logistic回归 c++ 实现

浏览数：13 / 时间：2015年06月09日

　　logistic回归是统计学习中经典的分类方法，他属于对数线性模型。本博文主要给出logistic的c++实现，具体理论请读者自行google。

　　本文用到的数据集来自于一个医学网站，具体出处不记得了（非常歉意）。数据的格式如下：

　　10009 1 0 0 1 0 1
　　10025 0 0 1 2 0 0
　　20035 0 0 1 0 0 1
　　20053 1 0 0 0 0 0
　　30627 1 0 1 2 0 0
　　30648 2 0 0 0 1 0

每行有7个列值，第一列是一个ID号，在具体操作中，忽略该列。之后的5列，每一个都表示一个特征的取值；最后一列是分类标记（0或1）。

　　在具体实现时，将分隔数据为训练数据和测试数据，并保存到文件中，文件组织形式如下：

　　

其中testdata.txt，保存测试数据；traindata.txt保存训练数据；logistic.cpp是代码源文件。三个文件保存在同一目录下。

实现代码如下：

?
/*********
    logistic回归（c++） by 姜富春
**********/
#include<iostream>
#include<fstream>
#include<vector>
#include<sstream>
#include<cmath>
using namespace std;
struct Data{
    vector<int> features;
    int cls;
    Data(vector<int> f,int c):features(f),cls(c){
         
    }
};
struct Param{
    vector<double> w;
    double d;
    Param(vector<double> w1,double d1):w(w1),d(d1){};
    Param():w(vector<double>()),d(0.0){}
};
class Logistic{
public:
    Logistic(){
        //载入traindata文件构造dataSet;
        loadDataSet(dataSet);
        //初始化Param，w的长度与数据特征的长度相同，初值为0.0。d的初值也为0.0
        vector<double> pw(dataSet[0].features.size(),0.0);
        Param pt(pw,0.0);
        param=pt;
 
    };
    void loadDataSet(vector<Data>& ds,string dataFile="./traindata.txt"){
        ifstream fin(dataFile.c_str());
        if(!fin){
            cout<<"文件打开失败"<<endl;
            exit(0);
        }
        while(fin){
            string line;
            getline(fin,line);
            if(line.size()>3){
                stringstream sin(line);
                int t;
                sin>>t;
                vector<int> fea;
                while(sin){
                    char c=sin.peek();
                    if(int(c)!=-1){
                        sin>>t;
                        fea.push_back(t);
                    }
                 
                }
                int cl=fea.back();
                fea.pop_back();
                ds.push_back(Data(fea,cl));
            }
        }
    }
     
    void displayDataSet(){
        for(int i=0;i<dataSet.size();i++){
            for(int j=0;j<dataSet[i].features.size();j++){
                cout<<dataSet[i].features[j]<<" ";
            }
            cout<<" 分类："<<dataSet[i].cls;
            cout<<endl;
        }
    }
    void logisticRegression(){
        //由目标函数为最大似然，因此最终求得的是目标函数的最大值，
        //因此迭代过程是梯度上升，而非梯度下降
        double lamda=0.1;//梯度下降的步长
        double delta=0.0001;//结束迭代的阈值
        //目标函数的值
        double objLw=Lw(param);
        //cout<<objLw<<endl;
        Param tpa(param.w,param.d);
        gradient(lamda);
        double newObjLw=Lw(param);
        int iter=0;
        cout<<"初始："<<endl;
        displayIterProcess(iter,objLw,newObjLw,1);
        while(fabs(newObjLw-objLw)>delta||!samewb(tpa,param,delta)){
            objLw=newObjLw;
            tpa=Param(param.w,param.d);
            gradient(lamda);
            newObjLw=Lw(param);
            ++iter;
            displayIterProcess(iter,objLw,newObjLw,5);
        }
        cout<<"迭代结束共迭代"<<iter<<"步"<<endl;
        displayIterProcess(iter,objLw,newObjLw,1);
 
