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一般線性模型的最小二次方擬合方法

考慮用有M個(gè)未定參數(shù)a_j(j=1,...,M)的模型來擬合N個(gè)數(shù)據(jù)點(diǎn)(x_i, y_i)，i= 1, 2, ..., N。
因變量與自變量的一個(gè)函數(shù)關(guān)系可以如下給出：
y(x) = y( x; a₁, a₂, ..., a_M)
如果所給的N個(gè)數(shù)據(jù)點(diǎn)（x_i, y_i）誤差都是獨(dú)立的，并且服從具有相同常數(shù)方差的正態(tài)分布，那么最小二乘方擬合就是擬合參數(shù)a_j的最大似然估計(jì)
X² = \sum_{i=1}^N [ y_i - y(x_i; a₁, a₂, ..., a_M)]²

如果這個(gè)模型是x的任意M個(gè)特定函數(shù)的線性組合，如
y(x) = a₀+ a₁cos(k₁x+2) + a₂sin(k₂x+3)
一般形式可以寫為
y(x) = \sum_{k=1}^N a_kX_k(x)

那么它的優(yōu)值函數(shù)為
X² = \sum_{i=1}^N [y_i- \sum_{k=1}^M a_kX_k(x_i)]²

將上式分別對(duì)M個(gè)參數(shù)a_k求偏導(dǎo)數(shù)，并令其等于0, 可以得到X²的最小值。
于是得到M個(gè)方程
\sum_{i=1}^N [y_i-\sum_{j=1}^M a_jX_j(x_i)]X_k(x_i) = 0 k= 1, 2, ..., M

用克萊姆法則求解這個(gè)M元一次方程組，可以得到 a₁, a_2,...,a_M的解，即為所求的a_j的最大似然估計(jì)。

接下來估計(jì)數(shù)據(jù)與模型的契合度
由平移伽瑪近似計(jì)算
ssq = gammq((N-2)/2, X²/2)

以下為c++實(shí)現(xiàn)
regression.h

  1 template<size_t SIZE, size_t N>
  2 class Regression
  3 {
  4 public:
  5     //ctor
  6     Regression()
  7     {
  8         Y.resize ( SIZE );
  9         X.resize ( SIZE * N );
10         Result.resize ( SIZE );
11         SSQ = 0.0L;
12     }
13
14     //dtor
15     ~Regression() {}
16     //get Y[SIZE]
17     void _y ( const std :: vector<long double>& y )
18     {
19         assert ( SIZE == y.size() );
20         Y = y;
21     }
22
23     //set vector x[][]
24     void _x ( const std :: vector<long double>& x )
25     {
26         assert ( SIZE * N == x.size() );
27         X = x;
28     }
29
30     //set vector x[][] cloumn index
31     void _x ( const std :: vector<long double>& x, const unsigned int& index )
32     {
33         assert ( index < N );
34         assert ( SIZE == x.size() );
35
36         for ( unsigned int i = 0; i < SIZE; ++i )
37         {
38             X[ i * N + index ] = x[i];
39         }
40     }
41
42     //execute least square fit
43     void fit()
44     {
45         //the model is
46         //             alfa[N][N] * result[N] = beta[N]
47         //
48         long double alfa[N][N];
49         long double beta[N];
50
51         for ( unsigned int k = 0; k < N; ++k )
52         {
53             for ( unsigned int j = 0; j < N; ++j )
54             {
55                 long double tmp = 0.0L;
56                 for ( unsigned int i = 0; i < SIZE; ++i )
57                 {
58                     tmp += X[i*N+k] * X[i*N+j];
59                 }//end of i loop
60                 alfa[k][j] = tmp;
61             }//end of j loop
62
63             long double tmp = 0.0L;
64             for ( unsigned int i = 0; i < SIZE; ++i )
65             {
66                 tmp += Y[i] * X[i*N+k];
67             }//end of i loop
68             beta[k] = tmp;
69         }//end of k loop
70
71         std::vector<long double> a;
72         a.clear();
73         for ( unsigned int i = 0; i < N; ++i )
74             for ( unsigned int j = 0; j < N; ++j )
75             {
76                 a.push_back ( alfa[i][j] );
77             }
78
79         std::vector<long double> b;
80         b.clear();
81         for ( unsigned int i = 0; i < N; ++i )
82             b.push_back ( beta[i] );
83
84         //calculate Result[N]
85         lq<long double> ( a, b, N, Result );
86
87         //calculate chi-square
88         long double chi = 0.0L;
89         for ( unsigned int i = 0; i < SIZE; ++i )
90         {
91             long double tmp = 0.0L;
92             for ( unsigned int j = 0; j < N; ++j )
93             {
94                 tmp += Result[j] * X[i*j+j];
95             }
96             tmp -= Y[i];
97             chi += tmp * tmp;
98         }
99
100         SSQ = gammq ( static_cast<long double> ( SIZE - 2 ) / 2.0L, chi / 2.0L );
101
102
103     }//end of k loop
104
105     std :: vector< long double> result() const
106     {
107         return Result;
108     }
109
110     long double ssq() const
111     {
112         return SSQ;
113     }
114
115 private:
116     std :: vector<long double> Y;       // -- Y[SIZE]
117     std :: vector<long double> X;       // -- X[N][SIZE]
118     std :: vector<long double> Result;  // -- Result[N]
119     long double SSQ;                      //fitness of the model
120 };
121
122

