在某些数据显示时，由于数据并不是均匀分布的，导致显示太难看；
所以不能使用简单分组=（最大值-最小值）/组；
后来发现这个算法！
自然均值算法：
/// <summary> /// 自然均值算法 !=（最大值-最小值）/分组数 /// </summary> /// <param name="list">排序后的数组</param> /// <param name="numclass">分成几组</param> /// <returns>返回一个数组，长度=分组数-1</returns> /// 用法：若返回[1,2] 则 (<1, 1<=x<2, x>=2) public double[] GetJenksBreaks(double[] list, int numclass) { int numdata = list.Length; double[,] mat1 = new double[numdata + 1, numclass + 1]; double[,] mat2 = new double[numdata + 1, numclass + 1]; double[] st = new double[numdata]; for (int i = 1; i <= numclass; i++) { mat1[1, i] = 1; mat2[1, i] = 0; for (int j = 2; j <= numdata; j++) { mat2[j, i] = Double.MaxValue; } } double v = 0.0; for (int l = 2; l <= numdata; l++) { double s1 = 0.0; double s2 = 0.0; double w = 0.0; for (int m = 1; m <= l; m++) { int i3 = l - m + 1; double val = list[i3 - 1]; s2 += val * val; s1 += val; w++; v = s2 - (s1 * s1) / w; int i4 = i3 - 1; if (i4 != 0) { for (int j = 2; j <= numclass; j++) { if (mat2[l, j] >= (v + mat2[i4, j - 1])) { mat1[l, j] = i3; mat2[l, j] = v + mat2[i4, j - 1]; } } } } mat1[l, 1] = 1; mat2[l, 1] = v; } int k = numdata; int[] kclass = new int[numclass]; kclass[numclass - 1] = list.Length - 1; double[] kvalues = new double[numclass - 1]; for (int j = numclass; j >= 2; j--) { int id = (int)(mat1[k, j]) - 2; kvalues[j - 2] = (double)list[id]; kclass[j - 2] = id; k = (int)mat1[k, j] - 1; } return kvalues; }