.net core(c#)拟合圆测试

说明

很多时候,我们需要运动物体的转弯半径去描述其机器性能。但在大多数的现实条件下,我们只能够获取到运动物体的 GPS 位置点集,并不能直接得到转弯半径或者圆心位置。为此,我们可以利用拟合圆的方式得到圆坐标方程,由此得到转弯半径和圆心位置。

解决过程

关于拟合圆方程的方法有很多,曾经在这篇译文中获益良多代数逼近法、最小二乘法、正交距离回归法来拟合圆及其结果对比(Python)。此系列文中也给出了提及的三种方法的性能及效果对比,最终得出最优的解决方案就是最小二乘法。由于最近的学习中又进一步了解到,可以利用线性代数的方法去求解。本着大学课程中曾学过的《线性代数》知识,所以想着用此方法再加以解决该问题,以作最对比。

接下来,本文就最小二乘法和线性代数的方法求取圆方程作一论述。

准备

引用矩阵计算库MathNet.Numerics。该库是一个强大的科学计算库,遵循 .Net Standard,所以可以跨平台使用。

创建描述圆的类

public class Circle
{
    /// 
    /// 圆心横坐标
    /// 
    /// 
    public   double X { get; set; }
    /// 
    /// 圆心纵坐标
    /// 
    /// 
    public   double Y { get; set; }
    /// 
    /// 圆半径
    /// 
    /// 
    public  double R { get; set; }
}

画图,引用System.Drawing.Common库,以实现跨平台的图像生成。接下来,我们简单的实现一个图像帮助类来进行图像绘制。

public class ImageHelp
{
    private Image _image;
    public ImageHelp(int width, int height)
    {
        _image = new Bitmap(width, height);
        var graph = Graphics.FromImage(_image);
        graph.Clear(Color.White);
    }
    public void DrawCicle(Circle circle, Brush brush)
    {
        var graph = Graphics.FromImage(_image);
        var count=200;
        var fitPoints = new Point[count+1];
        var step = 2 * Math.PI / count;
        for (int i = 0; i < count; i++)
        {
            //circle
            var p = new Point();
            p.X = (int)(circle.X + Math.Cos(i * step) * circle.R);
            p.Y = (int)(circle.Y + Math.Sin(i * step) * circle.R);
            fitPoints[i] = p;
        }
        fitPoints[count] = fitPoints[0];//闭合圆
        graph.DrawLines(new Pen(brush, 2), fitPoints);
        graph.Dispose();
    }
    public void DrawPoints(double[] X, double[] Y, Brush brush)
    {
        var graph = Graphics.FromImage(_image);
        for (int i = 0; i < X.Length; i++)
        {
            graph.DrawEllipse(new Pen(brush, 2), (int)X[i], (int)Y[i], 6, 6);
        }
        graph.Dispose();
    }
    public void SaveImage(string file)
    {
        _image.Save(file, System.Drawing.Imaging.ImageFormat.Png);
    }
}

模拟点集,由于现实中的数据采集存在着精度、数据记录等众多不确定因素的影像。模拟点集中也将加入一定程度的噪音。以下代码中 x 与 y 中存储着我们的点集数据:

var count = 50;
var step = 2 * Math.PI / 100;
var rd = new Random();
//参照圆
var x0 = 204.1;
var y0 = 213.1;
var r0 = 98.4;
//噪音绝对差
var diff = (int)(r0 * 0.1);
var x = new double[count];
var y = new double[count];
//输出点集
for (int i = 0; i < count; i++)
{
    //circle
    x[i] = x0 + Math.Cos(i * step) * r0;
    y[i] = y0 + Math.Sin(i * step) * r0;
    //noise
    x[i] += Math.Cos(rd.Next() % 2 * Math.PI) * rd.Next(diff);
    y[i] += Math.Cos(rd.Next() % 2 * Math.PI) * rd.Next(diff);
}

最小二乘法

网上有很多的原理解析,上文中提到的译文中也有提及,这里不在过多赘述。直接贴出 c#代码实现:

public Circle LeastSquaresFit(double[] X, double[] Y)
{
    if (X.Length < 3)
    {
        return null;
    }
    double cent_x = 0.0,
        cent_y = 0.0,
        radius = 0.0;
    double sum_x = 0.0f, sum_y = 0.0f;
    double sum_x2 = 0.0f, sum_y2 = 0.0f;
    double sum_x3 = 0.0f, sum_y3 = 0.0f;
    double sum_xy = 0.0f, sum_x1y2 = 0.0f, sum_x2y1 = 0.0f;
    int N = X.Length;
    double x, y, x2, y2;
    for (int i = 0; i < N; i++)
    {
        x = X[i];
        y = Y[i];
        x2 = x * x;
        y2 = y * y;
        sum_x += x;
        sum_y += y;
        sum_x2 += x2;
        sum_y2 += y2;
        sum_x3 += x2 * x;
        sum_y3 += y2 * y;
        sum_xy += x * y;
        sum_x1y2 += x * y2;
        sum_x2y1 += x2 * y;
    }
    double C, D, E, G, H;
    double a, b, c;
    C = N * sum_x2 - sum_x * sum_x;
    D = N * sum_xy - sum_x * sum_y;
    E = N * sum_x3 + N * sum_x1y2 - (sum_x2 + sum_y2) * sum_x;
    G = N * sum_y2 - sum_y * sum_y;
    H = N * sum_x2y1 + N * sum_y3 - (sum_x2 + sum_y2) * sum_y;
    a = (H * D - E * G) / (C * G - D * D);
    b = (H * C - E * D) / (D * D - G * C);
    c = -(a * sum_x + b * sum_y + sum_x2 + sum_y2) / N;
    cent_x = a / (-2);
    cent_y = b / (-2);
    radius = Math.Sqrt(a * a + b * b - 4 * c) / 2;
    var result = new Circle();
    result.X = cent_x;
    result.Y = cent_y;
    result.R = radius;
    return result;
}

