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
仅限飞镖:
import 'dart:math' show cos, sqrt, asin;
double calculateDistance(LatLng l1, LatLng l2) {
const p = 0.017453292519943295;
final a = 0.5 -
cos((l2.latitude - l1.latitude) * p) / 2 +
cos(l1.latitude * p) *
cos(l2.latitude * p) *
(1 - cos((l2.longitude - l1.longitude) * p)) /
2;
return 12742 * asin(sqrt(a));
}
其他回答
PIP安装haversine
Python实现
原产地是美国毗连的中心。
from haversine import haversine, Unit
origin = (39.50, 98.35)
paris = (48.8567, 2.3508)
haversine(origin, paris, unit=Unit.MILES)
要得到以千米为单位的答案,只需设置unit= unit。千米(这是默认值)。
下面是一个c#实现:
static class DistanceAlgorithm
{
const double PIx = 3.141592653589793;
const double RADIUS = 6378.16;
/// <summary>
/// Convert degrees to Radians
/// </summary>
/// <param name="x">Degrees</param>
/// <returns>The equivalent in radians</returns>
public static double Radians(double x)
{
return x * PIx / 180;
}
/// <summary>
/// Calculate the distance between two places.
/// </summary>
/// <param name="lon1"></param>
/// <param name="lat1"></param>
/// <param name="lon2"></param>
/// <param name="lat2"></param>
/// <returns></returns>
public static double DistanceBetweenPlaces(
double lon1,
double lat1,
double lon2,
double lat2)
{
double dlon = Radians(lon2 - lon1);
double dlat = Radians(lat2 - lat1);
double a = (Math.Sin(dlat / 2) * Math.Sin(dlat / 2)) + Math.Cos(Radians(lat1)) * Math.Cos(Radians(lat2)) * (Math.Sin(dlon / 2) * Math.Sin(dlon / 2));
double angle = 2 * Math.Atan2(Math.Sqrt(a), Math.Sqrt(1 - a));
return angle * RADIUS;
}
}
如果你正在使用python; PIP安装地质
from geopy.distance import geodesic
origin = (30.172705, 31.526725) # (latitude, longitude) don't confuse
destination = (30.288281, 31.732326)
print(geodesic(origin, destination).meters) # 23576.805481751613
print(geodesic(origin, destination).kilometers) # 23.576805481751613
print(geodesic(origin, destination).miles) # 14.64994773134371
下面是postgres SQL中的一个示例(以公里为单位,为英里版本,将1.609344替换为0.8684版本)
CREATE OR REPLACE FUNCTION public.geodistance(alat float, alng float, blat
float, blng float)
RETURNS float AS
$BODY$
DECLARE
v_distance float;
BEGIN
v_distance = asin( sqrt(
sin(radians(blat-alat)/2)^2
+ (
(sin(radians(blng-alng)/2)^2) *
cos(radians(alat)) *
cos(radians(blat))
)
)
) * cast('7926.3352' as float) * cast('1.609344' as float) ;
RETURN v_distance;
END
$BODY$
language plpgsql VOLATILE SECURITY DEFINER;
alter function geodistance(alat float, alng float, blat float, blng float)
owner to postgres;
function getDistanceFromLatLonInKm(lat1,lon1,lat2,lon2,units) {
var R = 6371; // Radius of the earth in km
var dLat = deg2rad(lat2-lat1); // deg2rad below
var dLon = deg2rad(lon2-lon1);
var a =
Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.cos(deg2rad(lat1)) * Math.cos(deg2rad(lat2)) *
Math.sin(dLon/2) * Math.sin(dLon/2)
;
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
var d = R * c;
var miles = d / 1.609344;
if ( units == 'km' ) {
return d;
} else {
return miles;
}}
查克的解决方案,也适用于英里。