    }
    bool samewb(const Param &tparam,const Param& param,double delta){
        for(int i=0;i<tparam.w.size();i++){
            if(fabs(tparam.w[i]-param.w[i])>delta){
                return false;
            }
        }
        if(fabs(tparam.d-param.d)>delta){
            return false;
        }
        return true;
    }
    void displayIterProcess(int iter,double objLw,double newObjLw,int mod){
        //每mod步打印一次迭代过程
        if(iter%mod==0){
            cout<<"迭代"<<iter<<"：目标函数值【"<<newObjLw<<"】，两次迭代目标函数差值【 "<<(newObjLw-objLw)<<"】"<<endl;
            cout<<"模型参数：";
            for(int i=0;i<param.w.size();i++){
                cout<<param.w[i]<<" ";
            }
            cout<<param.d<<endl<<endl;
        }
         
    }
    //梯度上升更新w和b
    void gradient(double lam){
        for(int i=0;i<param.w.size();i++){
            double tmp=0.0L;//保存梯度上升过程的中间值
            for(int j=0;j<dataSet.size();j++){
                tmp+=(dataSet[j].cls-logiFun(param,dataSet[j]))*dataSet[j].features[i]*lam;
            }
            param.w[i]+=(tmp);
        }
        double tmp=0.0L;
        for(int j=0;j<dataSet.size();j++){
            tmp+=(dataSet[j].cls-logiFun(param,dataSet[j]))*lam;
        }
        param.d+=tmp;
     
    }
    //计算logistic函数的值，即f(x)=exp(wx)/(1+exp(wx)),该表达式在求解梯度过程中出现，
    //因此计算这个值是为了辅助梯度上升计算过程
    inline double logiFun(const Param &p,const Data &d){
        double inner=innerWX(p,d);
        double le=exp(inner)/(1+exp(inner));
        return le;
    }
    //计算对数似然函数的值
    double Lw(Param p){
        double l=0.0L;
        for(int i=0;i<dataSet.size();i++){
            double inner=innerWX(p,dataSet[i]);
            l+=(dataSet[i].cls*inner-(log10(1+exp(inner))));
            //cout<<"l="<<l<<endl;
        }
         
        return l;
    }
    //计算wx+b的值
    inline double innerWX(const Param &p,const Data &data){
        if(p.w.size()!=data.features.size()){
            cout<<"参数与实例的维度不匹配，不能进行内积计算"<<endl;
            exit(0);
        }
        double innerP=0.0L;
        for(int i=0;i<p.w.size();i++){
            innerP+=(p.w[i]*data.features[i]);
        }
        innerP+=p.d;
        return innerP;
    }
    //给定测试集，预测分类
    void predictClass(){
        vector<Data> testDataSet;
        loadDataSet(testDataSet,"./testdata.txt");
        /*******************
        分别计算
            P(Y=1|x)=exp(w.x)/(1+exp(w.x))
        和
            P(Y=0|x)=1/(1+exp(w.x))
        然后取值大的作为x的分类
        *******************/
        cout<<endl<<"预测分类:"<<endl;
        for(int i=0;i<testDataSet.size();i++){
            double py1=0.0L;
            double py0=0.0L;
            double inner=innerWX(param,testDataSet[i]);
            py1=exp(inner)/(1+exp(inner));
            py0=1-py1;
            cout<<"实例: ";
            for(int j=0;j<testDataSet[i].features.size();j++){
                cout<<testDataSet[i].features[j]<<" ";
            }
            cout<<"标记分类【"<<testDataSet[i].cls<<"】，";
            if(py1>=py0){
                 
                cout<<"预测分类【"<<1<<"】"<<endl;
            }else{
                cout<<"预测分类【"<<0<<"】"<<endl;
            }
        }
    }
private:
    vector<Data> dataSet;
    Param param;
};
int main(){
    Logistic logist;
    //logist.displayDataSet();
    logist.logisticRegression();
    logist.predictClass();
    system("pause");
    return 0;
}

