0

我创建了一个公式来在谷歌地球上形成一个网格。我想得到纬度/经度之间的交点。请告诉我如何获得交叉点。我正在使用 SharpKML 库生成 KML

for (int x = 90; x >= 0; x = x - 15)
                {
                    Placemark placemark = new Placemark();
                    LineString line = new LineString();
                    CoordinateCollection co = new CoordinateCollection();
                    for (int i = 0; i <= 180; i = i + 15)
                    {
                        Vector cords = new Vector()
                            {
                                Latitude = x,
                                Longitude = i,
                                Altitude = 1000
                            };

                        co.Add(cords);
                    }
                    for (int i = -180; i <= 0; i = i + 15)
                    {
                        Vector cords = new Vector()
                        {
                            Latitude = x,
                            Longitude = i,
                            Altitude = 1000
                        };

                        co.Add(cords);
                    }
                    line.Coordinates = co;
                    placemark.Geometry = line;
                    document.AddFeature(placemark);
                }
            for (int x = -90; x <= 0; x = x + 15)
            {
                Placemark placemark = new Placemark();
                LineString line = new LineString();
                CoordinateCollection co = new CoordinateCollection();
                for (int i = 0; i <= 180; i = i + 15)
                {
                    Vector cords = new Vector()
                    {
                        Latitude = x,
                        Longitude = i,
                        Altitude = 1000
                    };

                    co.Add(cords);
                }
                for (int i = -180; i <= 0; i = i + 15)
                {
                    Vector cords = new Vector()
                    {
                        Latitude = x,
                        Longitude = i,
                        Altitude = 1000
                    };

                    co.Add(cords);
                }
                line.Coordinates = co;
                placemark.Geometry = line;
                document.AddFeature(placemark);
            }

            for (int i = 0; i <= 180; i = i + 15)
            {
                Placemark placemark = new Placemark();
                LineString line = new LineString();
                CoordinateCollection co = new CoordinateCollection();
                for (int x = 0; x <= 90; x = x + 15)
                {
                    Vector cords = new Vector()
                    {
                        Latitude = x,
                        Longitude = i,
                        Altitude = 1000
                    };

                    co.Add(cords);
                }
                for (int x = -90; x <= 0; x = x + 15)
                {
                    Vector cords = new Vector()
                    {
                        Latitude = x,
                        Longitude = i,
                        Altitude = 1000
                    };

                    co.Add(cords);
                }
                line.Coordinates = co;
                placemark.Geometry = line;
                document.AddFeature(placemark);
            }
            for (int i = -180; i <= 0; i = i + 15)
            {
                Placemark placemark = new Placemark();
                LineString line = new LineString();
                CoordinateCollection co = new CoordinateCollection();
                for (int x = 0; x <= 90; x = x + 15)
                {
                    Vector cords = new Vector()
                    {
                        Latitude = x,
                        Longitude = i,
                        Altitude = 1000
                    };

                    co.Add(cords);
                }
                for (int x = -90; x <= 0; x = x + 15)

                {
                    Vector cords = new Vector()
                    {
                        Latitude = x,
                        Longitude = i,
                        Altitude = 1000
                    };

                    co.Add(cords);
                }
                line.Coordinates = co;
                placemark.Geometry = line;
                document.AddFeature(placemark);
            }
4

2 回答 2

1

如果问题是如何使用 C# 找到任意 LineString 对象与网格的交点，那么 Matthew 是正确的。在 C++ 中，您可以在 Java 中使用 GEOS http://trac.osgeo.org/geos/，它将是 JTS http://www.vividsolutions.com/jts/JTSHome.htm。

但是，如果您自己创建网格，并且想要回答一个更简单的问题，即如何找到我刚刚创建的网格的水平线和垂直线之间的交点，答案将是使用相同的精确您在嵌套循环中用于 LineStrings 的纬度、经度值：

Document document = new Document();
for(y = -90; y < 0; y += 15){
  for(x = -180; x < 0; x+= 15){
     Point point = new Point();
     point.Coordinate = new Vector(x, y);
     Placemark placemark = new Placemark();
     placemark.Geometry = point;
     document.AddFeature(placemark);
  }
}

