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)。
其他回答
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
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Haversine公式的typescript实现
static getDistanceFromLatLonInKm(lat1: number, lon1: number, lat2: number, lon2: number): number {
var deg2Rad = deg => {
return deg * Math.PI / 180;
}
var r = 6371; // Radius of the earth in km
var dLat = deg2Rad(lat2 - lat1);
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; // Distance in km
return d;
}
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。