平滑曲线几种算法

1 .二次指数平滑法求预测值

 /**
     * 二次指数平滑法求预测值
     * @param list 基础数据集合
     * @param year 未来第几期
     * @param modulus 平滑系数
     * @return 预测值
     */
    private static Double getExpect(List list, int year, Double modulus ) {
        if (list.size() < 10 || modulus <= 0 || modulus >= 1) {
            return null;
        }
        Double modulusLeft = 1 - modulus;
        Double lastIndex = list.get(0);
        Double lastSecIndex = list.get(0);
        for (Double data :list) {
            lastIndex = modulus * data + modulusLeft * lastIndex;
            lastSecIndex = modulus * lastIndex + modulusLeft * lastSecIndex;
        }
        Double a = 2 * lastIndex - lastSecIndex;
        Double b = (modulus / modulusLeft) * (lastIndex - lastSecIndex);
        return a + b * year;
    } 

 2.最小二乘法曲线拟合

  /**
     * 最小二乘法曲线拟合
     * @param data
     * @return
     */
    public static List polynomial(List data,int degree){
        final WeightedObservedPoints obs = new WeightedObservedPoints();
        for (int i = 0; i < data.size(); i++) {
            obs.add(i,  data.get(i));
        }
        /**
         * 实例化一个2次多项式拟合器
         */
        final PolynomialCurveFitter fitter = PolynomialCurveFitter.create(degree);//degree 阶数，一般为 3
        /**
         * 实例化检索拟合参数(多项式函数的系数)
         */
        final double[] coeff = fitter.fit(obs.toList());//size 0-3  阶数
        List result = new ArrayList<>();
        for (int i = 0; i < data.size(); i++) {
        	double tmp=0.0;
        	/**
        	 * 多项式函数f(x) = a0 * x + a1 * pow(x, 2) + .. + an * pow(x, n).
                 */
        	 for (int j = 0; j<= degree; j++) {
        		 tmp+= coeff[j]* Math.pow(i,j);
        	 }	
            result.add(tmp);
        }
        return result;
    }

 3.5点3次平滑曲线

 public static  Double[] Mean5_3(Double[] a, int m) {
    	Double[] b = new Double[a.length];
        int n = a.length;
        for (int k = 0; k < m; k++) {
            b[0] = (69 * a[0] + 4 * (a[1] + a[3]) - 6 * a[2] - a[4]) / 70;
            b[1] = (2 * (a[0] + a[4]) + 27 * a[1] + 12 * a[2] - 8 * a[3]) / 35;
            for (int j = 2; j < n - 2; j++) {
                b[j] = (-3 * (a[j - 2] + a[j + 2]) + 12 * (a[j - 1] + a[j + 1]) + 17 * a[j]) / 35;
            }
            b[n - 2] = (2 * (a[n - 1] + a[n - 5]) + 27 * a[n - 2] + 12 * a[n - 3] - 8 * a[n - 4]) / 35;
            b[n - 1] = (69 * a[n - 1] + 4 * (a[n - 2] + a[n - 4]) - 6 * a[n - 3] - a[n - 5]) / 70;
        }
        return b;
    }