Question

我正在尝试使用简单的梯度下降算法进行线性回归。我正在使用Armadillo C ++线性代数库，而且我也是Armadillo的新手。这就是我想要做的事情：

void linRegression(mat &features, mat &targets, double alpha,double error){
    mat theta = ones(features.n_cols+1);
    mat temp = zeros(features.n_cols+1);
    mat features_new = join_horiz(ones(features.n_rows),features);
    mat predictions;
    double con = alpha*(1.0/features.n_rows);
    int j = 0;
    while(j<1000){
        mat step_error = (features_new*theta - targets);
        for(unsigned int i=0;i<theta.n_rows;i++){
            temp(i) = con*sum(step_error%features_new.col(i));
        }
        theta = theta-temp;
        mat pred = predict(theta,features_new);
        cout<<theta<<endl;
        j++;
    }
}

但theta的值不断增加并最终达到无穷大。我不确定我做错了什么。

Answer 1

我认为while循环中的计算不正确。至少你可以在没有for-loop的情况下做得更优雅。以下是1个功能问题的简短代码：

#include <iostream>
#include <armadillo>

using namespace std;
using namespace arma;

int main(int argc, char** argv)
{
    mat features(10, 1);

    features << 6.110100 << endr
         << 5.527700 << endr
         << 8.518600 << endr
         << 7.003200 << endr
         << 5.859800 << endr
         << 8.382900 << endr
         << 7.476400 << endr
         << 8.578100 << endr
         << 6.486200 << endr
         << 5.054600 << endr;

    mat targets(10, 1);

    targets << 17.59200 << endr
        << 9.130200 << endr
        << 13.66200 << endr
        << 11.85400 << endr
        << 6.823300 << endr
        << 11.88600 << endr
        << 4.348300 << endr
        << 12.00000 << endr
        << 6.598700 << endr
        << 3.816600 << endr;

    mat theta = ones(features.n_cols + 1);

    mat features_new = join_horiz(ones(features.n_rows), features);

    double alpha = 0.01;
    double con = alpha*(1.0 / features.n_rows);

    int j = 0;

    while (j < 20000){
        mat step_error = (features_new*theta - targets);
        theta = theta - con * (features_new.t() * step_error);
        j++;
    }

    theta.print("theta:");

    system("pause");

    return 0;
}

程序返回：

theta:
   0.5083
   1.3425

通过正规方程法得到的结果是：

theta:
   0.5071
   1.3427

修改

您的代码确实是正确的！问题可能出在功能标准化中。我将我的示例扩展为2个特征回归并添加规范化。没有标准化，它对我也不起作用。

#include <iostream> #include <armadillo> using namespace std; using namespace arma; int main(int argc, char** argv) { mat features(10, 2); features << 2104 << 3 << endr << 1600 << 3 << endr << 2400 << 3 << endr << 1416 << 2 << endr << 3000 << 4 << endr << 1985 << 4 << endr << 1534 << 3 << endr << 1427 << 3 << endr << 1380 << 3 << endr << 1494 << 3 << endr; mat m = mean(features, 0); mat s = stddev(features, 0, 0); int i, j; //normalization for (i = 0; i < features.n_rows; i++) { for (j = 0; j < features.n_cols; j++) { features(i, j) = (features(i, j) - m(j))/s(j); } } mat targets(10, 1); targets << 399900 << endr << 329900 << endr << 369000 << endr << 232000 << endr << 539900 << endr << 299900 << endr << 314900 << endr << 198999 << endr << 212000 << endr << 242500 << endr; mat theta = ones(features.n_cols + 1); mat features_new = join_horiz(ones(features.n_rows), features); double alpha = 0.01; double con = alpha*(1.0 / features.n_rows); while (j < 20000){ mat step_error = (features_new*theta - targets); theta = theta - con * (features_new.t() * step_error); j++; } cout << theta << endl; system("pause"); return 0; }

结果：

THETA：

3.1390e+005 9.9704e+004 -5.6835e+003

线性回归的梯度下降不起作用

1 个答案: