapis.com/maps/api/distancematrix/xml?',
    'origins=', lat1,
    ',', lon1,
    '&destinations=', lat2,
    ',', lon2,
    '&mode=', mode,
    '&sensor=false'
  )

  tryCatch({
    # Download the XML document with distance information.
    xml = readLines( url )

    # The <value> element immediately follows the distance element.
    value = xml[ grep( "<distance>", xml ) + 1 ]

    # Obtain the distance in meters.
    meters = sub( ".*>(.*?)<.*", "\\1", value )

    # Return the distance.
    as.numeric( meters )
  },
  warning = function( w ) {
    haversine( lat1, lon1, lat2, lon2 )
  },
  error = function( e ) {
    haversine( lat1, lon1, lat2, lon2 )
  })
}

# Computes distance using Haversine formula.
#
# Returns the result in meters.
haversine <- function( lat1, lon1, lat2, lon2, radius = 6371 ) {
  # Convert decimal degrees to radians
  lon1 = lon1 * pi / 180
  lon2 = lon2 * pi / 180
  lat1 = lat1 * pi / 180
  lat2 = lat2 * pi / 180

  # Haversine formula
  dlon = lon2 - lon1
  dlat = lat2 - lat1
  a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
  c = 2 * atan2(sqrt(a), sqrt(1-a))

  return( radius * c * 1000 )
}

输出：

distance( '44.5646', '-123.2620', '41.2861', '-124.0902' )
[1] 495892
distance( 44.5646, -123.2620, 41.2861, -124.0902, mode='walking' )
[1] 487715

从米到英尺的转换是读者的练习。

使用R和Google Map API获取两点之间的距离（lat，lon）

5 个答案: