如何计算由经纬度指定的两点之间的距离?
为了澄清,我想用千米来表示距离;这些点使用WGS84系统,我想了解可用方法的相对准确性。
如何计算由经纬度指定的两点之间的距离?
为了澄清,我想用千米来表示距离;这些点使用WGS84系统,我想了解可用方法的相对准确性。
当前回答
我已经创建了这个小Javascript LatLng对象,可能对某人有用。
var latLng1 = new LatLng(5, 3);
var latLng2 = new LatLng(6, 7);
var distance = latLng1.distanceTo(latLng2);
代码:
/**
* latLng point
* @param {Number} lat
* @param {Number} lng
* @returns {LatLng}
* @constructor
*/
function LatLng(lat,lng) {
this.lat = parseFloat(lat);
this.lng = parseFloat(lng);
this.__cache = {};
}
LatLng.prototype = {
toString: function() {
return [this.lat, this.lng].join(",");
},
/**
* calculate distance in km to another latLng, with caching
* @param {LatLng} latLng
* @returns {Number} distance in km
*/
distanceTo: function(latLng) {
var cacheKey = latLng.toString();
if(cacheKey in this.__cache) {
return this.__cache[cacheKey];
}
// the fastest way to calculate the distance, according to this jsperf test;
// http://jsperf.com/haversine-salvador/8
// http://stackoverflow.com/questions/27928
var deg2rad = 0.017453292519943295; // === Math.PI / 180
var lat1 = this.lat * deg2rad;
var lng1 = this.lng * deg2rad;
var lat2 = latLng.lat * deg2rad;
var lng2 = latLng.lng * deg2rad;
var a = (
(1 - Math.cos(lat2 - lat1)) +
(1 - Math.cos(lng2 - lng1)) * Math.cos(lat1) * Math.cos(lat2)
) / 2;
var distance = 12742 * Math.asin(Math.sqrt(a)); // Diameter of the earth in km (2 * 6371)
// cache the distance
this.__cache[cacheKey] = distance;
return distance;
}
};
其他回答
我在R中做了一个自定义函数,使用R基本包中可用的函数来计算两个空间点之间的距离(km)。
custom_hav_dist <- function(lat1, lon1, lat2, lon2) {
R <- 6371
Radian_factor <- 0.0174533
lat_1 <- (90-lat1)*Radian_factor
lat_2 <- (90-lat2)*Radian_factor
diff_long <-(lon1-lon2)*Radian_factor
distance_in_km <- 6371*acos((cos(lat_1)*cos(lat_2))+
(sin(lat_1)*sin(lat_2)*cos(diff_long)))
rm(lat1, lon1, lat2, lon2)
return(distance_in_km)
}
样例输出
custom_hav_dist(50.31,19.08,54.14,19.39)
[1] 426.3987
PS:要计算以英里为单位的距离,请将函数R(6371)替换为3958.756(海里使用3440.065)。
我通过简化公式来简化计算。
下面是Ruby版本:
include Math
earth_radius_mi = 3959
radians = lambda { |deg| deg * PI / 180 }
coord_radians = lambda { |c| { :lat => radians[c[:lat]], :lng => radians[c[:lng]] } }
# from/to = { :lat => (latitude_in_degrees), :lng => (longitude_in_degrees) }
def haversine_distance(from, to)
from, to = coord_radians[from], coord_radians[to]
cosines_product = cos(to[:lat]) * cos(from[:lat]) * cos(from[:lng] - to[:lng])
sines_product = sin(to[:lat]) * sin(from[:lat])
return earth_radius_mi * acos(cosines_product + sines_product)
end
下面是一个Scala实现:
def calculateHaversineDistance(lat1: Double, lon1: Double, lat2: Double, lon2: Double): Double = {
val long2 = lon2 * math.Pi / 180
val lat2 = lat2 * math.Pi / 180
val long1 = lon1 * math.Pi / 180
val lat1 = lat1 * math.Pi / 180
val dlon = long2 - long1
val dlat = lat2 - lat1
val a = math.pow(math.sin(dlat / 2), 2) + math.cos(lat1) * math.cos(lat2) * math.pow(math.sin(dlon / 2), 2)
val c = 2 * math.atan2(Math.sqrt(a), math.sqrt(1 - a))
val haversineDistance = 3961 * c // 3961 = radius of earth in miles
haversineDistance
}
Java实现在根据哈弗辛公式
double calculateDistance(double latPoint1, double lngPoint1,
double latPoint2, double lngPoint2) {
if(latPoint1 == latPoint2 && lngPoint1 == lngPoint2) {
return 0d;
}
final double EARTH_RADIUS = 6371.0; //km value;
//converting to radians
latPoint1 = Math.toRadians(latPoint1);
lngPoint1 = Math.toRadians(lngPoint1);
latPoint2 = Math.toRadians(latPoint2);
lngPoint2 = Math.toRadians(lngPoint2);
double distance = Math.pow(Math.sin((latPoint2 - latPoint1) / 2.0), 2)
+ Math.cos(latPoint1) * Math.cos(latPoint2)
* Math.pow(Math.sin((lngPoint2 - lngPoint1) / 2.0), 2);
distance = 2.0 * EARTH_RADIUS * Math.asin(Math.sqrt(distance));
return distance; //km value
}
下面是Haversine公式的java实现。
public final static double AVERAGE_RADIUS_OF_EARTH_KM = 6371;
public int calculateDistanceInKilometer(double userLat, double userLng,
double venueLat, double venueLng) {
double latDistance = Math.toRadians(userLat - venueLat);
double lngDistance = Math.toRadians(userLng - venueLng);
double a = Math.sin(latDistance / 2) * Math.sin(latDistance / 2)
+ Math.cos(Math.toRadians(userLat)) * Math.cos(Math.toRadians(venueLat))
* Math.sin(lngDistance / 2) * Math.sin(lngDistance / 2);
double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
return (int) (Math.round(AVERAGE_RADIUS_OF_EARTH_KM * c));
}
请注意,这里我们将答案四舍五入到最近的km。