WPF 折线/坐标点绘制 曲线抽稀 (Douglas-Peucker)道格拉斯-普克算法

这个算法经常用,例如GIS,数据保存,数据绘制都会用到。

算法是1973提出的,久经考验的算法,具体详情可以参考百度。

算法比较简单,大意是:

① 给出一个限定值表示距离

② 点集合或者坐标集合的首尾自动相连接成为直线,并会记录首尾两点到输出集合

③ 记录后寻找集合中距离这个直线最远的点,当这个点的距离超过限定值时,记录这个点,并由此将集合分为两段,首到点,点到尾

④ 对每个线段重复步骤②,步骤③,直至结束

 

也可以参考其他网友给出的算法推导. 

 

其中 找出最远点的点的算法,可以利用 三个点形成的面积,也就是三角形的面积。

其中求出点到线的距离,也就是三角形的高。

我们可以认为首尾连接的线是底部边,我们有底部边的起点和终点坐标,可以利用向量公式,尾坐标减去首坐标,求出向量坐标后,再利用向量的求模公式求出长度。

最后三角形面积除以底的长度乘与2就等于高度了。

 

 

截图

 

 

代码是现成的,虽然实现也不困难, 偷懒....

 public class Douglas_Peucker
    {
        /// <summary>
        /// Douglas-Peucker算法
        /// </summary>
        /// <param name="Points">坐标点集合</param>
        /// <param name="Tolerance">限定值</param>
        /// <returns></returns>
        public static List<Point> DouglasPeuckerReduction
            (List<Point> Points, Double Tolerance)
        {
            if (Points == null || Points.Count < 3)
                return Points;

            Int32 firstPoint = 0;
            Int32 lastPoint = Points.Count - 1;
            List<Int32> pointIndexsToKeep = new List<Int32>();

            //默认添加首尾两点
            pointIndexsToKeep.Add(firstPoint);
            pointIndexsToKeep.Add(lastPoint);

            //首尾两点不能相同
            while (Points[firstPoint].Equals(Points[lastPoint]))
            {
                lastPoint--;
            }
            //递归计算
            DouglasPeuckerReduction(Points, firstPoint, lastPoint,
            Tolerance, ref pointIndexsToKeep);
            //返回集合
            List<Point> returnPoints = new List<Point>();
            pointIndexsToKeep.Sort();
            foreach (Int32 index in pointIndexsToKeep)
            {
                returnPoints.Add(Points[index]);
            }

            return returnPoints;
        }

        /// <summary>
        /// 递归计算每个点到线段的长度,并分段递归重复计算
        /// </summary>
        /// <param name="points">点集合</param>
        /// <param name="firstPoint">首点</param>
        /// <param name="lastPoint">尾点</param>
        /// <param name="tolerance">限定值</param>
        /// <param name="pointIndexsToKeep">点集合下标</param>
        private static void DouglasPeuckerReduction(List<Point>
            points, Int32 firstPoint, Int32 lastPoint, Double tolerance,
            ref List<Int32> pointIndexsToKeep)
        {
            Double maxDistance = 0;
            Int32 indexFarthest = 0;
            //遍历每个点
            for (Int32 index = firstPoint; index < lastPoint; index++)
            {
                Double distance = PerpendicularDistance
                    (points[firstPoint], points[lastPoint], points[index]);
                //只寻找线段上最长的点
                if (distance > maxDistance)
                {
                    //替换值
                    maxDistance = distance;
                    //记录下标
                    indexFarthest = index;
                }
            }
            //确定最大值超过限定值且不为首点
            if (maxDistance > tolerance && indexFarthest != 0)
            {
                //记录最大距离的点的下标
                pointIndexsToKeep.Add(indexFarthest);
                //分段计算 Startpoint-MaxDistance 
                DouglasPeuckerReduction(points, firstPoint,
                indexFarthest, tolerance, ref pointIndexsToKeep);
                //分段计算 MaxDistance-Lastpoint
                DouglasPeuckerReduction(points, indexFarthest,
                lastPoint, tolerance, ref pointIndexsToKeep);
            }
        }

        /// <summary>
        /// 求出点到两点的距离
        /// </summary>
        /// <param name="pt1">线段的起点</param>
        /// <param name="pt2">线段的终点</param>
        /// <param name="p">计算的点</param>
        /// <returns></returns>
        public static Double PerpendicularDistance
            (Point Point1, Point Point2, Point Point)
        {
            //Area = |(1/2)(x1y2 + x2y3 + x3y1 - x2y1 - x3y2 - x1y3)|   *Area of triangle
            //Base = v((x1-x2)²+(x1-x2)²)                               *Base of Triangle*
            //Area = .5*Base*H                                          *Solve for height
            //Height = Area/.5/Base
            //求面积
            Double area = Math.Abs(.5 * (Point1.X * Point2.Y + Point2.X *
            Point.Y + Point.X * Point1.Y - Point2.X * Point1.Y - Point.X *
            Point2.Y - Point1.X * Point.Y));
            //求首尾两点的长度
            Double bottom = Math.Sqrt(Math.Pow(Point1.X - Point2.X, 2) +
            Math.Pow(Point1.Y - Point2.Y, 2));
            //三角形面积除以底*2=高
            //三角形面积除以高*2=底
            Double height = area / bottom * 2;

            return height;

            //Another option
            //Double A = Point.X - Point1.X;
            //Double B = Point.Y - Point1.Y;
            //Double C = Point2.X - Point1.X;
            //Double D = Point2.Y - Point1.Y;

            //Double dot = A * C + B * D;
            //Double len_sq = C * C + D * D;
            //Double param = dot / len_sq;

            //Double xx, yy;

            //if (param < 0)
            //{
            //    xx = Point1.X;
            //    yy = Point1.Y;
            //}
            //else if (param > 1)
            //{
            //    xx = Point2.X;
            //    yy = Point2.Y;
            //}
            //else
            //{
            //    xx = Point1.X + param * C;
            //    yy = Point1.Y + param * D;
            //}

            //Double d = DistanceBetweenOn2DPlane(Point, new Point(xx, yy));
        }
    }

 

使用这个算法后,能够将点减少很多,在视觉上差距不大,适用于很多点的时候,绘制困难,通过这个算法减少点的数量.

 

posted @ 2021-04-04 15:55  ARM830  阅读(63)  评论(0编辑  收藏