稳健回归 C++代码
时间: 2023-09-24 14:09:52 浏览: 45
以下是一个简单的稳健回归的 C++ 代码实现:
```cpp
#include <iostream>
#include <cmath>
using namespace std;
double median(double arr[], int n) {
sort(arr, arr + n);
if (n % 2 == 0) {
return (arr[n / 2 - 1] + arr[n / 2]) / 2.0;
}
else {
return arr[n / 2];
}
}
double mad(double arr[], int n, double med) {
double sum = 0.0;
for (int i = 0; i < n; i++) {
sum += abs(arr[i] - med);
}
return sum / n;
}
double robust_regression(double x[], double y[], int n, double& slope, double& intercept) {
double x_med = median(x, n);
double y_med = median(y, n);
double x_mad = mad(x, n, x_med);
double y_mad = mad(y, n, y_med);
double x_new[n], y_new[n];
int count = 0;
for (int i = 0; i < n; i++) {
double x_diff = abs(x[i] - x_med) / x_mad;
double y_diff = abs(y[i] - y_med) / y_mad;
if (x_diff <= 2.5 && y_diff <= 2.5) {
x_new[count] = x[i];
y_new[count] = y[i];
count++;
}
}
double x_med_new = median(x_new, count);
double y_med_new = median(y_new, count);
double num = 0.0, den = 0.0;
for (int i = 0; i < count; i++) {
num += (x_new[i] - x_med_new) * (y_new[i] - y_med_new);
den += pow(x_new[i] - x_med_new, 2);
}
slope = num / den;
intercept = y_med_new - slope * x_med_new;
double res = 0.0;
for (int i = 0; i < n; i++) {
res += pow(y[i] - slope * x[i] - intercept, 2);
}
return sqrt(res / n);
}
int main() {
double x[] = { 1, 2, 3, 4, 5 };
double y[] = { 1.2, 1.9, 3.2, 4.1, 5.2 };
int n = sizeof(x) / sizeof(x[0]);
double slope, intercept;
double rmse = robust_regression(x, y, n, slope, intercept);
cout << "Slope: " << slope << endl;
cout << "Intercept: " << intercept << endl;
cout << "RMSE: " << rmse << endl;
return 0;
}
```
该代码首先实现了计算中位数和中位绝对偏差(MAD)的函数。然后,它实现了稳健回归的函数。在该函数中,我们计算了每个点到中位数的绝对偏差,并将距离中位数不超过 2.5 MAD 的点用于回归计算。最后,该函数返回回归方程的斜率、截距和均方根误差(RMSE)。