我已经编写了以下Java程序来实现带有梯度下降的线性回归。代码执行，但结果不准确。 y的预测值与y的实际值不接近。例如，当x = 75时，预期的y = 208，但输出为y = 193.784。
class LinReg {
double theta0, theta1;
void buildModel(double[] x, double[] y) {
double x_avg, y_avg, x_sum = 0.0, y_sum = 0.0;
double xy_sum = 0.0, xx_sum = 0.0;
int n = x.length, i;
for( i = 0; i < n; i++ ) {
x_sum += x[i];
y_sum += y[i];
}
x_avg = x_sum/n;
y_avg = y_sum/n;
for( i = 0; i < n; i++) {
xx_sum += (x[i] - x_avg) * (x[i] - x_avg);
xy_sum += (x[i] - x_avg) * (y[i] - y_avg);
}
theta1 = xy_sum/xx_sum;
theta0 = y_avg - (theta1 * x_avg);
System.out.println(theta0);
System.out.println(theta1);
gradientDescent(x, y, 0.1, 1500);
}
void gradientDescent(double x[], double y[], double alpha, int maxIter) {
double oldtheta0, oldtheta1;
oldtheta0 = 0.0;
oldtheta1 = 0.0;
int n = x.length;
for(int i = 0; i < maxIter; i++) {
if(hasConverged(oldtheta0, theta0) && hasConverged(oldtheta1, theta1))
break;
oldtheta0 = theta0;
oldtheta1 = theta1;
theta0 = oldtheta0 - (alpha * (summ0(x, y, oldtheta0, oldtheta1)/(double)n));
theta1 = oldtheta1 - (alpha * (summ1(x, y, oldtheta0, oldtheta1)/(double)n));
System.out.println(theta0);
System.out.println(theta1);
}
}
double summ0(double x[], double y[], double theta0, double theta1) {
double sum = 0.0;
int n = x.length, i;
for( i = 0; i < n; i++ ) {
sum += (hypothesis(theta0, theta1, x[i]) - y[i]);
}
return sum;
}
double summ1(double x[], double y[], double theta0, double theta1) {
double sum = 0.0;
int n = x.length, i;
for( i = 0; i < n; i++ ) {
sum += (((hypothesis(theta0, theta1, x[i]) - y[i]))*x[i]);
}
return sum;
}
boolean hasConverged(double oldTheta, double newTheta) {
return ((newTheta - oldTheta) < (double)0);
}
double predict(double x) {
return hypothesis(theta0, theta1, x);
}
double hypothesis(double thta0, double thta1, double x) {
return (thta0 + thta1 * x);
}
}
public class LinearRegression {
public static void main(String[] args) {
//Height data
double x[] = {63.0, 64.0, 66.0, 69.0, 69.0, 71.0, 71.0, 72.0, 73.0, 75.0};
//Weight data
double y[] = {127.0, 121.0, 142.0, 157.0, 162.0, 156.0, 169.0, 165.0, 181.0, 208.0};
LinReg model = new LinReg();
model.buildModel(x, y);
System.out.println("----------------------");
System.out.println(model.theta0);
System.out.println(model.theta1);
System.out.println(model.predict(75.0));
}
}
最佳答案
没有什么是错的。
我在R中验证了解决方案：
x <- c(63.0, 64.0, 66.0, 69.0, 69.0, 71.0, 71.0, 72.0, 73.0, 75.0)
y <- c(127.0, 121.0, 142.0, 157.0, 162.0, 156.0, 169.0, 165.0, 181.0, 208.0)
mod <- lm(y~x)
summary(mod)
Call:
lm(formula = y ~ x)
Residuals:
Min 1Q Median 3Q Max
-13.2339 -4.0804 -0.0963 4.6445 14.2158
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -266.5344 51.0320 -5.223 8e-04 ***
x 6.1376 0.7353 8.347 3.21e-05 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 8.641 on 8 degrees of freedom
Multiple R-squared: 0.897, Adjusted R-squared: 0.8841
F-statistic: 69.67 on 1 and 8 DF, p-value: 3.214e-05
X值为75时计算出的y-hat：
-266.5344 +(6.1376 *75)
[1] 193.784
这是正确的预测。我认为混淆必须围绕回归如何工作。回归不会告诉您训练数据中与给定独立数据点相对应的数据点的精确实际值。那只是字典，而不是统计模型（在那种情况下，它将无法进行内插或外推）。
回归将最小二乘法拟合到您的数据以估计模型方程，然后使用模型方程预测给定独立变量值的因变量值。唯一能准确预测训练数据中数据点的情况是您过度拟合模型（不好）。
有关更多信息和链接：
https://en.wikipedia.org/wiki/Regression_analysis
关于java - 梯度下降的线性回归，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/41298008/