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| #include <math.h>
#ifndef PI
#define PI 3.14159265359
#endif
#define ACCURACY 1.0e-12
/*
***********************************************************
SOLVE LINEAR SYSTEM VIA GAUSSIAN ELIMINATION
( 10.01.1979 K. BANSE )
(SACLAY - HPCCD - 1982/84)
Ported to C
Jean Souviron 2015/12/13
***********************************************************
*/
int LNSYS(double *Array, double *X, int Order )
{
int i, j, k, MaxRow, Length ;
double Save, Rmax, Mid ;
Length = Order+1 ;
for ( k = 0 ; k < (Order-1) ; k++ )
{
/*
--- Get largest pivotal element
*/
Save = 0.0 ;
for ( j = k ; j < Order ; j++ )
{
Rmax = fabs(Array[j*Length+k]);
if( Save < Rmax )
{
Save = Rmax ;
MaxRow = j ;
}
}
/*
--- Exchange rows
*/
for ( j = k ; j < Length ; j++ )
{
Save = Array[k*Length+j] ;
Array[k*Length+j] = Array[MaxRow*Length+j] ;
Array[MaxRow*Length+j]= Save ;
}
/*
--- Test if array is singular
*/
Mid = Array[k*Length+k] ;
if ( fabs(Mid) < ACCURACY )
{
for ( j = 0 ; j < Order ; j++ )
X[j] = 0.0 ;
return 0 ;
}
/*
--- Substract row from lower rows
*/
for ( i = k+1 ; i < Order ; i++ )
{
Save = Array[i*Length+k] / Mid ;
Array[i*Length+k] = 0.0 ;
for ( j = k+1 ; j < Length ; j++ )
Array[i*Length+j] = Array[i*Length+j] - (Save*Array[k*Length+j]) ;
}
}
/*
================
Calculate vector X
================
*/
X[Order-1] = Array[Length*(Order-1)+Order] / Array[Length*(Order-1)+Order-1] ;
for ( k = 1 ; k < Order ; k++ )
{
i = Order + 1 - k ;
Save = Array[i*Length+Order+1] ;
for ( j = i+1 ; j < Order ; j++ )
Save = Save - Array[i*Length+j]*X[j] ;
X[i] = Save / Array[i*Length+i] ;
}
return 1 ;
}
/*
***********************************************************************
GENERALIZED LEAST-SQUARE FOR ANY NUMBER OF PARAMETERS
(Jean SOUVIRON - DAO Victoria - 1985)
(METHOD OF R. SEDGEWICK in ALGORITHMS
ADDISON-WESLEY PUBLISHING CO.
1984)
Modified 2015/12/12 Jean Souviron
Porting to C
Upgrading to take any number of parameters
-----------------------------------------------------------------------
INPUT :
NDims = NUMBER OF DIFFERENT PARAMETERS+1 (for the data to be fitted to)
NPts = TOTAL NUMBER OF POINTS
InputMatrix = ARRAY OF PARAMETERS FOR EACH POINT
(LAST VECTOR HAS TO BE THE DATA TO BE FITTED)
OUTPUT :
OutputCoeffs = ARRAY OF COEFFICIENTS (dimension NDims-1 )
Sigma
RETURN VALUE :
1 if everything is OK
0 if an error occurred (memory or singular array)
Example : if we want to fit an ellipse with a group of points according to
equation x2/a2 + y2/b2 = 1
the input matrix will be N pts long, and 3 columns wide, with for
each point the x2, the y2, and 1.
the output coeffs will be 2 items, 1/a2 and 1/b2
the sigma will be computed by applying these coeffs to the first 2 columns
for each point, and comparing with the result (1).
***********************************************************************
*/
int LeastSquares ( double *InputMatrix, int NDims, int NPts,
double **OutputCoeffs, double *Sigma )
{
double *Coeffs=NULL ;
double T, Z1, Z2 ;
int i,j, k, io = 0, ko, NParams ;
NParams = NDims-1 ;
/*
Allocations
*/
if ( (*OutputCoeffs = (double *)calloc(NParams, sizeof(double))) == NULL )
return 0 ;
if ( (Coeffs = (double *)calloc(NParams*NDims, sizeof(double))) == NULL )
{
free (*OutputCoeffs);
*OutputCoeffs = NULL ;
return 0 ;
}
/*
Initializations
*/
io = 0 ;
for ( i = 0 ; i < NParams ; i++ )
{
(*OutputCoeffs)[i] = 0.0 ;
for ( j = 0 ; j < NDims ; j++ )
{
T = 0.0 ;
for ( k=0 ; k < NPts ; k++ )
{
ko = k * NDims ;
Z1 = InputMatrix[ko+i] ;
Z2 = InputMatrix[ko+j] ;
T = T + (Z1*Z2) ;
}
Coeffs[io+j]= T ;
}
io = io + NDims ;
}
/*
-----------------
Solving by gaussian elimination
-----------------
*/
if ( LNSYS(Coeffs, *OutputCoeffs, NParams) != 1 )
{
free ( Coeffs );
free ( *OutputCoeffs ) ;
*OutputCoeffs = NULL ;
return 0 ;
}
free ( Coeffs );
/*
-----------------
Computes sigma
-----------------
*/
*Sigma = 0.0 ;
for ( i = 0 ; i < NPts ; i++ )
{
Z1 = InputMatrix[i*NDims+NParams]; /* The value to be compared with */
for ( j = 0 ; j < NParams ; j++ )
{
Z1= Z1 - ((*OutputCoeffs)[j]*InputMatrix[i*NDims+j]) ;
}
if( Z1 < 0.0 )
Z1 = -Z1;
*Sigma = *Sigma + Z1 ;
}
*Sigma = (*Sigma/(double)NPts)*sqrt(PI/2.0) ;
return 1 ;
} |
Résultats d'une régression linéaire
ton poste tu dois marquer quand la bonne réponse tu as obtenu.
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