.net中使用 道格拉斯-普特 抽希轨迹点

Douglas一Peukcer算法由D.Douglas和T.Peueker于1973年提出,简称D一P算法,是目前公认的线状要素化简经典算法。现有的线化简算法中,有相当一部分都是在该算法基础上进行改进产生的。它的优点是具有平移和旋转不变性,给定曲线与阂值后,抽样结果一定。

思路:对每一条曲线的首末点虚连一条直线,求所有点与直线的距离,并找出最大距离值dmax ,用dmax与限差D相比:若dmax < ,这条曲线上的中间点全部舍去;若dmax ≥,保留dmax 对应的坐标点,并以该点为界,把曲线分为两部分,对这两部分重复使用该方法。

 GpsPositionDto 

public class GpsPositionDto 
	{
		
		/// <summary>
		/// 转换后的纬度
		/// </summary>
		public decimal GpsWebLng { get; set; }
		/// <summary>
		/// 转换后的经度
		/// </summary>
		public decimal GpsWebLat { get; set; }

		/// <summary>
		/// 时间
		/// </summary>
		public DateTime Gpstime { get; set; }
                //其他参数自己加
}                

   

/// <summary>
	/// 道格拉斯-普特
	/// </summary>
	public class DouglasPeucker
	{
		/// <summary>
		/// 距离
		/// </summary>
		/// <param name="point1"></param>
		/// <param name="point2"></param>
		/// <returns></returns>
		public static double CalculationDistance(GpsPositionDto point1, GpsPositionDto point2)
		{
			var lat1 = (double)point1.GpsWebLat;
			var lat2 = (double)point2.GpsWebLat;
			var lng1 = (double)point1.GpsWebLng;
			var lng2 = (double)point2.GpsWebLng;
			var radLat1 = lat1 * Math.PI / 180.0;
			var radLat2 = lat2 * Math.PI / 180.0;
			var a = radLat1 - radLat2;
			var b = (lng1 * Math.PI / 180.0) - (lng2 * Math.PI / 180.0);
			var s = 2 * Math.Asin(Math.Sqrt(Math.Pow(Math.Sin(a / 2), 2)
				+ Math.Cos(radLat1) * Math.Cos(radLat2) * Math.Pow(Math.Sin(b / 2), 2)));
			return s * 6370996.81;
		}
		/// <summary>
		/// 直线距离
		/// </summary>
		/// <param name="start"></param>
		/// <param name="end"></param>
		/// <param name="center"></param>
		/// <returns></returns>
		public static double DistToSegment(GpsPositionDto start, GpsPositionDto end, GpsPositionDto center)
		{
			var a = Math.Abs(CalculationDistance(start, end));
			var b = Math.Abs(CalculationDistance(start, center));
			var c = Math.Abs(CalculationDistance(end, center));
			var p = (a + b + c) / 2.0;
			var s = Math.Sqrt(Math.Abs(p * (p - a) * (p - b) * (p - c)));
			return s * 2.0 / a;
		}
		/// <summary>
		/// 递归方式压缩轨迹
		/// </summary>
		/// <param name="coordinate"></param>
		/// <param name="result"></param>
		/// <param name="start"></param>
		/// <param name="end"></param>
		/// <param name="dMax"></param>
		/// <returns></returns>
		public static IList<GpsPositionDto> CompressLine(IList<GpsPositionDto> coordinate, IList<GpsPositionDto> result, int start, int end, double dMax)
		{
			if (start < end)
			{
				var maxDist = 0D;
				var currentIndex = 0;
				var startPoint = coordinate[start];
				var endPoint = coordinate[end];
				for (var i = start + 1; i < end; i++)
				{
					var currentDist = DistToSegment(startPoint, endPoint, coordinate[i]);
					if (currentDist > maxDist)
					{
						maxDist = currentDist;
						currentIndex = i;
					}
				}
				if (maxDist >= dMax)
				{
					//将当前点加入到过滤数组中
					result.Add(coordinate[currentIndex]);
					//将原来的线段以当前点为中心拆成两段,分别进行递归处理
					CompressLine(coordinate, result, start, currentIndex, dMax);
					CompressLine(coordinate, result, currentIndex, end, dMax);
				}
			}
			return result;
		}


		/// <summary>
		/// 抽希
		/// </summary>
		/// <param name="coordinate">原始轨迹Array</param>
		/// <param name="dMax">允许最大距离误差</param>
		/// <returns>抽稀后的轨迹</returns>
		public static IList<GpsPositionDto> Dilution(IList<GpsPositionDto> coordinate, double dMax = 10)
		{
			if (!(coordinate.Count > 2))
			{
				return null;
			}
			var result = CompressLine(coordinate, new List<GpsPositionDto>(), 0, coordinate.Count - 1, dMax);
			result.Add(coordinate[0]);
			result.Add(coordinate[coordinate.Count - 1]);
			//排序
			var resultLatLng = result.OrderBy(s => s.Gpstime).ToList();
			return resultLatLng;
		}
	}

  

原文地址:https://www.cnblogs.com/jiamiemie/p/10472503.html