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
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;
其他回答
在其他答案中,r中的实现是缺失的。
用地质圈包中的distm函数计算两点之间的距离非常简单:
distm(p1, p2, fun = distHaversine)
地点:
p1 = longitude/latitude for point(s)
p2 = longitude/latitude for point(s)
# type of distance calculation
fun = distCosine / distHaversine / distVincentySphere / distVincentyEllipsoid
由于地球不是完美的球形,所以椭球体的文森提公式可能是计算距离的最佳方法。因此,在地质圈包中,您可以使用:
distm(p1, p2, fun = distVincentyEllipsoid)
当然,你不一定要使用geosphere包,你也可以用一个函数来计算以R为基底的距离:
hav.dist <- function(long1, lat1, long2, lat2) {
R <- 6371
diff.long <- (long2 - long1)
diff.lat <- (lat2 - lat1)
a <- sin(diff.lat/2)^2 + cos(lat1) * cos(lat2) * sin(diff.long/2)^2
b <- 2 * asin(pmin(1, sqrt(a)))
d = R * b
return(d)
}
精确计算中长点之间距离所需的函数是复杂的,陷阱也很多。我不推荐哈弗辛或其他球形的解决方案,因为有很大的不准确性(地球不是一个完美的球体)。vincenty公式更好,但在某些情况下会抛出错误,即使编码正确。
与其自己编写函数,我建议使用geopy,它已经实现了非常精确的地理库来进行距离计算(论文来自作者)。
#pip install geopy
from geopy.distance import geodesic
NY = [40.71278,-74.00594]
Beijing = [39.90421,116.40739]
print("WGS84: ",geodesic(NY, Beijing).km) #WGS84 is Standard
print("Intl24: ",geodesic(NY, Beijing, ellipsoid='Intl 1924').km) #geopy includes different ellipsoids
print("Custom ellipsoid: ",geodesic(NY, Beijing, ellipsoid=(6377., 6356., 1 / 297.)).km) #custom ellipsoid
#supported ellipsoids:
#model major (km) minor (km) flattening
#'WGS-84': (6378.137, 6356.7523142, 1 / 298.257223563)
#'GRS-80': (6378.137, 6356.7523141, 1 / 298.257222101)
#'Airy (1830)': (6377.563396, 6356.256909, 1 / 299.3249646)
#'Intl 1924': (6378.388, 6356.911946, 1 / 297.0)
#'Clarke (1880)': (6378.249145, 6356.51486955, 1 / 293.465)
#'GRS-67': (6378.1600, 6356.774719, 1 / 298.25)
这个库的唯一缺点是它不支持向量化计算。 对于向量化计算,您可以使用新的gevectorslib。
#pip install geovectorslib
from geovectorslib import inverse
print(inverse(lats1,lons1,lats2,lons2)['s12'])
lat和lon是numpy数组。Geovectorslib是非常准确和非常快!我还没有找到改变椭球的方法。标准采用WGS84椭球,是大多数用途的最佳选择。
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)。
下面是移植到Java的已接受的答案实现,以备任何人需要。
package com.project529.garage.util;
/**
* Mean radius.
*/
private static double EARTH_RADIUS = 6371;
/**
* Returns the distance between two sets of latitudes and longitudes in meters.
* <p/>
* Based from the following JavaScript SO answer:
* http://stackoverflow.com/questions/27928/calculate-distance-between-two-latitude-longitude-points-haversine-formula,
* which is based on https://en.wikipedia.org/wiki/Haversine_formula (error rate: ~0.55%).
*/
public double getDistanceBetween(double lat1, double lon1, double lat2, double lon2) {
double dLat = toRadians(lat2 - lat1);
double dLon = toRadians(lon2 - lon1);
double a = Math.sin(dLat / 2) * Math.sin(dLat / 2) +
Math.cos(toRadians(lat1)) * Math.cos(toRadians(lat2)) *
Math.sin(dLon / 2) * Math.sin(dLon / 2);
double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
double d = EARTH_RADIUS * c;
return d;
}
public double toRadians(double degrees) {
return degrees * (Math.PI / 180);
}
FSharp版本,使用里程:
let radialDistanceHaversine location1 location2 : float =
let degreeToRadian degrees = degrees * System.Math.PI / 180.0
let earthRadius = 3959.0
let deltaLat = location2.Latitude - location1.Latitude |> degreeToRadian
let deltaLong = location2.Longitude - location1.Longitude |> degreeToRadian
let a =
(deltaLat / 2.0 |> sin) ** 2.0
+ (location1.Latitude |> degreeToRadian |> cos)
* (location2.Latitude |> degreeToRadian |> cos)
* (deltaLong / 2.0 |> sin) ** 2.0
atan2 (a |> sqrt) (1.0 - a |> sqrt)
* 2.0
* earthRadius