逻辑回归代价函数 Java实现

小鱼儿 2022-05-31 13:15 301阅读 0赞

数据集

0,1,2,3,4,5,6,7,8,9,10

0,0,0,0,0,1,1,1,1,1,1

这里可以看出 当X大于4时 Y等于1

逻辑回归代价函数计算公式

![70][]

右侧为正规化 但是这里我们并不加入正规化 因为已经足够明显了

daima

package ojama;
    
    import java.io.BufferedReader;
    import java.io.File;
    import java.io.FileReader;
    import java.io.IOException;
    import java.util.List;
    import java.util.Vector;
    
    public class CostFunction {
    	public static void main(String[] args) throws IOException {
    		Double[] x1 = CostFunction.read("C:/Users/BenQ/Desktop/X.txt");
    		Double[] y = CostFunction.read("C:/Users/BenQ/Desktop/Y.txt");
    		int m = y.length;
    		Double[] x0 = new Double[m];
    		for (int i = 0; i < x0.length; i++) {
    			x0[i] = 1.0;
    		}
    		List<Double[]> X = new Vector<Double[]>();
    		X.add(x0);
    		X.add(x1);
    		Double[] theta = GradientDescent.getTheta(X, y);
    		for (int i = 0; i < theta.length; i++) {
    			System.out.println(String.format("%.2f", theta[i]));
    		}
    		int num = 0;
    		for (int i = 0; i < m; i++) {
    			Double sum = 0.0;
    			for (int k = 0; k < theta.length; k++) {
    				// 在二元图形中,这里相当于k*x+b*1,三元相当于a*x+b*y+c*1,以此类推
    				sum += theta[k] * X.get(k)[i];
    			}
    			num += y[i] * Math.log(sigmoid(sum)) + (1 - y[i]) * Math.log(1 - sigmoid(sum));
    		}
    		System.out.println(-num / m);
    	}
    
    	public static double sigmoid(double z) {
    		return 1 / (1 + Math.pow(Math.E, -z));
    	}
    
    	public static Double[] read(String fileName) throws IOException {
    		File file = new File(fileName);
    		FileReader fileReader = new FileReader(file);
    		BufferedReader reader = new BufferedReader(fileReader);
    		StringBuilder sb = new StringBuilder();
    		String str = reader.readLine();
    		while (str != null) {
    			sb.append(str);
    			str = reader.readLine();
    		}
    		reader.close();
    		fileReader.close();
    		String[] X0 = sb.toString().replace(" ", "").split(",");
    		Double[] x0 = new Double[X0.length];
    		for (int i = 0; i < x0.length; i++) {
    			x0[i] = Double.parseDouble(X0[i]);
    		}
    		return x0;
    	}
    }

输出结果

![70 1][]

[70]: /images/20220531/698a334c513c4e88ada54c7b200f983f.png
[70 1]: /images/20220531/14c7f6dc1eb1432c9bc40c8d3f745a25.png