关于wpf:一种直线识别方案

本文介绍一种直线的辨认计划。

步骤

1. 应用最小二乘法回归直线：

$$\begin{cases}\frac{\sum_{i=1}^N(x_i,\overline{x})(y_i,\overline{y})}{\sum_{i=1}^N(x_i,\overline{x})^2}\\b=\overline{y}-k\overline{x}\end{cases}$$

1. 失去直线方程 y=kx+b 后，计算所有点到直线的间隔，若在阈值范畴内，认为是直线。

实现

/// <summary>
/// 最小二乘法求回归直线方程
/// </summary>
/// <param name="points"> 输出数据 </param>
/// <param name="k"> 直线斜率 </param>
/// <param name="b"> 直线截距 </param>
/// <param name="type"> 直线类型 1：水平线 2：垂直线 3：个别直线 </param>
/// <returns></returns>
public static bool IsLine(List<Point> points, out double k, out double b, out int type)
{
    k = 0;
    b = 0;
    type = 0;

    if (points.Count < 2) return false;

    double averageX = 0, averageY = 0, n = 0;
    n = points.Count;
    foreach (Point p in points)
    {
        averageX += p.X;
        averageY += p.Y;
    }
    averageX /= n;
    averageY /= n;

    double numerator = 0, denominator = 0;
    foreach (Point p in points)
    {numerator += (p.X - averageX) * (p.Y - averageY);
        denominator += (p.X - averageX) * (p.X - averageX);
    }

    if (numerator == 0) // 平行于 X 轴为水平线，返回纵坐标平均值
    {
        b = averageY;
        type = 1;
    }
    else if (denominator == 0)// 平行于 Y 轴为垂直线，返回横坐标平均值
    {
        b = averageX;
        type = 2;
    }
    else
    {type = 3;}

    k = numerator / denominator;
    b = averageY - k * averageX;

    foreach (Point p in points)
    {dis = GetPoint2LineDistance(p, k, b, type);
        if (dis > MAX_POINT_LINE_DIS) return false; // 点到拟合直线间隔过大
    }

    return true;
}

/// <summary>
/// 计算点到直线的间隔
/// </summary>
/// <param name="p"> 待计算点 </param>
/// <param name="k"> 直线斜率 </param>
/// <param name="b"> 直线截距 </param>
/// <param name="type"> 直线类型 1：水平线 2：垂直线 3：个别直线 </param>
/// <returns> 间隔 </returns>
private static double GetPoint2LineDistance(Point p, double k, double b, int type)
{if (type == 1)
    {return Math.Abs(p.Y - b);
    }
    else if (type == 2)
    {return Math.Abs(p.X - b);
    }
    else
    {
        double numerator = 0, denominator = 0;
        numerator = Math.Abs(k * p.X - p.Y + b);
        denominator = Math.Sqrt(k * k + 1);
        return numerator / denominator;
    }
}