【问题标题】:Representing and comparing geographic locations in Java在 Java 中表示和比较地理位置
【发布时间】:2015-04-25 10:03:04
【问题描述】:

我试图弄清楚如何表示地理位置,但我似乎无法在 SE 文档中找到任何相关的类。

我想做这样的事情:

Location locA = new Location(aa.aaaaaaa, bb.bbbbbbb); //lat/long coordinates
Location locB = .....
int meters = locA.distanceTo(locB);

理想情况下,我想要Android Location 之类的东西,因为无论如何我的大部分位置数据都是从安卓设备发送的。据我了解,它包含有关准确性、纬度和速度的信息,这些信息很有用(但不是必需的)。

我知道如何自己实现它,但开源替代方案可以节省我一些时间。

如果我可以进行一些 IP/位置查找,那也将非常有帮助。但我想这完全是另一个问题。

【问题讨论】:

    标签: java android geolocation


    【解决方案1】:

    这是一种计算2点之间距离的方法,取自AndroidLocation类,稍作修改。看起来有点复杂——这是因为它使用Vincenty's formulae对WGS84椭球进行迭代计算:

    /** distance between 2 geographic points on Earth, in km **/
    public static double geoDistance(GeoPoint gp1, GeoPoint gp2) {
            // Based on http://www.ngs.noaa.gov/PUBS_LIB/inverse.pdf
            // using the "Inverse Formula" (section 4)
    
            int MAXITERS = 20;
            // Convert lat/long to radians
            double lat1 = gp1.getLat() * Math.PI / 180.0;
            double lat2 = gp2.getLat() * Math.PI / 180.0;
            double lon1 = gp1.getLon() * Math.PI / 180.0;
            double lon2 = gp2.getLon() * Math.PI / 180.0;
    
            double a = 6378.137; // WGS84 major axis
            double b = 6356.7523142; // WGS84 semi-major axis
            double f = (a - b) / a;
            double aSqMinusBSqOverBSq = (a * a - b * b) / (b * b);
    
            double L = lon2 - lon1;
            double A = 0.0;
            double U1 = Math.atan((1.0 - f) * Math.tan(lat1));
            double U2 = Math.atan((1.0 - f) * Math.tan(lat2));
    
            double cosU1 = Math.cos(U1);
            double cosU2 = Math.cos(U2);
            double sinU1 = Math.sin(U1);
            double sinU2 = Math.sin(U2);
            double cosU1cosU2 = cosU1 * cosU2;
            double sinU1sinU2 = sinU1 * sinU2;
    
            double sigma = 0.0;
            double deltaSigma = 0.0;
            double cosSqAlpha = 0.0;
            double cos2SM = 0.0;
            double cosSigma = 0.0;
            double sinSigma = 0.0;
            double cosLambda = 0.0;
            double sinLambda = 0.0;
    
            double lambda = L; // initial guess
            for (int iter = 0; iter < MAXITERS; iter++) {
                double lambdaOrig = lambda;
                cosLambda = Math.cos(lambda);
                sinLambda = Math.sin(lambda);
                double t1 = cosU2 * sinLambda;
                double t2 = cosU1 * sinU2 - sinU1 * cosU2 * cosLambda;
                double sinSqSigma = t1 * t1 + t2 * t2; // (14)
                sinSigma = Math.sqrt(sinSqSigma);
                cosSigma = sinU1sinU2 + cosU1cosU2 * cosLambda; // (15)
                sigma = Math.atan2(sinSigma, cosSigma); // (16)
                double sinAlpha = (sinSigma == 0) ? 0.0 :
                    cosU1cosU2 * sinLambda / sinSigma; // (17)
                cosSqAlpha = 1.0 - sinAlpha * sinAlpha;
                cos2SM = (cosSqAlpha == 0) ? 0.0 :
                    cosSigma - 2.0 * sinU1sinU2 / cosSqAlpha; // (18)
    
                double uSquared = cosSqAlpha * aSqMinusBSqOverBSq; // defn
                A = 1 + (uSquared / 16384.0) * // (3)
                    (4096.0 + uSquared *
                     (-768 + uSquared * (320.0 - 175.0 * uSquared)));
                double B = (uSquared / 1024.0) * // (4)
                    (256.0 + uSquared *
                     (-128.0 + uSquared * (74.0 - 47.0 * uSquared)));
                double C = (f / 16.0) *
                    cosSqAlpha *
                    (4.0 + f * (4.0 - 3.0 * cosSqAlpha)); // (10)
                double cos2SMSq = cos2SM * cos2SM;
                deltaSigma = B * sinSigma * // (6)
                    (cos2SM + (B / 4.0) *
                     (cosSigma * (-1.0 + 2.0 * cos2SMSq) -
                      (B / 6.0) * cos2SM *
                      (-3.0 + 4.0 * sinSigma * sinSigma) *
                      (-3.0 + 4.0 * cos2SMSq)));
    
                lambda = L +
                    (1.0 - C) * f * sinAlpha *
                    (sigma + C * sinSigma *
                     (cos2SM + C * cosSigma *
                      (-1.0 + 2.0 * cos2SM * cos2SM))); // (11)
    
                double delta = (lambda - lambdaOrig) / lambda;
                if (Math.abs(delta) < 1.0e-12) {
                    break;
                }
            }
    
            return b * A * (sigma - deltaSigma);
        }
    

    GeoPoint 类如下所示:

    /**
     * Immutable point in geo coordinates (latitude, longitude) with accuracy in km
     */
    public class GeoPoint {
    
        private final double lat;
        private final double lon;
        private final double accuracy;
    
        /**
         * New geo point without accuracy
         */
        public GeoPoint(double lat, double lon){
            this(lat, lon, -1d);
        }
    
        /**
         * New geo point with specified accuracy
         * @param accuracy  accuracy in km
         */
        public GeoPoint(double lat, double lon, double accuracy){
            this.lat = lat;
            this.lon = lon;
            this.accuracy = accuracy < 0 ? -1d : accuracy;
        }
    
        public double getLat(){
            return this.lat;
        }
    
        public double getLon(){
            return this.lon;
        }
    
        /**
         * @return accuracy in km. If < 0, accuracy is not defined
         */
        public double getAccuracy(){
            return this.accuracy;
        }
    
        @Override
        public String toString(){
            return "lat = " + this.lat + "; lon = " + this.lon + (this.accuracy < 0 ? "" : ("; accuracy = " + this.accuracy));
        }
    
        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (!(o instanceof GeoPoint) || o == null) return false;
            GeoPoint g = (GeoPoint) o;
            return g.lat == this.lat && g.lon == this.lon && g.accuracy == this.accuracy; 
        }
    
    
    }
    

    【讨论】:

    • 您好,感谢您的回答。我最终做了类似的事情,但使用了另一种计算距离的方法,如this answer 所示。并实现了我自己的 GeoLoc 类。它没有我想要的所有功能,但暂时就足够了。
    • 这并没有花费太多时间,因为我已经准备好了代码。您提到的答案中的公式适用于短距离,因为它将地球视为球体,而不是椭球体。无论如何,不​​客气!
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