Hierarchical cluster算法介绍

　　突然想记录几个聚类算法，由于实力有限就先介绍一下层次聚类算法(Hierarchical cluster algorithm),这个聚类算法思想简单，但实现起来感觉复杂度挺大；以前看过《集体智慧编程》里介绍过，里面是用python实现的，由于python里面的列表和字典用起来方便，故实现该算法还行；这里我用c++重新写了一下，感觉代码蛮臃肿，可能是自己的c++没有学习好吧！！！对于容器的使用还不够熟练，这里贴出来的目的是希望哪位大牛看到了指导一二，这里感激不尽。废话不多说了，进入正题吧！

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Hierarchical cluster Algorithm的大致介绍

　　层次聚类算法有两种实现思想，一种是初始时将每个待聚类的数据样本视为一个cluster,采用合并的方式，每次合并两个"距离"最近的cluster，直到合并成一个cluster为止(当然可以在达到自己设定想得到的cluster个数时终止迭代)；另一种刚好与第一种相反，初始时将所有的数据样本视为一个cluster,采用分解的方式(这里没有实现就不说太多)。

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算法的步骤及相关问题

　　算法步骤:  (1)初始时，将每个数据样本视为一个cluster(选取一个度量两个cluster距离的方式)，

　　　　　　　(2)计算任意两个cluster之间的距离；每次选取距离最小的两个cluster，

　　　　　　　(3)合并(2)中选择的两个cluster,将合并产生的新cluster加入cluster set中，并删除被合并的两个cluster，

　　　　　　　(4)重复(2)(3)，知道cluster set中元素只剩下一个为止。

　　相关问题: (1)度量两个cluster之间的距离，应该选择哪种距离？？？《集体智慧编程》中选择的是Pearson,当然也可以直接选用欧氏距离

　　　　　　  (2)如何合并两个cluster,即新的cluster对应的属性值如何表示???这里是用被合并的两个cluster的平均值表示新的cluster

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/**
** Hierarchical cluster Algorithm
** step：(1)Firstly,regard each sample as a cluster, and
         (2)Each time merge two clusters if the distance between them is lowest.
         (3)then add the new cluster into cluster set, and delete two clusters merged from cluster set.
** method: (1)as to merging, here replace the old two clusters with their average;
           (2)measure the distance with the Pearson similarity.
** Time：2013/7/10 
**/
#include <iostream>
#include <map>
#include <vector>
#include <string>
#include <fstream> 
#include <cstring>
#include <sstream> 
#include <cmath>
#include <iterator>
using namespace std;
//cluster
typedef    struct bicluster{
    vector<double> attri;//attribute
    int  cid;//cluster id 
}Bicluster;
//a pair
typedef struct lowpair{
    int leftid;
    int rightid;
    double dist;
}Lpair;

/*****************************************************************
** convert string(char*) to double(or other type)
** here should be included <sstream> before using the stringstream
******************************************************************/
double str2double(char* str){
    stringstream ss;
    ss << str;
    double tmp;
    ss >> tmp;
    return tmp;    
}
/*****************************************************************
** split the string containing some special tokens
******************************************************************/
string split(string &str, vector<double>& dvec, const char* tok){
    char *pch = NULL;
    pch = strtok(const_cast<char*>(str.c_str()), tok);
    string stmp(pch);
    while( pch != NULL ){
        pch = strtok(NULL, tok);
        if( !pch )
            break;
        dvec.push_back(str2double(pch));
    }
    return stmp;
}
/******************************************************************
** read data from 'blogdata.txt'
** @is ------- a reference to ifstream object(input)
** @data ----- a map used to store the data (output)
******************************************************************/
bool readfile(ifstream &is, map<string, vector<double> >& mydata){
    if( is.fail() ){
        cerr << "can't open the file !!!" << endl;
        return false;
    }
    //ignore the first line of file
    string str;
    getline(is, str);
    
    //store the data read from file into mydata 
    while( !is.eof() ){
        vector<double> dtmp;
        string tmp;
        getline(is, str);
        tmp = split(str, dtmp, "\t");
        mydata.insert(pair<string,vector<double> >(tmp, dtmp));
    }
    return true;         
}
/*****************************************************************
** compute the distance between two clusters
** Note that Pearson value devotes to the similarity between 
    two clusters, that is, the greater the Pearson value, the 
    lower the distance between them.
*****************************************************************/ 
double distPearson(vector<double>& left, vector<double>& right){
    double sum1 = 0;
    double sum2 = 0;
    int len = left.size();
    for(int i=0; i<len; ++i){
        sum1 += left[i];
        sum2 += right[i];
    }
    
