神奇的SoftMax regression，搞了一晚上搞不定，凌晨3點(diǎn)起來繼續(xù)搞，剛剛終于調(diào)通。我算是徹底理解了，哈哈。代碼試驗(yàn)了Andrew Ng的第四課上提到的SoftMax regression算法，并參考了http://ufldl.stanford.edu/wiki/index.php/Softmax_Regression

最終收斂到這個(gè)結(jié)果，巨爽。

smaple 0: 0.983690,0.004888,0.011422,likelyhood:-0.016445
smaple 1: 0.940236,0.047957,0.011807,likelyhood:-0.061625
smaple 2: 0.818187,0.001651,0.180162,likelyhood:-0.200665
smaple 3: 0.000187,0.999813,0.000000,likelyhood:-0.000187
smaple 4: 0.007913,0.992087,0.000000,likelyhood:-0.007945
smaple 5: 0.001585,0.998415,0.000000,likelyhood:-0.001587
smaple 6: 0.020159,0.000001,0.979840,likelyhood:-0.020366
smaple 7: 0.018230,0.000000,0.981770,likelyhood:-0.018398

smaple 8: 0.025072,0.000000,0.974928,likelyhood:-0.025392

#include "stdio.h" #include "math.h"double matrix[9][4]={{1,47,76,24}, //include x0=1{1,46,77,23},{1,48,74,22},{1,34,76,21},{1,35,75,24},{1,34,77,25},{1,55,76,21},{1,56,74,22},{1,55,72,22},};double result[]={1,1,1,2,2,2,3,3,3,};double theta[2][4]={{0.3,0.3,0.01,0.01},{0.5,0.5,0.01,0.01}}; // include theta0double function_g(double x) {double ex = pow(2.718281828,x);return ex/(1+ex); }double function_e(double x) {return pow(2.718281828,x); }int main(void) {double likelyhood = 0.0;for(int j = 0;j<9;++j){double sum = 1.0; // this is very important, because exp(thetak x)=1for(int l = 0;l<2;++l){double xi = 0.0;for(int k=0;k<4;++k){xi += matrix[j][k]*theta[l][k];}sum += function_e(xi);}double xi = 0.0;for(int k=0;k<4;++k){xi += matrix[j][k]*theta[0][k];}double p1 = function_e(xi)/sum;xi = 0.0;for(int k=0;k<4;++k){xi += matrix[j][k]*theta[1][k];}double p2 = function_e(xi)/sum;double p3 = 1-p1-p2;double ltheta = 0.0;if(result[j]==1)ltheta = log(p1);else if(result[j]==2)ltheta = log(p2);else if(result[j]==3)ltheta = log(p3);else{}printf("smaple %d: %f,%f,%f,likelyhood:%f\n",j,p1,p2,p3,ltheta);}for(int i =0 ;i<1000;++i){for(int j=0;j<9;++j){double sum = 1.0; // this is very important, because exp(thetak x)=1for(int l = 0;l<2;++l){double xi = 0.0;for(int k=0;k<4;++k){xi += matrix[j][k]*theta[l][k];}sum += function_e(xi);}double xi = 0.0;for(int k=0;k<4;++k){xi += matrix[j][k]*theta[0][k];}double p1 = function_e(xi)/sum;xi = 0.0;for(int k=0;k<4;++k){xi += matrix[j][k]*theta[1][k];}double p2 = function_e(xi)/sum;double p3 = 1-p1-p2;for(int m = 0; m<4; ++m){if(result[j]==1){theta[0][m] = theta[0][m] + 0.001*(1-p1)*matrix[j][m];}else{theta[0][m] = theta[0][m] + 0.001*(-p1)*matrix[j][m];}if(result[j]==2){theta[1][m] = theta[1][m] + 0.001*(1-p2)*matrix[j][m];}else{theta[1][m] = theta[1][m] + 0.001*(-p2)*matrix[j][m];}}}double likelyhood = 0.0;for(int j = 0;j<9;++j){double sum = 1.0; // this is very important, because exp(thetak x)=1for(int l = 0;l<2;++l){double xi = 0.0;for(int k=0;k<4;++k){xi += matrix[j][k]*theta[l][k];}sum += function_e(xi);}double xi = 0.0;for(int k=0;k<4;++k){xi += matrix[j][k]*theta[0][k];}double p1 = function_e(xi)/sum;xi = 0.0;for(int k=0;k<4;++k){xi += matrix[j][k]*theta[1][k];}double p2 = function_e(xi)/sum;double p3 = 1-p1-p2;double ltheta = 0.0;if(result[j]==1)ltheta = log(p1);else if(result[j]==2)ltheta = log(p2);else if(result[j]==3)ltheta = log(p3);else{}printf("smaple %d: %f,%f,%f,likelyhood:%f\n",j,p1,p2,p3,ltheta);}}return 0; }

from:?http://blog.csdn.net/pennyliang/article/details/7048291

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