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
  }

这个脚本[在PHP中]计算两点之间的距离。

public static function getDistanceOfTwoPoints($source, $dest, $unit='K') {
        $lat1 = $source[0];
        $lon1 = $source[1];
        $lat2 = $dest[0];
        $lon2 = $dest[1];

        $theta = $lon1 - $lon2;
        $dist = sin(deg2rad($lat1)) * sin(deg2rad($lat2)) +  cos(deg2rad($lat1)) * cos(deg2rad($lat2)) * cos(deg2rad($theta));
        $dist = acos($dist);
        $dist = rad2deg($dist);
        $miles = $dist * 60 * 1.1515;
        $unit = strtoupper($unit);

        if ($unit == "K") {
            return ($miles * 1.609344);
        }
        else if ($unit == "M")
        {
            return ($miles * 1.609344 * 1000);
        }
        else if ($unit == "N") {
            return ($miles * 0.8684);
        } 
        else {
            return $miles;
        }
    }

在我的项目中，我需要计算很多点之间的距离，所以我继续尝试优化我在这里找到的代码。平均而言，在不同的浏览器中，我的新实现的运行速度比获得最多好评的答案快2倍。

function distance(lat1, lon1, lat2, lon2) {
  var p = 0.017453292519943295;    // Math.PI / 180
  var c = Math.cos;
  var a = 0.5 - c((lat2 - lat1) * p)/2 + 
          c(lat1 * p) * c(lat2 * p) * 
          (1 - c((lon2 - lon1) * p))/2;

  return 12742 * Math.asin(Math.sqrt(a)); // 2 * R; R = 6371 km
}

您可以在这里使用我的jsPerf并查看结果。

最近我需要在python中做同样的事情，所以这里是一个python实现:

from math import cos, asin, sqrt, pi

def distance(lat1, lon1, lat2, lon2):
    p = pi/180
    a = 0.5 - cos((lat2-lat1)*p)/2 + cos(lat1*p) * cos(lat2*p) * (1-cos((lon2-lon1)*p))/2
    return 12742 * asin(sqrt(a)) #2*R*asin...

为了完整起见:维基百科上的Haversine。

非常感谢这一切。我在Objective-C iPhone应用程序中使用了以下代码:

const double PIx = 3.141592653589793;
const double RADIO = 6371; // Mean radius of Earth in Km

double convertToRadians(double val) {

   return val * PIx / 180;
}

-(double)kilometresBetweenPlace1:(CLLocationCoordinate2D) place1 andPlace2:(CLLocationCoordinate2D) place2 {

        double dlon = convertToRadians(place2.longitude - place1.longitude);
        double dlat = convertToRadians(place2.latitude - place1.latitude);

        double a = ( pow(sin(dlat / 2), 2) + cos(convertToRadians(place1.latitude))) * cos(convertToRadians(place2.latitude)) * pow(sin(dlon / 2), 2);
        double angle = 2 * asin(sqrt(a));

        return angle * RADIO;
}

纬度和经度是十进制的。我没有在asin()调用中使用min()，因为我使用的距离非常小，以至于它们不需要min()。

它给出了错误的答案，直到我传入弧度的值-现在它几乎与从苹果地图应用程序中获得的值相同:-)

额外的更新:

如果你使用的是iOS4或更高版本，那么苹果会提供一些方法来实现相同的功能:

-(double)kilometresBetweenPlace1:(CLLocationCoordinate2D) place1 andPlace2:(CLLocationCoordinate2D) place2 {

    MKMapPoint  start, finish;


    start = MKMapPointForCoordinate(place1);
    finish = MKMapPointForCoordinate(place2);

    return MKMetersBetweenMapPoints(start, finish) / 1000;
}

如果你想要驾驶距离/路线(张贴在这里，因为这是谷歌上两点之间距离的第一个结果，但对大多数人来说，驾驶距离更有用)，你可以使用谷歌地图距离矩阵服务:

getDrivingDistanceBetweenTwoLatLong(origin, destination) {

 return new Observable(subscriber => {
  let service = new google.maps.DistanceMatrixService();
  service.getDistanceMatrix(
    {
      origins: [new google.maps.LatLng(origin.lat, origin.long)],
      destinations: [new google.maps.LatLng(destination.lat, destination.long)],
      travelMode: 'DRIVING'
    }, (response, status) => {
      if (status !== google.maps.DistanceMatrixStatus.OK) {
        console.log('Error:', status);
        subscriber.error({error: status, status: status});
      } else {
        console.log(response);
        try {
          let valueInMeters = response.rows[0].elements[0].distance.value;
          let valueInKms = valueInMeters / 1000;
          subscriber.next(valueInKms);
          subscriber.complete();
        }
       catch(error) {
        subscriber.error({error: error, status: status});
       }
      }
    });
});
}

下面是Erlang实现

lat_lng({Lat1, Lon1}=_Point1, {Lat2, Lon2}=_Point2) ->
  P = math:pi() / 180,
  R = 6371, % Radius of Earth in KM
  A = 0.5 - math:cos((Lat2 - Lat1) * P) / 2 +
    math:cos(Lat1 * P) * math:cos(Lat2 * P) * (1 - math:cos((Lon2 - Lon1) * P))/2,
  R * 2 * math:asin(math:sqrt(A)).