    /**
    ** maybe you will feel it's complex, 
    **  and here we could replace Pearson with Euclidean distance
    **/
    double sum1Sq = 0;
    double sum2Sq = 0;
    for(int j=0; j<len; ++j){
        sum1Sq += pow(left[j], 2);
        sum2Sq += pow(right[j], 2);
    }
    
    double pSum = 0, num, den;
    for(int k=0; k<len; ++k)
        pSum += left[k]*right[k];
    num = pSum - sum1*sum2 / len;
    den = sqrt((sum1Sq - pow(sum1,2)/len) * (sum1Sq - pow(sum2,2)/len));
    if( den == 0 )
        return 0;
    return 1.0 - num/den;
}
/*************************************************************
** Given two clusters, the distance between them 
    should be checked whether it exists before compute it.
**************************************************************/
bool isExist(vector<Lpair> &lp, int leftid, int rightid, double &d){
    vector<Lpair>::iterator it = lp.begin();
    for(; it!=lp.end(); ++it){
        if( (it->leftid==leftid) && (it->rightid==rightid) ){
            d = it->dist;//if the distance has been computed, assign its value to d
            return true;
        }        
    }
    d = 0;
    return false;
}
/*************************************************************
** Given a cluster's id, delete the cluster from cluster set
**************************************************************/
void Del(vector<Bicluster> &cvec, int clusterid){
    vector<Bicluster>::iterator it = cvec.begin();
    for(; it!=cvec.end(); ++it){
        if( it->cid == clusterid )
            break;
    }
    cvec.erase(it);
} 
/*************************************************************
** Hierarchical Cluster Algorithm
**************************************************************/
void HierarchicalCluster(map<string, vector<double> > &mydata){
    vector<Lpair> distances;//used to store the distance
     
    //firstly,regard each sample as a cluster
    vector<Bicluster> cvec;
    map<string, vector<double> >::iterator it = mydata.begin();
    int myid = 0;
    for(; it!= mydata.end(); ++it){
        Bicluster btmp;
        btmp.attri = it->second;
        btmp.cid = myid++;
        cvec.push_back(btmp);
    } 
    myid = -1;
    //search the pair
    while( cvec.size()>1 ){
        Lpair lowp;
        double closedis = distPearson(cvec[0].attri,cvec[1].attri);
        lowp.leftid = cvec[0].cid, lowp.rightid = cvec[1].cid;
        lowp.dist = closedis;
        
        int leftps = 0, rightps = 1;
        for(int ix=0; ix<cvec.size(); ++ix){
            for(int iy=ix+1; iy<cvec.size(); ++iy){
                double d;
                int lid = cvec[ix].cid, rid = cvec[iy].cid;
                if( !isExist(distances,lid,rid,d) ){
                    Lpair lptmp;
                    lptmp.dist = distPearson(cvec[ix].attri, cvec[iy].attri);
                    lptmp.leftid = lid;
                    lptmp.rightid= rid;
                    distances.push_back(lptmp);
                    d = lptmp.dist;
                  } 
                 if( d < lowp.dist ){
                     lowp.leftid = lid;
                     lowp.rightid = rid;
                     leftps = ix;
                     rightps = iy;
                     lowp.dist = d;
                 }
            }
        }
        //create a new cluster
        Bicluster ncluster;
        for(int i=0; i<cvec[0].attri.size(); ++i){
            double av;
            av = (cvec[leftps].attri[i] + cvec[rightps].attri[i]) / 2.0;
            ncluster.attri.push_back(av);
        }
        ncluster.cid = myid--;//assign negative to the new cluster's id
        cout << "leftid: " << lowp.leftid <<  ", rightid: " << lowp.rightid << endl;
        //delete the pair
        Del(cvec, lowp.leftid); 
        Del(cvec, lowp.rightid);
        cvec.push_back(ncluster);
    } 
} 
int main()
{
    ifstream is("blogdata.txt");
    if( is.fail() ){
        cerr << "error!!!" << endl;
        exit(-1);
    }
    map<string, vector<double> > mydata;
    if(readfile(is, mydata))
        HierarchicalCluster(mydata);
    return 0;
}

　　代码写的有点乱且复杂，最后显示的结果不是树状图（python很易实现），只是简单的显示了每次被合并的两个cluster的id.代码中用到的数据可以从http://kiwitobes.com/clusters/blog.txt下载得到。