假如给定的实验数据点为（Xi,Yi),其中i=0,1,...n,那么 直线与数据点的偏差平方和为
要使得
取到极小值，则要求:
，
这两个式子是取得极小值的必要条件，具体运算的过程如下:
对该式求解得到:
public class LeastSquare
{
/// <summary>
/// To Draw Linear Least Square Fitting Curve
/// </summary>
/// <param name="g">The device on which to draw the curve</param>
/// <param name="lp">A list of point used to do the approximation</param>
public static void LeastSquare2(Graphics g, List<PointF> lp)
{
// Clear the client window with the white color
g.Clear(Color.White);
// Move the drawing origin to the point(200,200)
g.TranslateTransform(200, 200);
// Use FillEllipse method to draw each point as an ellipse
foreach (PointF p in lp)
{
g.FillEllipse(Brushes.Red, new RectangleF(p.X - 5.0f, p.Y - 5.0f, 10.0f, 10.0f));
}
int i;
float a, b, sumX, sumY2, sumY, sumXY;
sumX = 0.0f;
sumY2 = 0.0f;
sumY = 0.0f;
sumXY = 0.0f;
// To calculate as per the description of the Mathematical Formular
for (i = 0; i < lp.Count; i++)
{
sumY += lp[i].Y;
sumY2 += lp[i].Y * lp[i].Y;
sumX += lp[i].X;
sumXY += lp[i].X * lp[i].Y;
}
// Deduct the coefficients required to do the approximation using the mathematical formular
a = (lp.Count * sumXY - sumX * sumY) / (lp.Count * sumY2 - sumY * sumY);
b = (sumY2 * sumX - sumY * sumXY) / (lp.Count * sumY2 - sumY * sumY);
Pen newPen = new Pen(Color.Blue, 3.0f);
g.DrawLine(newPen, new PointF(0, -b / a), new PointF(360, (360 - b) / a));
}
}
{
/// <summary>
/// To Draw Linear Least Square Fitting Curve
/// </summary>
/// <param name="g">The device on which to draw the curve</param>
/// <param name="lp">A list of point used to do the approximation</param>
public static void LeastSquare2(Graphics g, List<PointF> lp)
{
// Clear the client window with the white color
g.Clear(Color.White);
// Move the drawing origin to the point(200,200)
g.TranslateTransform(200, 200);
// Use FillEllipse method to draw each point as an ellipse
foreach (PointF p in lp)
{
g.FillEllipse(Brushes.Red, new RectangleF(p.X - 5.0f, p.Y - 5.0f, 10.0f, 10.0f));
}
int i;
float a, b, sumX, sumY2, sumY, sumXY;
sumX = 0.0f;
sumY2 = 0.0f;
sumY = 0.0f;
sumXY = 0.0f;
// To calculate as per the description of the Mathematical Formular
for (i = 0; i < lp.Count; i++)
{
sumY += lp[i].Y;
sumY2 += lp[i].Y * lp[i].Y;
sumX += lp[i].X;
sumXY += lp[i].X * lp[i].Y;
}
// Deduct the coefficients required to do the approximation using the mathematical formular
a = (lp.Count * sumXY - sumX * sumY) / (lp.Count * sumY2 - sumY * sumY);
b = (sumY2 * sumX - sumY * sumXY) / (lp.Count * sumY2 - sumY * sumY);
Pen newPen = new Pen(Color.Blue, 3.0f);
g.DrawLine(newPen, new PointF(0, -b / a), new PointF(360, (360 - b) / a));
}
}
下面则是调用上述代码的程序:
private void linearToolStripMenuItem_Click(object sender, EventArgs e)
{
// Declare a list of points
List<PointF> lp = new List<PointF>();
// PointF array
PointF[] pf = new PointF[]{
new PointF(0.0f,68.0f),
new PointF(10.0f,73.1f),new PointF(20.0f,66.4f),
new PointF(30.0f,70.6f),new PointF(40.0f,64.6f),
new PointF(50.0f,68.8f),new PointF(60.0f,61.0f),
new PointF(70.0f,65.8f),new PointF(80.0f,60.4f),
new PointF(90.0f,61.0f)
};
// Using AddRange method of the list to add the pointf array to the end of the list
lp.AddRange(pf);
// Call the static metod LeastSquare2 of LeastSquare Class to proceed
LeastSquare.LeastSquare2(this.CreateGraphics(), lp);
}
下面是本程序运行结果的屏幕截图(Screen Shot):{
// Declare a list of points
List<PointF> lp = new List<PointF>();
// PointF array
PointF[] pf = new PointF[]{
new PointF(0.0f,68.0f),
new PointF(10.0f,73.1f),new PointF(20.0f,66.4f),
new PointF(30.0f,70.6f),new PointF(40.0f,64.6f),
new PointF(50.0f,68.8f),new PointF(60.0f,61.0f),
new PointF(70.0f,65.8f),new PointF(80.0f,60.4f),
new PointF(90.0f,61.0f)
};
// Using AddRange method of the list to add the pointf array to the end of the list
lp.AddRange(pf);
// Call the static metod LeastSquare2 of LeastSquare Class to proceed
LeastSquare.LeastSquare2(this.CreateGraphics(), lp);
}
大家如果有什么问题的话请给我留言吧。
