我已经成功使用了DotSpace。包含用于测试我的每个点(180万)是否位于我的shapefile中的函数。然而,该算法非常缓慢,因为我正在针对大量点和非常复杂的多边形进行测试。以下是一个示例:http://picload.org/image/igararl/pointshapesel.png
图像中德国的边界是我的shapefile,它用于选择点(简化了,但仍然有140000个顶点),红色矩形是我180万个点所在的区域。
为了在空间经纬度坐标的多边形测试中寻找更快的点,我遇到了光线投射算法:http://alienryderflex.com/polygon/
我把代码翻译成VB.Net,它运行时没有错误,但没有找到任何点/shapefile的交集。我知道纬度/经度坐标的困难,但在德国地区,纬度/经度的坐标与标准笛卡尔坐标系相匹配。
这是我的密码。出于速度优先的原因,我声明全局变量:
Public polyCorners As Integer
Public polyX() As Double
Public polyY() As Double
Public xP, yP As Double
Public constant() As Double
Public multiple() As Double
然后我将我的Shapefile顶点添加到polyCorners列表中(这很有效):
Dim ShapefilePoly As Shapefile = Shapefile.OpenFile(TextBox4.Text)
Dim x As Long = 1
For Each MyShapeRange As ShapeRange In ShapefilePoly.ShapeIndices
For Each MyPartRange As PartRange In MyShapeRange.Parts
For Each MyVertex As Vertex In MyPartRange
If MyVertex.X > 0 AndAlso MyVertex.Y > 0 Then
pointsShape.Add(New PointLatLng(MyVertex.Y, MyVertex.X))
ReDim Preserve polyY(x)
ReDim Preserve polyX(x)
polyY(x) = MyVertex.Y
polyX(x) = MyVertex.X
x = x + 1
End If
Next
Next
Next
ReDim constant(x)
ReDim multiple(x)
在实际搜索之前,我调用了作者建议的prelc_values():
Private Sub precalc_values()
Dim i As Integer, j As Integer = polyCorners - 1
For i = 0 To polyCorners - 1
If polyY(j) = polyY(i) Then
constant(i) = polyX(i)
multiple(i) = 0
Else
constant(i) = polyX(i) - (polyY(i) * polyX(j)) / (polyY(j) - polyY(i)) + (polyY(i) * polyX(i)) / (polyY(j) - polyY(i))
multiple(i) = (polyX(j) - polyX(i)) / (polyY(j) - polyY(i))
End If
j = i
Next
End Sub
最后,我为我的每个lat/lng点调用pointInPolygon():
Function LiesWithin(latP As Double, lngP As Double) As Boolean
LiesWithin = False
xP = lngP
yP = latP
If pointInPolygon() = True Then LiesWithin = True
End Function
Private Function pointInPolygon() As Boolean
Dim i As Integer, j As Integer = polyCorners - 1
Dim oddNodes As Boolean = False
For i = 0 To polyCorners - 1
If (polyY(i) < yP AndAlso polyY(j) >= yP OrElse polyY(j) < yP AndAlso polyY(i) >= yP) Then
oddNodes = oddNodes Xor (yP * multiple(i) + constant(i) < xP)
End If
j = i
Next
Return oddNodes
End Function
所有变量似乎都得到了正确的填充,数组包含我的多边形角,点列表从第一个点到最后一个点都得到了准确的检查。它在不到20秒内完成了180万点的完整列表(相比之下,使用DotSpatial.Contains函数需要1小时30分钟)。有人知道为什么它没有找到任何相交点吗?
好吧,我发现问题的速度比预期的要快:我忘记将Shapefile顶点的数量指定给polyCorners。在我上面的代码中,只需在后添加polyCorners=x
ReDim constant(x)
ReDim multiple(x)
polyCorners = x
也许有人觉得这个代码很有用。我真的很惊讶它有这么快!