最小二乘法拟合曲线C语言实现：四次函数

公式原理，请移步二阶回归曲线拟合

以下是我改编的4阶实现

从data.txt中，读入指定数量(x,y)，进行拟合四阶曲线，打印输出各系数值。

改编成n阶，仅需要修改do{}while();循环即可。

#include 
#include
#include 
#define N 1e-13	
#define T 4		//阶数
#define NUM 101  //数据数量
#define OUTPUT
int main() {

	int i, j, k;

	//初始化数组
	long double x[NUM + 1] = { 0 };
	long double y[NUM + 1] = { 0 };

  //读入数据
	FILE *fp;
	fopen_s(&fp, "data.txt", "r");
	int checkNum;
	for (i = 0; i <= NUM; i++) {
		if (!fscanf_s(fp, "%lf", &x[i]))break;
		if (!fscanf_s(fp, "%lf", &y[i]))break;
		
	}
	fclose(fp);
#ifdef OUTPUT 
	for (i = 0; i < 10; i++) {
		printf("%16.9lf ", x[i]);
		printf("%16.9lf\n", y[i]);
	}
	printf("\n\n");
#endif 

	long double a[T + 1] = { 0 }; // 参数
	long double m[T + 1] = { 0 };// 中间变量
	long double z[T + 1] = { 0 };// 中间变量

	long double sumXY[T * 2 + 1][2] = { 0 };//存储各幂值累加和，如sumxy[3][3] 存储 （x ^ 3 * y ^ 3）的累加和

	for (j = 0; j <= T * 2 + 1; j++) {
		for (k = 0; k <= 1; k++) {
			for (i = 0; i < NUM; i++)
			{
				sumXY[j][k] += (pow(x[i], j) * pow(y[i], k));
							
			}
		}
	}
#ifdef OUTPUT 
	for (i = 0; i < 5; i++) {
		printf("%16.9lf ", sumXY[i][0]);
		printf("%16.9lf\n", sumXY[i][1]);
	}
	printf("\n\n");
#endif 
	int countNum = 0;
	do {
		m[4] = a[4]; 
		a[4] = (sumXY[4][1] - a[0] * sumXY[4][0] - a[1] * sumXY[5][0] - a[2] * sumXY[6][0] - a[3] * sumXY[7][0])/ sumXY[8][0]; 
		z[4] = (a[4] - m[4])*(a[4] - m[4]);


		m[3] = a[3]; 
		a[3] = (sumXY[3][1] - a[0] * sumXY[3][0] - a[1] * sumXY[4][0] - a[2] * sumXY[5][0] - a[4] * sumXY[7][0]) / sumXY[6][0];
		z[3] = (a[3] - m[3])*(a[3] - m[3]);
		
		m[2] = a[2];
		a[2] = (sumXY[2][1] - a[0] * sumXY[2][0] - a[1] * sumXY[3][0] - a[3] * sumXY[5][0] - a[4] * sumXY[6][0]) / sumXY[4][0];
		z[2] = (a[2] - m[2])*(a[2] - m[2]);
		
		m[1] = a[1];
		a[1] = (sumXY[1][1] - a[0] * sumXY[1][0] - a[2] * sumXY[3][0] - a[3] * sumXY[4][0] - a[4] * sumXY[5][0]) / sumXY[2][0];
		z[1] = (a[1] - m[1])*(a[1] - m[1]);

		m[0] = a[0];
		a[0] = (sumXY[0][1] - a[1] * sumXY[1][0] - a[2] * sumXY[2][0] - a[3] * sumXY[3][0] - a[4] * sumXY[4][0]) / NUM;
		z[0] = (a[0] - m[0])*(a[0] - m[0]);


		printf("%d\n", countNum++);//统计运算次数
	} while ((z[0] > N || z[1] > N || z[2] > N || z[3] > N || z[4] > N ));
  //打印输出
	printf("a=%9.6f,\nb=%9.6f,\nc=%9.6f\nd=%9.6f\ne=%9.6f\n", a[4], a[3], a[2], a[1], a[0]);
	printf("四阶拟合方程为  y=%9.6fx^4 + %9.6fx^3 + %9.6fx^2 + %9.6fx^1 + %9.6f ", a[4], a[3], a[2], a[1], a[0]);


	getchar();
	return 0;
}