　　程序运行结果如下：

  1. 迭代训练模型过程如下

　　2. 测试分类过程如下：

本例中分类测试的效果并不是太好，在调试的时候，看到这样的结果我也仔细地审查代码，并未发现程序的错误（也希望读者帮忙审查）。考虑可能的原因：

　　（1）数据太少

　　（2）数据特征不能提供足够信息

　　（3）线性模型，无法很好地划分数据

附件：

训练数据：traindata.txt

10009 1 0 0 1 0 1
10025 0 0 1 2 0 0
10038 1 0 0 1 1 0
10042 0 0 0 0 1 0
10049 0 0 1 0 0 0
10113 0 0 1 0 1 0
10131 0 0 1 2 1 0
10160 1 0 0 0 0 0
10164 0 0 1 0 1 0
10189 1 0 1 0 0 0
10215 0 0 1 0 1 0
10216 0 0 1 0 0 0
10235 0 0 1 0 1 0
10270 1 0 0 1 0 0
10282 1 0 0 0 1 0
10303 2 0 0 0 1 0
10346 1 0 0 2 1 0
10380 2 0 0 0 1 0
10429 2 0 1 0 0 0
10441 0 0 1 0 1 0
10443 0 0 1 2 0 0
10463 0 0 0 0 0 0
10475 0 0 1 0 1 0
10489 1 0 1 0 1 1
10518 0 0 1 2 1 0
10529 1 0 1 0 0 0
10545 0 0 1 0 0 0
10546 0 0 0 2 0 0
10575 1 0 0 0 1 0
10579 2 0 1 0 0 0
10581 2 0 1 1 1 0
10600 1 0 1 1 0 0
10627 1 0 1 2 0 0
10653 1 0 0 1 1 0
10664 0 0 0 0 1 0
10691 1 1 0 0 1 0
10692 1 0 1 2 1 0
10711 0 0 0 0 1 0
10714 0 0 1 0 0 0
10739 1 0 1 1 1 0
10750 1 0 1 0 1 0
10764 2 0 1 2 0 0
10770 0 0 1 2 1 0
10780 0 0 1 0 1 0
10784 2 0 1 0 1 0
10785 0 0 1 0 1 0
10788 1 0 0 0 0 0
10815 1 0 0 0 1 0
10816 0 0 0 0 1 0
10818 0 0 1 2 1 0
11095 0 1 1 0 0 0
11146 0 1 0 0 1 0
11206 2 1 0 0 0 0
11223 2 1 0 0 0 0
11236 1 1 0 2 0 0
11244 1 1 0 0 0 1
11245 0 1 0 0 0 0
11278 2 1 0 0 1 0
11322 0 1 0 0 1 0
11326 2 1 0 2 1 0
11329 2 1 0 2 1 0
11344 1 1 0 2 1 0
11358 0 1 0 0 0 1
11417 2 1 1 0 1 0
11421 2 1 0 1 1 0
11484 1 1 0 0 0 1
11499 2 1 0 0 0 0
11503 1 1 0 0 1 0
11527 1 1 0 0 0 0
11540 2 1 0 1 1 0
11580 1 1 0 0 1 0
11583 1 0 1 1 0 1
11592 2 1 0 1 1 0
11604 0 1 0 0 1 0
11625 1 0 1 0 0 0
20035 0 0 1 0 0 1
20053 1 0 0 0 0 0
20070 0 0 0 2 1 0
20074 1 0 1 2 0 1
20146 1 0 0 1 1 0
20149 2 0 1 2 1 0
20158 2 0 0 0 1 0
20185 1 0 0 1 1 0
20193 1 0 1 0 1 0
20194 0 0 1 0 0 0
20205 1 0 0 2 1 0
20206 2 0 1 1 1 0
20265 0 0 1 0 1 0
20311 0 0 0 0 1 0
20328 2 0 0 1 0 1
20353 0 0 1 0 0 0
20372 0 0 0 0 0 0
20405 1 0 1 1 1 1
20413 2 0 1 0 1 0
