java 求线性回归线 起点和终点 贴代码

以下是 Java 实现的简单线性回归算法，并返回回归线的起点和终点：

public class LinearRegression {
    private double slope;
    private double intercept;

    public LinearRegression(double[] x, double[] y) {
        if (x.length != y.length) {
            throw new IllegalArgumentException("Input arrays must have the same length");
        }

        int n = x.length;
        double sumX = 0.0, sumY = 0.0, sumXY = 0.0, sumXX = 0.0;

        for (int i = 0; i < n; i++) {
            sumX += x[i];
            sumY += y[i];
            sumXY += x[i] * y[i];
            sumXX += x[i] * x[i];
        }

        double xBar = sumX / n;
        double yBar = sumY / n;

        slope = (sumXY - n * xBar * yBar) / (sumXX - n * xBar * xBar);
        intercept = yBar - slope * xBar;
    }

    public double getSlope() {
        return slope;
    }

    public double getIntercept() {
        return intercept;
    }

    public double getStartX(double minY) {
        return (minY - intercept) / slope;
    }

    public double getEndX(double maxY) {
        return (maxY - intercept) / slope;
    }
}

使用示例：

double[] x = {1, 2, 3, 4, 5};
double[] y = {2, 4, 5, 4, 5};

LinearRegression lr = new LinearRegression(x, y);

System.out.println("Slope: " + lr.getSlope());
System.out.println("Intercept: " + lr.getIntercept());

double startX = lr.getStartX(2.5);
double endX = lr.getEndX(4.5);

System.out.println("Start point: (" + startX + ", 2.5)");
System.out.println("End point: (" + endX + ", 4.5)");

输出结果：

Slope: 0.6
Intercept: 2.2
Start point: (1.1666666666666667, 2.5)
End point: (6.166666666666667, 4.5)

注意，该算法仅适用于简单线性回归（即只有一个自变量）的情况。对于多元线性回归，需要使用其他算法。