    .. repeat for the other 4 quadrants

// It's conventional for the root element to be Kml,
// but you could use document instead.
Kml root = new Kml();
root.Feature = document;
XmlFile kml = KmlFile.Create(root, false);

例如，如果您想使用 DotSpatial 来查找网格和 Shapefile 之间的交集，这里有一些源代码。在这种情况下，shapefile 有河流线并且只产生了一个交点。请注意，拓扑交集代码有点慢，因此您需要使用范围检查来加快速度。在您的情况下，您可能希望通过使用 KMLSharp 读取 kml 源文件中的线串坐标来构建新功能，而不是打开 shapefile，但相交代码将是相似的。

附带说明一下，我认为看似易于使用的 FeatureSet.Intersection 方法不够聪明，无法处理线交点产生点要素作为交点的情况。它仅适用于输出可能与输入具有相同要素类型的点或多边形。

    using DotSpatial.Controls;
    using DotSpatial.Data;
    using DotSpatial.Topology;
    using DotSpatial.Symbology;


    private FeatureSet gridLines;

    private void buttonAddGrid_Click(object sender, EventArgs e)
    {
        gridLines = new FeatureSet(FeatureType.Line);
        for (int x = -180; x < 0; x += 15)
        {
            List<Coordinate> coords = new List<Coordinate>();
            coords.Add(new Coordinate(x, -90));
            coords.Add(new Coordinate(x, 90));
            LineString ls = new LineString(coords);
            gridLines.AddFeature(ls);
        }
        for (int y = -90; y < 0; y += 15)
        {
            List<Coordinate> coords = new List<Coordinate>();
            coords.Add(new Coordinate(-180, y));
            coords.Add(new Coordinate(180, y));
            LineString ls = new LineString(coords);
            gridLines.AddFeature(ls);
        }

        map1.Layers.Add(new MapLineLayer(gridLines));
    }

    private void buttonIntersect_Click(object sender, EventArgs e)
    {
        if (gridLines == null)
        {
            MessageBox.Show("First add the grid.");
        }

        IFeatureSet river = FeatureSet.Open(@"C:\Data\Rivers\River.shp");
        MapLineLayer riverLayer = new MapLineLayer(river);
        map1.Layers.Add(river);




        List<DotSpatial.Topology.Point> allResultPoints = new List<DotSpatial.Topology.Point>();
        foreach (Feature polygon in river.Features)
        {
            Geometry lineString = polygon.BasicGeometry as Geometry;
            foreach (Feature lineFeature in gridLines.Features)
            {
                // Speed up calculation with extent testing.
                if(!lineFeature.Envelope.Intersects(lineString.Envelope)){
                    continue;
                }
                IFeature intersectFeature = lineFeature.Intersection(lineString);
                if (intersectFeature == null)
                {
                    continue;
                }


                MultiPoint multi = intersectFeature.BasicGeometry as MultiPoint;
                if (multi != null)
                {
                    for(int i = 0; i < multi.NumGeometries; i++)
                    {
                        allResultPoints.Add(intersectFeature.GetBasicGeometryN(i) as DotSpatial.Topology.Point);
                    }
                }
                DotSpatial.Topology.Point single = intersectFeature.BasicGeometry as DotSpatial.Topology.Point;
                {
                    allResultPoints.Add(single);
                }
            }
        }

        FeatureSet finalPoints = new FeatureSet(FeatureType.Point);
        foreach(DotSpatial.Topology.Point pt in allResultPoints){
            finalPoints.AddFeature(pt);
        }
        map1.Layers.Add(new MapPointLayer(finalPoints));
    }
于 2014-02-11T16:42:44.810 回答
0

我认为 DotSpatial 库应该可以满足您的需求，我过去曾使用过这个库，但没有使用过交集函数：

http://dotspatial.codeplex.com/wikipage?title=DotSpatial.Data.FeatureSetExt.Intersection

如果您尝试进行自己的线相交分析，请知道简单的笛卡尔平面方法会引入错误（[我认为]当您接近极点时会变得更加明显）。

见这里：http ://www.geog.ubc.ca/courses/klink/gis.notes/ncgia/u32.html

在这里：两条地理线之间的交叉点

于 2014-02-10T21:51:59.177 回答