我创建了一个公式来在Google地球上形成一个网格。我想得到lat / long之间的交点。请告诉我如何获得交集。我使用SharpKML lib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);
}
答案 0 :(得分：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，并想要回答更简单的问题，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叉点，答案就是使用您在嵌套循环中用于LineStrings的相同精确纬度，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);
如果您想使用DotSpatial来查找网格和Shapefile之间的交集，可以使用以下源代码。在这种情况下，shapefile具有河流线并且仅产生一个交叉点。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来读取kml源文件中的线串坐标而不是打开shapefile来构建新功能，但交叉代码将类似。
作为旁注，我不认为看似容易使用的FeatureSet.Intersection方法足够聪明，可以处理线交点产生点特征作为交叉点的情况。它仅适用于输出可能与输入具有相同特征类型的点或多边形。
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));
}
答案 1 :(得分：0)
我认为DotSpatial库应该满足您的需求，我以前使用过这个库但没有使用交叉点函数：
http://dotspatial.codeplex.com/wikipage?title=DotSpatial.Data.FeatureSetExt.Intersection
如果你尝试进行自己的线交叉分析，要知道简单的笛卡尔平面方法会引入误差（当我接近极点时，[我认为]会变得更加明显。）
见这里：http://www.geog.ubc.ca/courses/klink/gis.notes/ncgia/u32.html