其中SIZE即為所要擬合的a_j的數(shù)量M，而N為原始數(shù)據(jù)點(diǎn)數(shù)（x_i, y_i）的數(shù)量N。

給出gamma函數(shù)計(jì)算代碼

#include <cmath>
#include <cstdlib>

static long double
gammln ( const long double &xx )
{
    long double     x,
    tmp,
    ser;
    static long double cof[6] = { 76.18009173, -86.50532033, 24.01409822,
                                  -1.231739516, 0.120858003e-2, -0.536382e-5
                                };
    int             j;

    x = xx - 1.0;
    tmp = x + 5.5;
    tmp -= ( x + 0.5 ) * log ( tmp );
    ser = 1.0;
    for ( j = 0; j <= 5; j++ )
    {
        x += 1.0;
        ser += cof[j] / x;
    }
    return -tmp + log ( 2.50662827465 * ser );
}

static void
gcf ( long double *gammcf,
      const long double &a, const long double &x, long double *gln )
{
    const unsigned int ITMAX = 100;
    const long double EPS = 3.0e-7;
    unsigned int    n = 0;
    long double     gold = 0.0,
                           g,
                           fac = 1.0,
                                 b1 = 1.0;
    long double     b0 = 0.0,
                         anf,
                         ana,
                         an,
                         a1,
                         a0 = 1.0;

    *gln = gammln ( a );
    a1 = x;
    for ( n = 1; n <= ITMAX; n++ )
    {
        an = ( float ) n;
        ana = an - a;
        a0 = ( a1 + a0 * ana ) * fac;
        b0 = ( b1 + b0 * ana ) * fac;
        anf = an * fac;
        a1 = x * a0 + anf * a1;
        b1 = x * b0 + anf * b1;
        if ( a1 )
        {
            fac = 1.0 / a1;
            g = b1 * fac;
            if ( fabs ( ( g - gold ) / g ) < EPS )
            {
                *gammcf = exp ( -x + a * log ( x ) - ( *gln ) ) * g;
                return;
            }
            gold = g;
        }
    }
    if ( "a too large, ITMAX too small in routine GCF" )
        exit ( EXIT_FAILURE );
}

static void
gser ( long double *gamser,
       const long double &a, const long double &x, long double *gln )
{
    const unsigned int ITMAX = 100;
    const long double EPS = 3.0e-7;
    unsigned int    n = 0;
    long double     sum,
    del,
    ap;

    *gln = gammln ( a );
    if ( x <= 0.0 )
    {
        if ( x < 0.0 )
            if ( "x less than 0 in routine GSER" )
                exit ( EXIT_FAILURE );
        *gamser = 0.0;
        return;
    }
    else
    {
        ap = a;
        del = sum = 1.0 / a;
        for ( n = 1; n <= ITMAX; n++ )
        {
            ap += 1.0;
            del *= x / ap;
            sum += del;
            if ( fabs ( del ) < fabs ( sum ) * EPS )
            {
                *gamser = sum * exp ( -x + a * log ( x ) - ( *gln ) );
                return;
            }
        }
        if ( "a too large, ITMAX too small in routine GSER" )
            exit ( EXIT_FAILURE );

        return;
    }
}

long double
gammq ( const long double &a, const long double &x )
{
    long double     gamser,
    gammcf,
    gln;

    if ( x < 0.0 || a <= 0.0 )
        if ( "Invalid arguments in routine gammq." )
            exit ( EXIT_FAILURE );

    if ( x < ( a + 1.0 ) )
    {
        gser ( &gamser, a, x, &gln );
        return 1.0 - gamser;
    }
    else
    {
        gcf ( &gammcf, a, x, &gln );
        return gammcf;
    }
}

