Lars 线性回归 C++ 代码
时间: 2023-09-07 17:15:01 浏览: 34
以下是一个简单的用C++实现的线性回归代码示例,其中使用了Lars算法:
```cpp
#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
const double EPSILON = 1e-10;
void lars(vector<vector<double>> &x, vector<double> &y, vector<double> &w) {
int n = x.size(), p = x[0].size();
w.resize(p);
vector<double> r = y, mu(p), c(p);
vector<int> index(p), active_set;
for (int i = 0; i < p; i++) {
mu[i] = 0.0;
index[i] = i;
}
double lambda = 0.0;
while (active_set.size() < p) {
double max_c = -1.0;
int max_index = -1;
for (int i = 0; i < p; i++) {
if (abs(mu[i]) < EPSILON && index[i] >= 0) {
c[i] = 0.0;
for (int j = 0; j < n; j++) {
c[i] += x[j][i] * r[j];
}
c[i] /= n;
if (abs(c[i]) > max_c) {
max_c = abs(c[i]);
max_index = i;
}
}
}
if (max_index < 0) {
break;
}
active_set.push_back(index[max_index]);
index[max_index] = -1;
double gamma = 0.0;
for (int i = 0; i < active_set.size(); i++) {
gamma += c[index[active_set[i]]];
}
gamma /= active_set.size();
double s = (lambda - gamma) / max_c;
for (int i = 0; i < p; i++) {
if (index[i] >= 0) {
mu[i] += s * c[i];
}
}
if (abs(lambda - gamma) < EPSILON) {
break;
}
lambda = gamma;
for (int i = 0; i < n; i++) {
r[i] = y[i];
for (int j = 0; j < active_set.size(); j++) {
r[i] -= x[i][index[active_set[j]]] * mu[index[active_set[j]]];
}
}
}
for (int i = 0; i < active_set.size(); i++) {
w[active_set[i]] = mu[index[active_set[i]]];
}
}
int main() {
vector<vector<double>> x{{1.0, 2.0, 3.0}, {4.0, 5.0, 6.0}, {7.0, 8.0, 9.0}};
vector<double> y{10.0, 11.0, 12.0};
vector<double> w;
lars(x, y, w);
for (int i = 0; i < w.size(); i++) {
cout << "w[" << i << "] = " << w[i] << endl;
}
return 0;
}
```
需要注意的是,这只是一个简单的示例代码,实际使用时需要根据具体情况进行调整和改进。