多元线性回归批量梯度下降java代码

多元假
设函数
h θ ( x ) = θ T X = θ 0 x 0 + θ 1 x 1 + . . . + θ n x n , x 0 = 1
多元代价函数
J ( θ 0 , θ 1 , . . . , θ n ) = 1 2 m i = 1 m ( h θ ( x i ) y i ) 2
梯度下降
Repeat{
θ j = θ j α θ j 1 2 m i = 1 m ( h θ ( x i ) y i ) 2
}

θ 0 = θ 0 α 1 m i = 1 m ( h θ ( x i ) y i ) x 0 i , x 0 = 1
θ 1 = θ 1 α 1 m i = 1 m ( h θ ( x i ) y i ) x 1 i
θ 2 = θ 2 α 1 m i = 1 m ( h θ ( x i ) y i ) x 2 i

α 学习率 太大可能导致不能收敛,太小会很慢,经过太多次迭代才能到底
怎么判断收敛,一种是循环很大的次数,反正约到最后移动很慢,第二个是如果后面一次值比前面一次值变化很小(比如0.0001)就认为收敛。

多元函数是每次同时更新每个 θ

下面是实现代码

package com.zy.ml;

import java.io.File;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;

import org.apache.commons.io.FileUtils;
import org.apache.commons.io.LineIterator;
/**
 * 多元线性回归批量梯度下降
 * @author yzhang
 *
 */
public class LinearRegressionBGD {

    public static void main(String[] args) {
        LinearRegressionBGD lr = new LinearRegressionBGD(0.0001,100000,0.0001);
        lr.getTrainDate("/home/zy/Desktop/test.txt");
        lr.getTheta();
    }

    static List<String> trainData = new ArrayList<String>();
    static double alpha = 0.01;
    static Double[] theta = new Double[] { 1.0, 1.0, 1.0 };
    static Double[] tmptheta = new Double[] { 1.0, 1.0, 1.0 };
    static int m = 0;
    static int time = 100000;
    static double end = 0.0001;

    public LinearRegressionBGD() {

    }

    public LinearRegressionBGD(double alpha, int time, double end) {

    }

    public void getTrainDate(String file) {
        LineIterator li = null;

        try {
            li = FileUtils.lineIterator(new File(file));
        } catch (IOException e) {
            e.printStackTrace();
        }
        for (; li.hasNext();) {
            String line = li.nextLine();
            trainData.add(line);
        }
        m = trainData.size();
    }

    /**
     * 每一个偏导数的值 theta_j偏导数= 1/m*[sum(h(x)-y)*x_j]
     * 
     * @param i
     *            训练集行数time
     * @param j
     *            参数下标
     */
    private Double computerDerivativeofCostFunction(int j) {
        double sum = 0;
        for (int i = 0; i < trainData.size(); i++) {
            // h(x_i)值
            double x = 0.0;
            // 计算h(x)
            for (int k = 0; k < theta.length; k++) {

                double n = 1;
                if (k != 0) {
                    n = Double.valueOf(trainData.get(i).split(",")[k - 1]);
                }

                x += (theta[k] * n);
            }
            // 求导后后面乘以的x_i theta0没有所以j=0时,是1
            double d = 1;
            if (j != 0) {
                d = Double.valueOf(trainData.get(i).split(",")[j - 1]);
            }
            // (h(x_i)-y_i)*x_i
            double z = (x - Double.valueOf(trainData.get(i).split(",")[trainData.get(i).split(",").length - 1])) * d;
            // ∑(h(x_i)-y_i)*x_i
            sum = sum + z;
        }
        // 1/m × ∑(h(x_i)-y_i)*x_i
        double costfun = sum / m;
        return costfun;
    }

    public void getTheta() {
        int c = 1;
        boolean flg = false;
        while ((time--) > 0) {
            System.out.println("第" + (c++) + "次");

            tmptheta = theta.clone();
            for (int i = 0; i < theta.length; i++) {
                theta[i] -= alpha * computerDerivativeofCostFunction(i);

                System.out.println("theta" + i + ":" + theta[i]);
            }
            //判断终止条件
            for (int i = 0; i < theta.length; i++) {
                if (Math.abs(theta[i] - tmptheta[i]) < end) {
                    flg = true;
                } else {
                    flg = false;
                }
            }
            if (flg) {
                return;
            }
        }

    }
}

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转载自blog.csdn.net/jshazhang/article/details/80410069
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