lsqe.c

来自「掌握如何用C来实现各种算法」· C语言 代码 · 共 58 行

/*****************************************************************************
*                                                                            *
*  -------------------------------- lsqe.c --------------------------------  *
*                                                                            *
*****************************************************************************/

#include <math.h>

#include "nummeths.h"

/*****************************************************************************
*                                                                            *
*  --------------------------------- lsqe ---------------------------------  *
*                                                                            *
*****************************************************************************/

void lsqe(const double *x, const double *y, int n, double *b1, double *b0) {

double             sumx,
                   sumy,
                   sumx2,
                   sumxy;

int                i;

/*****************************************************************************
*                                                                            *
*  Compute the required summations.                                          *
*                                                                            *
*****************************************************************************/

sumx = 0.0;
sumy = 0.0;
sumx2 = 0.0;
sumxy = 0.0;

for (i = 0; i < n; i++) {

   sumx = sumx + x[i];
   sumy = sumy + y[i];
   sumx2 = sumx2 + pow(x[i], 2.0);
   sumxy = sumxy + (x[i] * y[i]);

}

/*****************************************************************************
*                                                                            *
*  Compute the least-squares estimators.                                     *
*                                                                            *
*****************************************************************************/

*b1 = (sumxy - ((sumx * sumy)/(double)n)) / (sumx2-(pow(sumx,2.0)/(double)n));
*b0 = (sumy - ((*b1) * sumx)) / (double)n;

return;

}