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

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)
}

其他回答

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。

//JAVA
    public Double getDistanceBetweenTwoPoints(Double latitude1, Double longitude1, Double latitude2, Double longitude2) {
    final int RADIUS_EARTH = 6371;

    double dLat = getRad(latitude2 - latitude1);
    double dLong = getRad(longitude2 - longitude1);

    double a = Math.sin(dLat / 2) * Math.sin(dLat / 2) + Math.cos(getRad(latitude1)) * Math.cos(getRad(latitude2)) * Math.sin(dLong / 2) * Math.sin(dLong / 2);
    double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
    return (RADIUS_EARTH * c) * 1000;
    }

    private Double getRad(Double x) {
    return x * Math.PI / 180;
    }

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)。

下面是SQL实现,以km为单位计算距离,

SELECT UserId, ( 3959 * acos( cos( radians( your latitude here ) ) * cos( radians(latitude) ) * 
cos( radians(longitude) - radians( your longitude here ) ) + sin( radians( your latitude here ) ) * 
sin( radians(latitude) ) ) ) AS distance FROM user HAVING
distance < 5  ORDER BY distance LIMIT 0 , 5;

要获得通过编程语言实现的更多细节,您可以浏览这里给出的php脚本

下面是另一个转换为Ruby代码的代码:

include Math
#Note: from/to = [lat, long]

def get_distance_in_km(from, to)
  radians = lambda { |deg| deg * Math.PI / 180 }
  radius = 6371 # Radius of the earth in kilometer
  dLat = radians[to[0]-from[0]]
  dLon = radians[to[1]-from[1]]

  cosines_product = Math.sin(dLat/2) * Math.sin(dLat/2) + Math.cos(radians[from[0]]) * Math.cos(radians[to[1]]) * Math.sin(dLon/2) * Math.sin(dLon/2)

  c = 2 * Math.atan2(Math.sqrt(cosines_product), Math.sqrt(1-cosines_product)) 
  return radius * c # Distance in kilometer
end