36

我正在尝试使用 Haversine 距离公式（如在此处找到：http ://www.movable-type.co.uk/scripts/latlong.html ）但我无法让它工作，请参阅以下代码

    function test() { 
    var lat2 = 42.741; 
    var lon2 = -71.3161; 
    var lat1 = 42.806911; 
    var lon1 = -71.290611; 

    var R = 6371; // km 
    //has a problem with the .toRad() method below.
    var dLat = (lat2-lat1).toRad();  
    var dLon = (lon2-lon1).toRad();  
    var a = Math.sin(dLat/2) * Math.sin(dLat/2) + 
                    Math.cos(lat1.toRad()) * Math.cos(lat2.toRad()) * 
                    Math.sin(dLon/2) * Math.sin(dLon/2);  
    var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a)); 
    var d = R * c; 

    alert(d); 
}

错误是：

Uncaught TypeError: Object -0.06591099999999983 has no method 'toRad'

我理解是因为它需要执行以下操作：

Number.prototype.toRad = function() {
return this * Math.PI / 180;
}

但是当我把它放在函数下面时，它仍然会返回相同的错误消息。如何使它使用辅助方法？或者是否有另一种方法来编写代码以使其工作？谢谢！

4

9 回答 9

54

此代码有效：

Number.prototype.toRad = function() {
   return this * Math.PI / 180;
}

var lat2 = 42.741; 
var lon2 = -71.3161; 
var lat1 = 42.806911; 
var lon1 = -71.290611; 

var R = 6371; // km 
//has a problem with the .toRad() method below.
var x1 = lat2-lat1;
var dLat = x1.toRad();  
var x2 = lon2-lon1;
var dLon = x2.toRad();  
var a = Math.sin(dLat/2) * Math.sin(dLat/2) + 
                Math.cos(lat1.toRad()) * Math.cos(lat2.toRad()) * 
                Math.sin(dLon/2) * Math.sin(dLon/2);  
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a)); 
var d = R * c; 

alert(d);

注意我是如何定义 x1 和 x2 的。在以下位置玩它：https ://tinker.io/3f794

于 2013-01-28T11:51:52.843 回答
28

这是基于其他 3 个答案的重构函数！

请注意，坐标参数是 [longitude, latitude]。

function haversineDistance(coords1, coords2, isMiles) {
  function toRad(x) {
    return x * Math.PI / 180;
  }

  var lon1 = coords1[0];
  var lat1 = coords1[1];

  var lon2 = coords2[0];
  var lat2 = coords2[1];

  var R = 6371; // km

  var x1 = lat2 - lat1;
  var dLat = toRad(x1);
  var x2 = lon2 - lon1;
  var dLon = toRad(x2)
  var a = Math.sin(dLat / 2) * Math.sin(dLat / 2) +
    Math.cos(toRad(lat1)) * Math.cos(toRad(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;

  if(isMiles) d /= 1.60934;

  return d;
}
于 2015-05-19T03:57:42.837 回答
21

ES6 JavaScript/NodeJS 重构版本：

   /**
     * Calculates the haversine distance between point A, and B.
     * @param {number[]} latlngA [lat, lng] point A
     * @param {number[]} latlngB [lat, lng] point B
     * @param {boolean} isMiles If we are using miles, else km.
     */
    const haversineDistance = ([lat1, lon1], [lat2, lon2], isMiles = false) => {
      const toRadian = angle => (Math.PI / 180) * angle;
      const distance = (a, b) => (Math.PI / 180) * (a - b);
      const RADIUS_OF_EARTH_IN_KM = 6371;

      const dLat = distance(lat2, lat1);
      const dLon = distance(lon2, lon1);

      lat1 = toRadian(lat1);
      lat2 = toRadian(lat2);

      // Haversine Formula
      const a =
        Math.pow(Math.sin(dLat / 2), 2) +
        Math.pow(Math.sin(dLon / 2), 2) * Math.cos(lat1) * Math.cos(lat2);
      const c = 2 * Math.asin(Math.sqrt(a));

      let finalDistance = RADIUS_OF_EARTH_IN_KM * c;

      if (isMiles) {
        finalDistance /= 1.60934;
      }

      return finalDistance;
    };