利用克萊姆法則求解線性方程組
lq.h

1 #include "det.h"
2
3 #include <vector>
4 #include <cassert>
5 #include <cstdlib>
6
7 //
8 //for solution of the linear equation
9 //                  A X = B
10 //
11 template<class T>
12 void lq ( const std :: vector<T>& a,    //store A[size][size]
13           const std :: vector<T>& b,     //store B[size]
14           const unsigned int& order, //store size
15           std :: vector<T>& result       //store X[size]
16         )
17 {
18     unsigned int Order = order;
19     //if order is set to default, calculate
20     if ( 0 == Order )
21     {
22         Order = b.size();
23     }
24
25     assert ( Order * Order == a.size() );
26     assert ( Order == b.size() );
27
28     result.clear();
29
30     const T _D = det ( a, Order );  //store D
31
32     if ( 0 == _D )
33     {
34         if ( "Failed to solve the linear equation" )
35             exit ( EXIT_FAILURE );
36     }
37
38     for ( unsigned int i = 0; i < Order; ++i )
39     {
40         std :: vector<T> A = a;
41         for ( unsigned int j = 0; j < Order; ++j )
42         {
43             A[i+j*Order] = b[j];
44         }
45         const T D = det ( A, Order );   //store D[i]
46
47         result.push_back ( D / _D );    //get X[i]
48     }
49 }
50

在lq.h中有提到一個(gè)det函數(shù)，這個(gè)是用來計(jì)算矩陣的行列式值的，因?yàn)榭巳R姆法則就是幾個(gè)行列式的值之間除來除去罷了，代碼如下
det.h

1 //return the determinant of a square matrix arr whose size is order by order
2 template< class T >
3 T det ( const std :: vector<T>& arr, const unsigned int& order = 0 )
4 {
5     unsigned int Order = order;
6
7     //if order is set to default, calculate it
8     if ( 0 == Order )
9     {
10         Order = sqrt ( arr.size() );
11     }
12
13     assert ( Order * Order == arr.size() );
14
15
16     if ( 1 == Order )
17         return ( arr[0] ) ;
18
19     T sum = 0;
20     std ::vector<T> arr2 ;
21     int sign = 1;
22
23     for ( unsigned int i = 0 ; i < Order ; ++i, sign *= -1 )
24     {
25         /* copy n-1 by n-1 array into another array */
26         const unsigned int newsize = ( Order - 1 ) * ( Order - 1 ) ;
27
28         arr2.resize ( newsize );
29         for ( unsigned int j = 1 ; j < Order ; ++j )
30         {
31             for ( unsigned int k = 0, count = 0 ; k < Order ; ++k )
32             {
33                 if ( k == i )
34                     continue ;//jump out of k loop
35
36                 const unsigned int pos = j * Order + k ;
37                 const unsigned int newpos = ( j - 1 ) *
38                                             ( Order - 1 ) + count ;
39                 arr2[newpos] = arr[pos] ;
40                 count++ ;
41             }//end of k loop
42         }//end of j loop
43
44         /*  find determinant value of n-1 by n-1 array and  add it to sum */
45         sum += arr[i] * sign * det ( arr2, Order - 1 ) ;
46     }
47     return sum;
48 }
49

下邊給出一個(gè)示例
要擬合的參數(shù)方程為
a₁x₁ + a₂x₂ + a₃x₃ + a₄x₄ + a₅x₅ = y
給出5組數(shù)據(jù)進(jìn)行擬合，分別是（1 0 0 0 0 1）（0 1 0 0 0 2）（0 0 1 0 0 3）（0 0 0 1 0 4）（0 0 0 0 1 5）

1 #include "regression.h"
2 #include <iostream>
3 #include <vector>
4
5 using namespace std;
6
7 int main()
8 {
9     const unsigned int N = 5;
10
11     Regression<N, N> r;
12     vector<long double> y;
13     vector<long double> x;
14     vector<long double> result;
15
16
17     for ( unsigned int i = 0; i < N; ++i )
18         y.push_back( 1.0L + i );
19     r._y( y );
20
21     for ( unsigned int i = 0; i < N; ++i )
22     {
23         x.clear();
24         for ( unsigned int j = 0; j < N; ++j )
25         {
26             if ( i == j )
27                 x.push_back( 1.0L );
28             else
29                 x.push_back( 0.0L );
30         }
31         r._x( x, i );
32     }
33
34     r.fit();
35
36     result = r.result();
37
38     for ( unsigned int i = 0; i < N; ++i )
39         cout << result[i] << endl;
40
41
42     return 0;
43 }

posted on 2008-04-15 22:38 Wang Feng 閱讀(4916) 評(píng)論(2) 編輯收藏引用所屬分類: Numerical C++

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