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How to Calculate Centroid

I am working with geospatial shapes and looking at the centroid algorithm here,

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

I have implemented the code in C# like this (which is just this adapted),

Finding the centroid of a polygon?

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; }
}

The problem is that this algorithm when I have this shape (which is an L shape),

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

It gives me the result (3.62, 3.62). Which is OK, except that point is outside the shape. Is there another algorithm around that takes this into account?

Basically a person is going to be drawing a shape on a map. This shape might span multiple roads (so could be an L shape) and I want to work out the centre of the shape. This is so I can work out the road name at that point. It doesn't make sense to me for it to be outside the shape if they have drawn a long skinny L shape.

like image 743
peter Avatar asked Mar 22 '12 02:03

peter


People also ask

What is the formula of centroid?

The centroid of a triangle is used for the calculation of the centroid when the vertices of the triangle are known. The centroid of a triangle with coordinates (x1 x 1 , y1 y 1 ), (x2 x 2 , y2 y 2 ), and (x3 x 3 , y3 y 3 ) is given as, G = ((x1 x 1 + x2 x 2 + x3 x 3 )/3, (y1 y 1 + y2 y 2 + y3 y 3 )/3).

What is the easiest way to find the centroid?

To find the centroid, follow these steps: Step 1: Identify the coordinates of each vertex. Step 2: Add all the x values from the three vertices coordinates and divide by 3. Step 3: Add all the y values from the three vertices coordinates and divide by 3.

What is centroid how the centroid is calculated?

The centroid of the triangle separates the median in the ratio of 2: 1. It can be found by taking the average of x- coordinate points and y-coordinate points of all the vertices of the triangle.


3 Answers

This answer is inspired by the answer by Jer2654 and this source: 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);
  }
like image 83
RenniePet Avatar answered Sep 30 '22 12:09

RenniePet


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 );
}
like image 28
Jer2654 Avatar answered Sep 30 '22 13:09

Jer2654


You may check if .NET 4.5 DbSpatialServices functions, like DbSpatialServices.GetCentroid

like image 42
ITmeze Avatar answered Sep 30 '22 13:09

ITmeze