多元最小二乘回归 C++有截距的带类实现 及案例

以下是一个简单的多元最小二乘回归的C++实现，包括截距。我们将使用一个类来实现这个回归。

#include <iostream>
#include <vector>
#include <cmath>

using namespace std;

class MultipleLinearRegression {
private:
    vector<vector<double>> x;   // 自变量
    vector<double> y;           // 因变量
    int n;                      // 样本数
    int k;                      // 自变量数
    vector<double> beta;        // 系数
    double b0;                  // 截距

public:
    MultipleLinearRegression(vector<vector<double>> x, vector<double> y) {
        this->x = x;
        this->y = y;
        n = x.size();
        k = x[0].size();
    }

    void fit() {
        vector<vector<double>> x_transpose(k, vector<double>(n));  // 自变量矩阵的转置
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < k; j++) {
                x_transpose[j][i] = x[i][j];
            }
        }
        vector<vector<double>> x_transpose_x(k, vector<double>(k));  // 自变量矩阵的转置乘自变量矩阵
        for (int i = 0; i < k; i++) {
            for (int j = 0; j < k; j++) {
                for (int l = 0; l < n; l++) {
                    x_transpose_x[i][j] += x_transpose[i][l] * x[l][j];
                }
            }
        }
        vector<double> x_transpose_y(k);  // 自变量矩阵的转置乘因变量向量
        for (int i = 0; i < k; i++) {
            for (int j = 0; j < n; j++) {
                x_transpose_y[i] += x_transpose[i][j] * y[j];
            }
        }
        beta.resize(k);
        for (int i = 0; i < k; i++) {
            beta[i] = 0;
        }
        for (int i = 0; i < k; i++) {  // 解方程 Ax=b
            int max_row = i;
            for (int j = i + 1; j < k; j++) {
                if (abs(x_transpose_x[j][i]) > abs(x_transpose_x[max_row][i])) {
                    max_row = j;
                }
            }
            swap(x_transpose_x[i], x_transpose_x[max_row]);
            swap(x_transpose_y[i], x_transpose_y[max_row]);
            for (int j = i + 1; j < k; j++) {
                double ratio = x_transpose_x[j][i] / x_transpose_x[i][i];
                for (int l = i; l < k; l++) {
                    x_transpose_x[j][l] -= ratio * x_transpose_x[i][l];
                }
                x_transpose_y[j] -= ratio * x_transpose_y[i];
            }
        }
        for (int i = k - 1; i >= 0; i--) {  // 回代求解
            double sum = 0;
            for (int j = i + 1; j < k; j++) {
                sum += x_transpose_x[i][j] * beta[j];
            }
            beta[i] = (x_transpose_y[i] - sum) / x_transpose_x[i][i];
        }
        double sum = 0;
        for (int i = 0; i < n; i++) {
            sum += y[i];
        }
        double y_mean = sum / n;
        sum = 0;
        for (int i = 0; i < n; i++) {
            double y_pred = b0;
            for (int j = 0; j < k; j++) {
                y_pred += beta[j] * x[i][j];
            }
            sum += pow(y_pred - y_mean, 2);
        }
        double ssr = 0;
        for (int i = 0; i < n; i++) {
            double y_pred = b0;
            for (int j = 0; j < k; j++) {
                y_pred += beta[j] * x[i][j];
            }
            ssr += pow(y_pred - y_mean, 2);
        }
        double sse = 0;
        for (int i = 0; i < n; i++) {
            double y_pred = b0;
            for (int j = 0; j < k; j++) {
                y_pred += beta[j] * x[i][j];
            }
            sse += pow(y[i] - y_pred, 2);
        }
        double sst = ssr + sse;
        double r_squared = ssr / sst;
        b0 = y_mean - beta[0] * x_transpose[0][0] / x_transpose_x[0][0];
        cout << "系数：";
        for (int i = 0; i < k; i++) {
            cout << beta[i] << " ";
        }
        cout << endl << "截距：" << b0 << endl;
        cout << "R²：" << r_squared << endl;
    }
};

int main() {
    vector<vector<double>> x{{1, 2}, {2, 3}, {3, 4}, {4, 5}, {5, 6}};
    vector<double> y{1, 2, 3, 4, 5};
    MultipleLinearRegression mlr(x, y);
    mlr.fit();
    return 0;
}

在这个实现中，我们通过 $Ax=b$ 解多元线性方程组，其中 $A$ 是自变量矩阵的转置乘自变量矩阵，$b$ 是自变量矩阵的转置乘因变量向量。我们使用高斯消元法来解这个方程组。

最后，我们计算回归系数和截距，以及计算 $R^2$ 来评估回归模型的拟合程度。注意，在计算截距时，我们使用了第一个自变量的均值。