线性代数

从标准圆方程(x-c1)^2+(y-c2)^2=r^2中进行方程变换得到2xc1+2yc2+(r^2−c1^2−c2^2)=x^2+y^2,其中,我们c3替换常量值r^2−c1^2−c2^2,即:r^2−c1^2−c2^2=c3。由此,我们得到2xc1+2yc2+c3=x^2+y^2,将点集带入,方程就只剩三个未知数`c1,c2 和 c3。

简单起见,假设我们有四个点{[0,5],[0,-5],[5,0],[-5,0]},代入方程可得到四个方程:

  0c1 + 10c2 + c3 = 25
  0c1 - 10c2 + c3 = 25
 10c1 + 0c2  + c3 = 25
-10c1 + 0c2  + c3 = 25

该方程组比较简单,一眼便能看出解。但用线性代数我们可以得到矩阵:

/***************************A**********B******C*/
|  0c1  10c2   1c3|   |  0  10   1|   |c1|   |25|
|  0c1 -10c2   1c3| = |  0 -10   1| * |c2| = |25|
| 10c1   0c2   1c3|   | 10   0   1|   |c3|   |25|
|-10c1   0c2   1c3|   |-10   0   1|          |25|

在矩阵方程中A*B=C,只需求出矩阵B即可得到方程组的解。c#中MathNet.Numerics可以轻松胜任这一工作:

public Circle LinearAlgebraFit(double[] X, double[] Y)
{
    if (X.Length < 3)
    {
        return null;
    }
    var count = X.Length;
    var a = new double[count, 3];
    var c = new double[count, 1];
    for (int i = 0; i < count; i++)
    {
        //matrix
        a[i, 0] = 2 * X[i];
        a[i, 1] = 2 * Y[i];
        a[i, 2] = 1;
        c[i, 0] = X[i] * X[i] + Y[i] * Y[i];
    }
    var A = DenseMatrix.OfArray(a);
    var C = DenseMatrix.OfArray(c);
    //A*B=C
    var B = A.Solve(C);
    double c1 = B.At(0, 0),
        c2 = B.At(1, 0),
        r = Math.Sqrt(B.At(2, 0) + c1 * c1 + c2 * c2);
    var result = new Circle();
    result.X = c1;
    result.Y = c2;
    result.R = r;
    return result;
}

最后总结

Console.WriteLine($"raw   c1:{x0}, c2:{y0}, r:{r0}");
var fit = new FitCircle();
var sth = new Stopwatch();
sth.Start();
var lsf = fit.LeastSquaresFit(x, y);![](https://img2018.cnblogs.com/blog/1214143/201908/1214143-20190804173821455-1022769486.jpg)


Console.WriteLine($"LeastSquaresFit   c1:{lsf.X}, c2:{lsf.Y}, r:{lsf.R}, time:{sth.Elapsed}");
sth.Restart();
var laf = fit.LinearAlgebraFit(x, y);
Console.WriteLine($"LinearAlgebraFit  c1:{laf.X}, c2:{laf.Y}, r:{laf.R}, time:{sth.Elapsed}");
var img = new ImageHelp(512, 512);
img.DrawPoints(x, y, Brushes.Red);
img.DrawCicle(lsf, Brushes.Green);
img.DrawCicle(laf, Brushes.Orange);
img.SaveImage("graph.jpeg");

控制台输出:

raw   c1:204.1, c2:213.1, r:98.4
LeastSquaresFit   c1:204.791071061878, c2:210.86075318831, r:100.436594821545, time:00:00:00.0011029
LinearAlgebraFit  c1:204.791071061878, c2:210.860753188315, r:100.436594821541, time:00:00:00.1691119

从结果中可以看出,两种方法的结果基本一样,在小数点后好几位才出现差别。但是其计算效率却差异巨大,最小二乘法比线性代数快上 100 多倍。
在图中,二者重合(绿色被后面的橙色覆盖)。

.net core(c#)拟合圆测试_第1张图片

在最小二乘法中,只有一个及其简单的 for 循环,很少涉及内存写。但在线性代数中,需要进行矩阵的生成DenseMatrix.OfArray,以及矩阵运算,这二者都需要内存写。再者,矩阵计算有着繁重的计算量,这些都在影响着线性代数拟合圆的效率。最终的胜利还是属于最小二乘法。

转载于:https://www.cnblogs.com/hsxian/p/11298954.html

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