Using the Haversine Formula in Javascript

问题

I\'m trying to use the Haversine Distance Formula (as found here: http://www.movable-type.co.uk/scripts/latlong.html) but I can\'t get it to work, please see the following code

    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); 
}

And the error is:

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

Which I understand to be because it needs to do the following:

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

But when I put this below the function, it still comes back with the same error message. How do I make it use the helper method? Or is there an alternative way to code this to get it to work? Thanks!

回答1:

This code is working:

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);

Notice how I defined x1 and x2. Play with it at: https://tinker.io/3f794

回答2:

Here's a refactored function based on 3 of the other answers!

Please note that the coords arguments are [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;
}

回答3:

Why not try the straight forward solution? Instead of extending Number prototype, just define toRad as a regular function:

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

and then call toRad everywhere:

var dLat = toRad(lat2-lat1);

Extending the Number prototype does not always work as expected. For example calling 123.toRad() does not work. I think that if you do var x1 = lat2 - lat1; x1.toRad(); works better than doing (lat2-lat1).toRad()

回答4:

ES6 JavaScript/NodeJS refactored version:

/**
 * 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 = (latlngA, latlngB, isMiles) => {
  const toRadian = angle => (Math.PI / 180) * angle;
  const distance = (a, b) => (Math.PI / 180) * (a - b);
  const RADIUS_OF_EARTH_IN_KM = 6371;

  let lat1 = latlngA[0];
  let lat2 = latlngB[0];
  const lon1 = latlngA[1];
  const lon2 = latlngB[1];

  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;
};

回答5:

when I put this below the function

You only need to put it above the point where you call test(). Where the test function itself is declared does not matter.

回答6:

You need to extend the Number prototype, before calling those extensions in a function.

So just ensure

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

is called before your function is called.

回答7:

Another variant to reduce redundancy and also compatible with Google LatLng objects:

  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));
}

回答8:

This is a java implemetation of talkol's solution above. His or her solution worked very well for us. I'm not trying to answer the question, since the original question was for javascript. I'm just sharing our java implementation of the given javascript solution in case others find it of use.

// 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;
}

回答9:

Here's another refactored answer in JavaScript:

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

来源：https://stackoverflow.com/questions/14560999/using-the-haversine-formula-in-javascript

标签

javascript

haversine