计算C#中不规则多边形的面积

Question

我设法写了一个'for dummies'如何计算C#中不规则多边形的面积,但我需要它对任何数量的顶点都是动态的.

有人可以帮忙吗？

类:

public class Vertex
{
    private int _vertexIdx;
    private double _coordX;
    private double _coordY;
    private double _coordZ;

    public Vertex()
    { }

    public Vertex(int vertexIdx, double coordX, double coordY, double coordZ)
    {
        _vertexIdx = vertexIdx;
        _coordX = coordX;
        _coordY = coordY;
        _coordZ = coordZ;
    }

    public int VertexIdx
    {
        get { return _vertexIdx; }
        set { _vertexIdx = value; }
    }

    public double X
    {
        get { return _coordX; }
        set { _coordX = value; }
    }

    public double Y
    {
        get { return _coordY; }
        set { _coordY = value; }
    }

    public double Z
    {
        get { return _coordZ; }
        set { _coordZ = value; }
    }
}

的Form_Load:

List<Vertex> verticies = new List<Vertex>();

verticies.Add(new Vertex(1, 930.9729, 802.8789, 0));
verticies.Add(new Vertex(2, 941.5341, 805.662, 0));
verticies.Add(new Vertex(3, 946.5828, 799.271, 0));
verticies.Add(new Vertex(4, 932.6215, 797.0548, 0));

dataGridView1.DataSource = verticies;

按下按钮时计算的代码:(硬编码为4点多边形 - 应该是任何数量......)

        // X-coords
        double x1;
        double x2;
        double x3;
        double x4;
        double x5;

        // Y-coords
        double y1;
        double y2;
        double y3;
        double y4;
        double y5;

        // Xn * Yn++
        double x1y2;
        double x2y3;
        double x3y4;
        double x4y5;

        // Yn * Xn++
        double y1x2;
        double y2x3;
        double y3x4;
        double y4x5;

        // XnYn++ - YnXn++
        double x1y2my1x2;
        double x2y3my2x3;
        double x3y4my3x4;
        double x4y5my4x5;

        double result;
        double area;

        x1 = Convert.ToDouble(dataGridView1.Rows[0].Cells[1].Value.ToString());
        y1 = Convert.ToDouble(dataGridView1.Rows[0].Cells[2].Value.ToString());
        txtLog.Text += String.Format("X1 = {0}\tY1 = {1}\r\n", x1, y1);

        x2 = Convert.ToDouble(dataGridView1.Rows[1].Cells[1].Value.ToString());
        y2 = Convert.ToDouble(dataGridView1.Rows[1].Cells[2].Value.ToString());
        txtLog.Text += String.Format("X2 = {0}\tY2 = {1}\r\n", x2, y2);

        x3 = Convert.ToDouble(dataGridView1.Rows[2].Cells[1].Value.ToString());
        y3 = Convert.ToDouble(dataGridView1.Rows[2].Cells[2].Value.ToString());
        txtLog.Text += String.Format("X3 = {0}\tY3 = {1}\r\n", x3, y3);

        x4 = Convert.ToDouble(dataGridView1.Rows[3].Cells[1].Value.ToString());
        y4 = Convert.ToDouble(dataGridView1.Rows[3].Cells[2].Value.ToString());
        txtLog.Text += String.Format("X4 = {0}\tY4 = {1}\r\n", x4, y4);

        // add the start point again
        x5 = Convert.ToDouble(dataGridView1.Rows[0].Cells[1].Value.ToString());
        y5 = Convert.ToDouble(dataGridView1.Rows[0].Cells[2].Value.ToString());
        txtLog.Text += String.Format("X5 = {0}\tY5 = {1}\r\n", x5, y5);
        txtLog.Text += "\r\n";

        // Multiply 
        x1y2 = x1 * y2;
        x2y3 = x2 * y3;
        x3y4 = x3 * y4;
        x4y5 = x4 * y5;

        y1x2 = y1 * x2;
        y2x3 = y2 * x3;
        y3x4 = y3 * x4;
        y4x5 = y4 * x5;

        // Subtract from each other
        x1y2my1x2 = x1y2 - y1x2;
        x2y3my2x3 = x2y3 - y2x3; 
        x3y4my3x4 = x3y4 - y3x4;
        x4y5my4x5 = x4y5 - y4x5;

        // Sum all results
        result = x1y2my1x2 + x2y3my2x3 + x3y4my3x4 + x4y5my4x5;
        area = Math.Abs(result / 2);

        txtLog.Text += String.Format("Area = {0}\r\n", area);

示例输出:

X1 = 930.9729 Y1 = 802.8789

X2 = 941.5341 Y2 = 805.662

X3 = 946.5828 Y3 = 799.271

X4 = 932.6215 Y4 = 797.0548

X5 = 930.9729 Y5 = 802.8789

面积= 83.2566504099523

Answer 1

使用lambda表达式这变得微不足道!

var points = GetSomePoints();

points.Add(points[0]);
var area = Math.Abs(points.Take(points.Count - 1)
   .Select((p, i) => (points[i + 1].X - p.X) * (points[i + 1].Y + p.Y))
   .Sum() / 2);

该算法解释在这里:

[此方法添加]由多边形边缘定义的梯形区域下降到X轴.当程序考虑多边形的下边缘时,计算给出负面积,因此减去多边形和轴之间的空间,留下多边形的面积.

如果多边形顺时针方向,则总计算面积为负[所以]函数只返回绝对值.

此方法为非简单多边形(边交叉)提供奇怪的结果.

Answer 2

类似于平面多边形的东西（用记事本编译）：

static double GetDeterminant(double x1, double y1, double x2, double y2)
{
    return x1 * y2 - x2 * y1;
}

static double GetArea(IList<Vertex> vertices)
{
    if(vertices.Count < 3)
    {
        return 0;
    }
    double area = GetDeterminant(vertices[vertices.Count - 1].X, vertices[vertices.Count - 1].Y, vertices[0].X, vertices[0].Y);
    for (int i = 1; i < vertices.Count; i++)
    {
        area += GetDeterminant(vertices[i - 1].X, vertices[i - 1].Y, vertices[i].X, vertices[i].Y);
    }
    return area / 2;
}

虽然你的方法没有关注Z轴。因此，我建议应用一些变换来消除它：如果多边形不是平面，您将无法获得面积，而如果它是平面，您就可以消除第三维。

Answer 3

public float Area(List<PointF> vertices)
{
    vertices.Add(vertices[0]);
    return Math.Abs(vertices.Take(vertices.Count - 1).Select((p, i) => (p.X * vertices[i + 1].Y) - (p.Y * vertices[i + 1].X)).Sum() / 2);
}