最小二乘法JAVA代码

在网上找到的代码是错的，自己写了一个

最小二乘法的代码，二个构造方法，一个是不带权的，一个是带权的

/** * 最小二乘法计算类 * * @author Administrator * */public class LeastSquareMethod {private double[] x;private double[] y;private double[] weight;private int m;private double[] coefficient;public LeastSquareMethod(double[] x, double[] y, int m) {if (x == null || y == null || x.length < 2 || x.length != y.length|| m < 2)throw new IllegalArgumentException("无效的参数");this.x = x;this.y = y;this.m = m;weight = new double[x.length];for (int i = 0; i < x.length; i++) {weight[i] = 1;}}public LeastSquareMethod(double[] x, double[] y, double[] weight, int m) {if (x == null || y == null || weight == null || x.length < 2|| x.length != y.length || x.length != weight.length || m < 2)throw new IllegalArgumentException("无效的参数");this.x = x;this.y = y;this.m = m;this.weight = weight;}public double[] getCoefficient() {if (coefficient == null)compute();return coefficient;}public double fit(double v) {if (coefficient == null)compute();if (coefficient == null)return 0;double sum = 0;for (int i = 0; i < coefficient.length; i++) {sum += Math.pow(v, i) * coefficient[i];}return sum;}private void compute() {if (x == null || y == null || x.length <= 1 || x.length != y.length|| x.length < m || m < 2)return;double[] s = new double[(m - 1) * 2 + 1];for (int i = 0; i < s.length; i++) {for (int j = 0; j < x.length; j++)s[i] += Math.pow(x[j], i) * weight[j];}double[] f = new double[m];for (int i = 0; i < f.length; i++) {for (int j = 0; j < x.length; j++)f[i] += Math.pow(x[j], i) * y[j] * weight[j];}double[][] a = new double[m][m];for (int i = 0; i < m; i++) {for (int j = 0; j < m; j++) {a[i][j] = s[i + j];}}coefficient = Algorithm.multiLinearEquationGroup(a, f);}/** * @param args */public static void main(String[] args) {LeastSquareMethod l = new LeastSquareMethod(new double[] { 2000, 2001, 2002, 2003, 2004, 2005, 2006, 2007, 2008 }, new double[] { 37.84, 44.55, 45.74, 63.8, 76.67, 105.59, 178.48, 355.27, 409.92 }, new double[] { 11, 12, 13, 14, 15, 16, 17, 18, 19 },2);double[] x = l.getCoefficient();for (double xx : x) {System.out.println(xx);}System.out.println(l.fit(2009));}}

用到了多元一次方程的解法：

public class Algorithm {/** * 多元一次方程求解 * @param a 系数数组 * @param b 结果数组 * @return 解答 */public static double[] multiLinearEquationGroup(double[][] a,double[] b){if (a==null||b==null||a.length==0||a.length!=b.length)return null;for (double[] x:a){if (x==null||x.length!=a.length)return null;}int len=a.length-1;double[] result=new double[a.length];if (len==0){result[0]=b[0]/a[0][0];return result;}double[][] aa=new double[len][len];double[] bb=new double[len];int posx=-1,posy=-1;for (int i=0;i<=len;i++){for (int j=0;j<=len;j++)if (a[i][j]!=0.0d){posy=j;break;}if (posy!=-1){posx=i;break;}}if (posx==-1)return null;int count=0;for (int i=0;i<=len;i++){if (i==posx)continue;bb[count]=b[i]*a[posx][posy]-b[posx]*a[i][posy];int count2=0;for (int j=0;j<=len;j++){if (j==posy)continue;aa[count][count2]=a[i][j]*a[posx][posy]-a[posx][j]*a[i][posy];count2++;}count++;}double[] result2=multiLinearEquationGroup(aa,bb);double sum=b[posx];count=0;for (int i=0;i<=len;i++){if (i==posy)continue;sum-=a[posx][i]*result2[count];result[i]=result2[count];count++;}result[posy]=sum/a[posx][posy];return result;}/** * @param args */public static void main(String[] args) {double[] r=multiLinearEquationGroup(new double[][]{{1,2,3,1},{2,0,1,0},{5,2,0,0},{7,1,1,0}},new double[]{18,5,9,12});for (int i=0;i<r.length;i++){System.out.println(r[i]);}}}