Question

我正在处理地理空间形状并在此处查看质心算法

http://en.wikipedia.org/wiki/Centroid#Centroid_of_polygon

我已经在C＃中实现了这样的代码（只是这个改编），

class Program
{
    static void Main(string[] args)
    {
        List<Point> vertices = new List<Point>();

        vertices.Add(new Point() { X = 1, Y = 1 });
        vertices.Add(new Point() { X = 1, Y = 10 });
        vertices.Add(new Point() { X = 2, Y = 10 });
        vertices.Add(new Point() { X = 2, Y = 2 });
        vertices.Add(new Point() { X = 10, Y = 2 });
        vertices.Add(new Point() { X = 10, Y = 1 });
        vertices.Add(new Point() { X = 1, Y = 1 });

        Point centroid = Compute2DPolygonCentroid(vertices);
    }

    static Point Compute2DPolygonCentroid(List<Point> vertices)
    {
        Point centroid = new Point() { X = 0.0, Y = 0.0 };
        double signedArea = 0.0;
        double x0 = 0.0; // Current vertex X
        double y0 = 0.0; // Current vertex Y
        double x1 = 0.0; // Next vertex X
        double y1 = 0.0; // Next vertex Y
        double a = 0.0;  // Partial signed area

        // For all vertices except last
        int i=0;
        for (i = 0; i < vertices.Count - 1; ++i)
        {
            x0 = vertices[i].X;
            y0 = vertices[i].Y;
            x1 = vertices[i+1].X;
            y1 = vertices[i+1].Y;
            a = x0*y1 - x1*y0;
            signedArea += a;
            centroid.X += (x0 + x1)*a;
            centroid.Y += (y0 + y1)*a;
        }

        // Do last vertex
        x0 = vertices[i].X;
        y0 = vertices[i].Y;
        x1 = vertices[0].X;
        y1 = vertices[0].Y;
        a = x0*y1 - x1*y0;
        signedArea += a;
        centroid.X += (x0 + x1)*a;
        centroid.Y += (y0 + y1)*a;

        signedArea *= 0.5;
        centroid.X /= (6*signedArea);
        centroid.Y /= (6*signedArea);

        return centroid;
    }
}

public class Point
{
    public double X { get; set; }
    public double Y { get; set; }
}

问题是当我有这个形状（这是一个L形）时，这个算法，

（1,1）（1,10）（2,10）（2,2）（10,2）（10,1）（1,1）

它给了我结果（3.62,3.62）。哪个是好的，除了那个点在形状之外。是否有其他算法考虑到这一点？

基本上，一个人将在地图上绘制一个形状。这种形状可能跨越多条道路（因此可能是L形），我想要计算出形状的中心。这样我就可以在那时找出道路名称。如果它们画出一个长的瘦L形状，那么它在形状之外是没有意义的。

Answer 1

这个答案的灵感来自于Jer2654及其来源的回答： http://coding-experiments.blogspot.com/2009/09/xna-quest-for-centroid-of-polygon.html

  /// <summary>
  /// Method to compute the centroid of a polygon. This does NOT work for a complex polygon.
  /// </summary>
  /// <param name="poly">points that define the polygon</param>
  /// <returns>centroid point, or PointF.Empty if something wrong</returns>
  public static PointF GetCentroid(List<PointF> poly)
  {
     float accumulatedArea = 0.0f;
     float centerX = 0.0f;
     float centerY = 0.0f;

     for (int i = 0, j = poly.Count - 1; i < poly.Count; j = i++)
     {
        float temp = poly[i].X * poly[j].Y - poly[j].X * poly[i].Y;
        accumulatedArea += temp;
        centerX += (poly[i].X + poly[j].X) * temp;
        centerY += (poly[i].Y + poly[j].Y) * temp;
     }

     if (Math.Abs(accumulatedArea) < 1E-7f)
        return PointF.Empty;  // Avoid division by zero

     accumulatedArea *= 3f;
     return new PointF(centerX / accumulatedArea, centerY / accumulatedArea);
  }

Answer 2

您可以检查.NET 4.5 DbSpatialServices是否正常运行，例如DbSpatialServices.GetCentroid

Answer 3

public static Point GetCentroid( Point[ ] nodes, int count )
{
    int x = 0, y = 0, area = 0, k;
    Point a, b = nodes[ count - 1 ];

    for( int i = 0; i < count; i++ )
    {
        a = nodes[ i ];

        k = a.Y * b.X - a.X * b.Y;
        area += k;
        x += ( a.X + b.X ) * k;
        y += ( a.Y + b.Y ) * k;

        b = a;
    }
    area *= 3;

    return ( area == 0 ) ? Point.Empty : new Point( x /= area, y /= area );
}

Answer 4

我建议不要自己编写，而是使用SharpMap

http://sharpmap.codeplex.com/

他们做了很好的工作，给你的质心（甚至必须在里面）以及你想要的更多功能。

Answer 5

对于3d点，我在C＃中创建了一个方法，希望对您有所帮助：

public static double[] GetCentroid(List<double[]> listOfPoints)
    {
        // centroid[0] = X
        // centroid[1] = Y
        // centroid[2] = Z
        double[] centroid = new double[3];

        // List iteration
        // Link reference:
        // https://en.wikipedia.org/wiki/Centroid
        foreach (double[] point in listOfPoints)
        {
            centroid[0] += point[0];
            centroid[1] += point[1];
            centroid[2] += point[2];
        }

        centroid[0] /= listOfPoints.Count;
        centroid[1] /= listOfPoints.Count;
        centroid[2] /= listOfPoints.Count;

        return centroid;
    }

如何计算质心

5 个答案: