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线性回归 c++

程序员文章站 2024-01-02 15:26:40
线性回归 c++ 多元线性回归指的是根据多个自变量拟合一条曲线可以根据给定的自变量输出尽可能准确的因变量。 #include #include #include #i...

线性回归 c++
多元线性回归指的是根据多个自变量拟合一条曲线可以根据给定的自变量输出尽可能准确的因变量。

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using namespace std;

const int counts = 3; //系数的个数-（x+1）

//解线性方程组-高斯列主元素消元法
void swap_row(int row1, int row2, double matrix[][counts+1]) {  //交换两行元素
    double *d = (double*)malloc((counts + 1) * sizeof(double));

    memcpy(d, matrix[row1], (counts + 1) * sizeof(double));
    memcpy(matrix[row1], matrix[row2], (counts + 1) * sizeof(double));
    memcpy(matrix[row2], d, (counts + 1) * sizeof(double));
}
void find_max(int k, double matrix[][counts+1]) {//列元素中找最大值
    int r = k;
    double maxs = abs(matrix[k][k]);
    for(int i = k; i < counts; i++) {
        if(abs(matrix[i][k]) > maxs) {
            r = i;
            maxs = abs(matrix[i][k]);
        }
    }
    swap_row(k, r, matrix);
}
void LinearEquations(double matrix[][counts+1], double Answer[]) {
    //消元过程
    for(int i = 0; i < counts-1; i++) {
        find_max(i, matrix);//使主元素选取相较更大的值

        for(int j = i+1; j < counts; j++) {//第j行
            double l = matrix[j][i] / matrix[i][i];
            for(int k = i; k <= counts; k++)//j行的每一个元素
                matrix[j][k] = matrix[j][k] - l*matrix[i][k];
        }
    }

    //回代过程
    for(int i = counts-1; i >= 0; i--) {
        double sum = 0;
        for(int j = i+1; j < counts; j++)//求累加和
            sum += matrix[i][j] * Answer[j];
        Answer[i] = (matrix[i][counts] - sum) / matrix[i][i];
    }
}

void MultipleRegression(double data[][counts], int rows, int cols, double Answer[], double SquarePoor[]) {
    int count = cols - 1;

    double a, b;

    //打表回归系数方程组
    double dat[counts][counts+1];
    dat[0][0] = rows;
    for (int n = 0; n < count; n++) {                     // n = 0 to cols - 2,不包括cols-1这一列的Y
        a = b = 0.0;
        for (int m = 0; m < rows; m++) {//遍历每一行的第n列
            a += data[m][n];
            b += (data[m][n] * data[m][n]);
        }
        dat[0][n+1] = a;                              // dat[0, n+1] = Sum(Xn)
        dat[n+1][0] = a;                               // dat[n+1, 0] = Sum(Xn)
        dat[n+1][n+1] = b;                            // dat[n+1,n+1] = Sum(Xn * Xn)
        for (int i = n + 1; i < count; i ++) {            // i = n+1 to cols - 2
            a = 0;
            for (int m = 0; m < rows; m++)
                a += data[m][n] * data[m][i];
            dat[n+1][i+1] = a;        // dat[n+1, i+1] = Sum(Xn * Xi)
            dat[i+1][n+1] = a;   // dat[i+1, n+1] = Sum(Xn * Xi)
        }
    }
    b = 0;
    for (int m = 0; m < rows; m++)
        b += data[m][count];
    dat[0][cols] = b;                                   // dat[0, cols] = Sum(Y)
    for (int n = 0; n < count; n++) {
        a = 0;
        for (int m = 0; m < rows; m++)
            a += data[m][n] * data[m][count];
        dat[n+1][cols] = a;        // dat[n+1, cols] = Sum(Xn * Y)
    }

    LinearEquations(dat, Answer);          // 计算方程式
    // 方差分析
    if (SquarePoor) {
        b = b / rows;                                // b = Y的平均值
        SquarePoor[0] = SquarePoor[1] = 0.0;
        for (int m = 0; m < rows; m++) {
            a = Answer[0];
            for (int i = 1; i < cols; i++)
                a += (data[m][i-1] * Answer[i]);               // a = Ym的估计值
            SquarePoor[0] += ((a - b) * (a - b));    // U(回归平方和)
            SquarePoor[1] += ((data[m][count] - a) * (data[m][count] - a));  // Q(剩余平方和)(*p = Ym)
        }
        SquarePoor[2] = SquarePoor[0] / count;       // 回归方差
        if(rows - cols > 0.0)
            SquarePoor[3] = SquarePoor[1] / (rows - cols); // 剩余方差
        else
            SquarePoor[3] = 0.0;
    }
}

double calculate(double x[counts-1], double Answer[]) {
    double ans = Answer[0];
    for(int i = 1; i < counts; i++)
        ans += Answer[i] * x[i-1];
    return ans;
}

int main() {
//    freopen("/Users/really/Documents/code/input","r",stdin);
    double data[50][counts];

    for(int i = 0; i < 40; i++) {
        for(int j = 0; j < counts; j++) {
            cin >> data[i][j];
            getchar();
        }
    }

    double Answer[counts], SquarePoor[4];
    MultipleRegression(data, 40, counts, Answer, SquarePoor);

    double x[10];
    for(int i = 0; i < 7; i++) {
        for(int j = 0; j < counts-1; j++) {
            cin >> x[j];
            getchar();
        }

        cout << calculate(x, Answer) << endl;;
    }

    return 0;
}

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