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使用 NetTopologySuite 计算点之间的地理距离

[英]Calculate geographic distance between points using NetTopologySuite

我正在尝试使用 NetTopologySuite 计算两点之间的距离。 由于我引用了 Microsoft 文档,因此我想出了以下 GeometryExtension 和 GeometryHelper 类:

        public static class GeometryExtensions
    {
        private static readonly CoordinateSystemServices _coordinateSystemServices = new CoordinateSystemServices(new Dictionary<int, string>
            {
                // Coordinate systems:

                [4326] = GeographicCoordinateSystem.WGS84.WKT,

                // CRS for Denmark ESPG 25832. Source: https://epsg.io/25832 and https://sdfe.dk/
                [25832] = @"PROJCS[""ETRS89 / UTM zone 32N"",
                                GEOGCS[""ETRS89"",
                                    DATUM[""European_Terrestrial_Reference_System_1989"",
                                        SPHEROID[""GRS 1980"",6378137,298.257222101,
                                            AUTHORITY[""EPSG"",""7019""]],
                                        TOWGS84[0,0,0,0,0,0,0],
                                        AUTHORITY[""EPSG"",""6258""]],
                                    PRIMEM[""Greenwich"",0,
                                        AUTHORITY[""EPSG"",""8901""]],
                                    UNIT[""degree"",0.0174532925199433,
                                        AUTHORITY[""EPSG"",""9122""]],
                                    AUTHORITY[""EPSG"",""4258""]],
                                PROJECTION[""Transverse_Mercator""],
                                PARAMETER[""latitude_of_origin"",0],
                                PARAMETER[""central_meridian"",9],
                                PARAMETER[""scale_factor"",0.9996],
                                PARAMETER[""false_easting"",500000],
                                PARAMETER[""false_northing"",0],
                                UNIT[""metre"",1,
                                    AUTHORITY[""EPSG"",""9001""]],
                                AXIS[""Easting"",EAST],
                                AXIS[""Northing"",NORTH],
                                AUTHORITY[""EPSG"",""25832""]]"
            }
        );
        


        /// <summary>
        /// Projects a geometry to another SRID
        /// </summary>
        /// <param name="geometry"></param>
        /// <param name="targetSrid"></param>
        /// <param name="defaultSourceSrid">If the geometry SRID has not been specified (i.e. equals 0) defaultSourceSrid is used</param>
        /// <returns></returns>
        public static Geometry ProjectTo(this Geometry geometry, int targetSrid, int? defaultSourceSrid = null)
        {
            if (geometry == null)
                throw new Exception("Geometry is null, cannot project");

            var sourceSrid = geometry.SRID == 0 && defaultSourceSrid.HasValue ? defaultSourceSrid.Value : geometry.SRID;
            var transformation = _coordinateSystemServices.CreateTransformation(sourceSrid, targetSrid);

            var result = geometry.Copy();
            result.Apply(new MathTransformFilter(transformation.MathTransform));

            return result;
        }
}




     public  class GeometryHelper
    {
        private static readonly int _longLatSRID = 4326;
        private static readonly int _targetSRID = 25832;

        /// <summary>
        /// In order to get the distance in meters, we need to project to an appropriate
        /// coordinate system. If no SRID is provided 25832, which covers Denmark is used.
        /// If the provided Points have no SRID, 4326 (longitude/latitude) is assumed.
        /// </summary>
        /// <param name="a"></param>
        /// <param name="b"></param>
        /// <param name="targetSrid"></param>
        /// <returns></returns>
        public static double DistanceInMeters(Point a, Point b, int? targetSrid = null)
        {
            targetSrid ??= _targetSRID;

            try
            {
                //If SRID is not set, assume long/lat, ie. 4326
                return a.ProjectTo(targetSrid.Value, _longLatSRID).Distance(b.ProjectTo(targetSrid.Value, _longLatSRID));
            }
            catch (Exception e)
            {
                throw new Exception("Failed to calculate distance", e);
            }
        }
}

为了测试我的距离计算是否正确,我打开了谷歌地图并选择了计算 3 对不同点之间的距离。 然后我复制粘贴测试中的值以查看它们是否匹配。 我从代码中获得的值是从谷歌地图获得的值的 2 倍。 我究竟做错了什么? 这是我的测试:

public class GeometryHelperTests
{
    [Theory]
    [InlineData(55.676518126293466, 12.567203858066554, 55.67645068023813, 12.56698376863065, 15.68)]//I am getting 36.1
    [InlineData(55.659368700924475, 12.546625254609248, 55.65940085355421, 12.546601114728679, 3.77)]//I am getting 6.2
    [InlineData(55.65896978705746, 12.546674114514795, 55.6596855501795, 12.547258269821455, 87.09)]//I am getting 173.7
    public void DistanceInMeters(double x1, double y1, double x2, double y2, double expected)
    {
        // setup
        Point point1 = new Point(x1, y1) { SRID = 4326 };
        Point point2 = new Point(x2, y2) { SRID = 4326 };

        //CoordinateSystemFactory csFact = new CoordinateSystemFactory();
        

        // act
        var result = GeometryHelper.DistanceInMeters(point1, point2);

        

        // assert
        result.Should().Be(expected);
    }

}

你需要计算great circle distance NetTopologySuite Point.Distance方法返回cartesian distance

尝试以下操作:

public static double Radians(double x)
{
    return x * Math.PI / 180;
}

public static double GreatCircleDistance(double lon1, double lat1, double lon2, double lat2)
{
    double R = 6371e3; // m

    double sLat1 = Math.Sin(Radians(lat1));
    double sLat2 = Math.Sin(Radians(lat2));
    double cLat1 = Math.Cos(Radians(lat1));
    double cLat2 = Math.Cos(Radians(lat2));
    double cLon = Math.Cos(Radians(lon1) - Radians(lon2));

    double cosD = sLat1*sLat2 + cLat1*cLat2*cLon;

    double d = Math.Acos(cosD);

    double dist = R * d;

    return dist;
}

您还可以使用内置的GeoCoordinate.GetDistanceTo() ,它实现了对小距离更准确的Haversine公式。

尽管 saeedkazemi 的答案是正确的并且给出的结果更接近真实值,但我的代码给出的结果也非常接近,但我发现我必须翻转谷歌地图给出的坐标。 所以如果给我 55.6765, 12.5672 我需要给公式 12.5672,55.6765。 话虽如此,saeedkazemi 的答案提供的结果更接近实际价值。

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