20427 0 0 0 0 0 0
20455 1 0 1 0 1 0
20462 0 0 0 0 1 0
20472 0 0 0 2 0 0
20485 0 0 0 0 0 0
20523 0 0 1 2 0 0
20539 0 0 1 0 1 0
20554 0 0 1 0 0 1
20565 0 0 0 2 1 0
20566 1 0 1 1 1 0
20567 1 0 0 1 1 0
20568 0 0 1 0 1 0
20569 1 0 0 0 0 0
20571 1 0 1 0 1 0
20581 2 0 0 0 1 0
20583 1 0 0 0 1 0
20585 2 0 0 1 1 0
20586 0 0 1 2 1 0
20591 1 0 1 2 0 0
20595 0 0 1 2 1 0
20597 1 0 0 0 0 0
20599 0 0 1 0 1 0
20607 0 0 0 1 1 0
20611 1 0 0 0 1 0
20612 2 0 0 1 1 0
20614 1 0 0 1 1 0
20615 1 0 1 0 0 0
21017 1 1 0 1 1 0
21058 2 1 0 0 1 0
21063 0 1 0 0 0 0
21084 1 1 0 1 0 1
21087 1 1 0 2 1 0
21098 0 1 0 0 0 0
21099 1 1 0 2 0 0
21113 0 1 0 0 1 0
21114 1 1 0 0 1 1
21116 1 1 0 2 1 0
21117 1 0 0 2 1 0
21138 2 1 1 1 1 0
21154 0 1 0 0 1 0
21165 0 1 0 0 1 0
21181 2 1 0 0 0 1
21183 1 1 0 2 1 0
21231 1 1 0 0 1 0
21234 1 1 1 0 0 0
21286 2 1 0 2 1 0
21352 2 1 1 1 0 0
21395 0 1 0 0 1 0
21417 1 1 0 2 1 0
21423 0 1 0 0 1 0
21426 1 1 0 1 1 0
21433 0 1 0 0 1 0
21435 0 1 0 0 0 0
21436 1 1 0 0 0 0
21439 1 1 0 2 1 0
21446 1 1 0 0 0 0
21448 0 1 1 2 0 0
21453 2 1 0 0 1 0
30042 2 0 1 0 0 1
30080 0 0 1 0 1 0
301003 1 0 1 0 0 0
301009 0 0 1 2 1 0
301017 0 0 1 0 0 0
30154 1 0 1 0 1 0
30176 0 0 1 0 1 0
30210 0 0 1 0 1 0
30239 1 0 1 0 1 0
30311 0 0 0 0 0 1
30382 0 0 1 2 1 0
30387 0 0 1 0 1 0
30415 0 0 1 0 1 0
30428 0 0 1 0 0 0
30479 0 0 1 0 0 1
30485 0 0 1 2 1 0
30493 2 0 1 2 1 0
30519 0 0 1 0 1 0
30532 0 0 1 0 1 0
30541 0 0 1 0 1 0
30567 1 0 0 0 0 0
30569 2 0 1 1 1 0
30578 0 0 1 0 0 1
30579 1 0 1 0 0 0
30596 1 0 1 1 1 0
30597 1 0 1 1 0 0
30618 0 0 1 0 0 0
30622 1 0 1 1 1 0
30627 1 0 1 2 0 0
30648 2 0 0 0 1 0
30655 0 0 1 0 0 1
30658 0 0 1 0 1 0
30667 0 0 1 0 1 0
30678 1 0 1 0 0 0
30701 0 0 1 0 0 0
30703 2 0 1 1 0 0
30710 0 0 1 2 0 0
30713 1 0 0 1 1 1
30716 0 0 0 0 1 0
30721 0 0 0 0 0 1
30723 0 0 1 0 1 0
30724 2 0 1 2 1 0
30733 1 0 0 1 0 0
30734 0 0 1 0 0 0
30736 2 0 0 1 1 1
30737 0 0 1 0 0 0
30740 0 0 1 0 1 0