有关针对已接受答案的测试，请参阅 codepen：https ://codepen.io/harrymt/pen/dyYvLpJ?editors=1011

于 2018-02-15T10:43:53.743 回答
13

为什么不尝试直接的解决方案？无需扩展 Number 原型，只需将 toRad 定义为常规函数：

function toRad(x) {
   return x * Math.PI / 180;
}

然后toRad到处打电话：

var dLat = toRad(lat2-lat1);

扩展 Number 原型并不总是按预期工作。例如调用 123.toRad() 不起作用。我认为如果你做var x1 = lat2 - lat1; x1.toRad();的比做的更好(lat2-lat1).toRad()

于 2013-01-28T11:38:04.987 回答
1

在函数中调用这些扩展之前，您需要扩展 Number 原型。

所以只要确保

Number.prototype.toRad = function() {
  return this * Math.PI / 180;
}

在你的函数被调用之前被调用。

于 2013-01-28T11:42:30.110 回答
1

当我把它放在函数下面时

你只需要把它放在你调用的点之上test()。函数test本身的声明位置无关紧要。

于 2013-01-28T11:46:55.343 回答
0

另一个减少冗余并且与 Google LatLng 对象兼容的变体：

  function haversine_distance(coords1, coords2) {

     function toRad(x) {
         return x * Math.PI / 180;
    }

  var dLat = toRad(coords2.latitude - coords1.latitude);
  var dLon = toRad(coords2.longitude - coords1.longitude)

  var a = Math.sin(dLat / 2) * Math.sin(dLat / 2) +
          Math.cos(toRad(coords1.latitude)) * 
          Math.cos(toRad(coords2.latitude)) *
          Math.sin(dLon / 2) * Math.sin(dLon / 2);

  return 12742 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
}
于 2016-07-24T06:22:15.690 回答
0

这是 JavaScript 中另一个重构的答案：

getHaversineDistance = (firstLocation, secondLocation) => {
    const earthRadius = 6371; // km 

    const diffLat = (secondLocation.lat-firstLocation.lat) * Math.PI / 180;  
    const diffLng = (secondLocation.lng-firstLocation.lng) * Math.PI / 180;  

    const arc = Math.cos(
                    firstLocation.lat * Math.PI / 180) * Math.cos(secondLocation.lat * Math.PI / 180) 
                    * Math.sin(diffLng/2) * Math.sin(diffLng/2)
                    + Math.sin(diffLat/2) * Math.sin(diffLat/2);
    const line = 2 * Math.atan2(Math.sqrt(arc), Math.sqrt(1-arc));

    const distance = earthRadius * line; 

    return distance;
}

const philly = { lat: 39.9526, lng: -75.1652 }
const nyc = { lat: 40.7128, lng: -74.0060 }
const losAngeles = { lat: 34.0522, lng: -118.2437 }

console.log(getHaversineDistance(philly, nyc)) //129.61277152662188
console.log(getHaversineDistance(philly, losAngeles)) //3843.4534005980404
于 2018-12-14T16:32:26.407 回答
-1

这是上面talkol解决方案的java实现。他或她的解决方案对我们非常有效。我不想回答这个问题，因为最初的问题是针对 javascript 的。我只是分享给定 javascript 解决方案的 java 实现，以防其他人发现它有用。

// this was a pojo class we used internally...
public class GisPostalCode {

    private String country;
    private String postalCode;
    private double latitude;
    private double longitude;

    // getters/setters, etc.
}


public static double distanceBetweenCoordinatesInMiles2(GisPostalCode c1, GisPostalCode c2) {

    double lat2 = c2.getLatitude();
    double lon2 = c2.getLongitude();
    double lat1 = c1.getLatitude();
    double lon1 = c1.getLongitude();

    double R = 6371; // km
    double x1 = lat2 - lat1;
    double dLat = x1 * Math.PI / 180;
    double x2 = lon2 - lon1;
    double dLon = x2 * Math.PI / 180;

    double a = Math.sin(dLat/2) * Math.sin(dLat/2) +
        Math.cos(lat1*Math.PI/180) * Math.cos(lat2*Math.PI/180) *
        Math.sin(dLon/2) * Math.sin(dLon/2);

    double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
    double d = R * c;

    // convert to miles
    return d / 1.60934;
}
于 2013-12-17T14:27:29.327 回答