30742 2 0 1 0 1 0
30743 0 0 1 0 1 0
30745 2 0 0 0 1 0
30754 1 0 1 0 1 0
30758 1 0 0 0 1 0
30764 0 0 1 0 0 1
30765 2 0 0 0 0 0
30769 2 0 0 1 1 0
30772 0 0 1 0 1 0
30774 0 0 0 0 1 0
30784 2 0 1 0 0 0
30786 1 0 1 0 1 0
30787 0 0 0 0 1 0
30789 1 0 1 0 1 0
30800 0 0 1 0 0 0
30801 1 0 1 0 1 0
30803 1 0 1 0 1 0
30806 1 0 1 0 1 0
30817 0 0 1 2 0 0
30819 2 0 1 0 1 1
30822 0 0 1 0 1 0
30823 0 0 1 2 1 0
30834 0 0 0 0 0 0
30836 0 0 1 0 1 0
30837 1 0 1 0 1 0
30840 0 0 1 0 1 0
30841 1 0 1 0 0 0
30844 0 0 1 0 1 0
30845 0 0 1 0 0 0
30847 1 0 1 0 0 0
30848 0 0 1 0 1 0
30850 0 0 1 0 1 0
30856 1 0 0 0 1 0
30858 0 0 1 0 0 0
30860 0 0 0 0 1 0
30862 1 0 1 1 1 0
30864 0 0 0 2 0 0
30867 0 0 1 0 1 0
30869 0 0 1 0 1 0
30887 0 0 1 0 1 0
30900 1 0 0 1 1 0
30913 2 0 0 0 1 0
30914 1 0 0 0 0 0
30922 2 0 0 2 1 0
30923 0 0 1 2 1 0
30927 1 0 1 0 0 1
30929 0 0 1 2 1 0
30933 0 0 1 2 1 0
30940 0 0 1 0 1 0
30943 1 0 1 2 1 0
30945 0 0 0 2 0 0
30951 1 0 0 0 0 0
30964 0 0 0 2 1 0
30969 0 0 1 0 1 0
30979 2 0 0 0 1 0
30980 1 0 0 0 0 0
30982 1 0 0 1 1 0
30990 1 0 1 1 1 0
30991 1 0 1 0 1 1
30999 0 0 1 0 1 0
31056 1 1 0 2 1 0
31068 1 1 0 1 0 0
31108 2 1 0 2 1 0
31168 1 1 1 0 0 0
31191 0 1 1 0 0 0
31229 0 1 1 0 0 1
31263 0 1 0 0 1 0
31281 1 1 1 0 0 0
31340 1 1 1 0 1 0
31375 0 1 0 0 1 0
31401 0 1 1 0 0 1
31480 1 1 1 1 1 0
31501 1 1 0 2 1 0
31514 0 1 0 2 0 0
31518 1 1 0 2 1 0
31532 0 0 1 2 1 0
31543 2 1 1 1 1 0
31588 0 1 0 0 1 0
31590 0 0 1 0 1 0
31591 2 1 0 1 1 0
31595 0 1 0 0 1 0
31596 1 1 0 0 0 0
31598 1 1 0 0 1 0
31599 0 1 0 0 0 0
31605 0 1 1 0 0 0
31612 2 1 0 0 1 0
31615 2 1 0 0 0 0
31628 1 1 0 0 1 0
31640 2 1 0 1 1 0

测试数据：testdata.txt

10009 1 0 0 1 0 1
10025 0 0 1 2 0 0
20035 0 0 1 0 0 1
20053 1 0 0 0 0 0
30627 1 0 1 2 0 0
30648 2 0 0 0 1 0

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标签： log c++ ti c 实现 gis log c++ ti c 实现 gis

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