Apriopri算法的简单实现代码,C/C++代码库,德仔网

代码语言
代码分类
【C/C++】 Apriopri算法的简单实现
作者:欣雨 / 发布于2016/2/17/ 718
剪枝算法，在Apriori算法中,寻找最大项目集的基本思想是:算法需要对数据集进行多步处理.第一步,简单统计所有含一个元素项目集出现的频率,并找出那些不小于最小支持度的项目集,即一维最大项目集.从第二步开始循环处理直到再没有最大项目集生成.循环过程是:第k步中,根据第k-1步生成的（k-1）维最大项目集产生k维候选项目集,然后对数据库进行搜索,得到候选项目集的项集支持度,与最小支持度比较,从而找到k维最大项目集.
#include<iostream>
#include<string>
#include <vector>
#include <map>
#include <algorithm>
using namespace std;
 
class Apriori
{
public:
    Apriori(size_t is = 0, unsigned int mv = 0)
    {
        item_size = is;
        min_value = mv;
    }
    //~Apriori() {};
    void getItem();
    map< vector<string>, unsigned int> find_freitem();//求事务的频繁项
    //连接连个k-1级频繁项，得到第k级频繁项
    map< vector<string>, unsigned int > apri_gen(unsigned int K, map< vector<string>, unsigned int > K_item);
    //展示频繁项集
    void showAprioriItem(unsigned int K, map< vector<string>, unsigned int > showmap);
private:
    map< int, vector<string> > item;//存储所有最开始的事务及其项
    map< vector<string>, unsigned int > K_item;//存储频繁项集
    size_t item_size;//事务数目
    unsigned  int min_value;//最小阈值
};
 
void Apriori::getItem()//用户输入最初的事务集
{
    int ci = item_size;
    for (int i = 0; i<ci; i++)
    {
        string str;
        vector<string> temp;
        cout << "请输入第　" << i + 1 << "个事务的项集(123 end)：";
        while (cin >> str && str != "123")
        {
            temp.push_back(str);
        }
        sort(temp.begin(), temp.end());
        pair< map<int, vector<string> >::iterator, bool> ret = item.insert(make_pair(i + 1, temp));
        if (!ret.second)
        {
            --i;
            cout << "你输入的元素已存在！请重新输入！" << endl;
        }
    }
    cout << "-------------运行结果如下：--------------" << endl;
}
 
map< vector<string>, unsigned int> Apriori::find_freitem()
{
    unsigned int i = 1;
    bool isEmpty = false;
    map< int, vector<string> >::iterator mit;
    for (mit = item.begin(); mit != item.end(); mit++)
    {
        vector<string> vec = mit->second;
        if (vec.size() != 0)
            break;
    }
    if (mit == item.end())//事务集为空
    {
        isEmpty = true;
        cout << "事务集为空！程序无法进行..." << endl;
        map< vector<string>, unsigned int> empty;
        return empty;
    }
    while (1)
    {
        map< vector<string>, unsigned int > K_itemTemp = K_item;
 
        K_item = apri_gen(i++, K_item);
 
        if (K_itemTemp == K_item)
        {
            i = UINT_MAX;
            break;
        }
        //判断是否需要进行下一次的寻找
        map< vector<string>, unsigned int > pre_K_item = K_item;
        size_t Kitemsize = K_item.size();
        //存储应该删除的第K级频繁项集，不能和其他K级频繁项集构成第K+1级项集的集合
        if (Kitemsize != 1 && i != 1)
        {
            vector< map< vector<string>, unsigned int >::iterator > eraseVecMit;
            map< vector<string>, unsigned int >::iterator pre_K_item_it1 = pre_K_item.begin(), pre_K_item_it2;
            while (pre_K_item_it1 != pre_K_item.end())
            {
                map< vector<string>, unsigned int >::iterator mit = pre_K_item_it1;
                bool isExist = true;
                vector<string> vec1;
                vec1 = pre_K_item_it1->first;
                vector<string> vec11(vec1.begin(), vec1.end() - 1);
                while (mit != pre_K_item.end())
                {
                    vector<string> vec2;
                    vec2 = mit->first;
                    vector<string> vec22(vec2.begin(), vec2.end() - 1);
                    if (vec11 == vec22)
                        break;
                    ++mit;
                }
                if (mit == pre_K_item.end())
                    isExist = false;
                if (!isExist && pre_K_item_it1 != pre_K_item.end())
                    eraseVecMit.push_back(pre_K_item_it1);//该第K级频繁项应该删除
                ++pre_K_item_it1;
            }
            size_t eraseSetSize = eraseVecMit.size();
            if (eraseSetSize == Kitemsize)
                break;
            else
            {
                vector< map< vector<string>, unsigned int >::iterator >::iterator currentErs = eraseVecMit.begin();
                while (currentErs != eraseVecMit.end())//删除所有应该删除的第K级频繁项
                {
                    map< vector<string>, unsigned int >::iterator eraseMit = *currentErs;
                    K_item.erase(eraseMit);
                    ++currentErs;
                }
            }
        }
        else
        if (Kitemsize == 1)
            break;
    }
    cout << endl;
    showAprioriItem(i, K_item);
    return K_item;
}
 
map< vector<string>, unsigned int > Apriori::apri_gen(unsigned int K, map< vector<string>, unsigned int > K_item)
{
    if (1 == K)//求候选集C1
    {
        size_t c1 = item_size;
        map< int, vector<string> >::iterator mapit = item.begin();
        vector<string> vec;
        map<string, unsigned int> c1_itemtemp;
        while (mapit != item.end())//将原事务中所有的单项统计出来
        {
 
            vector<string> temp = mapit->second;
            vector<string>::iterator vecit = temp.begin();
            while (vecit != temp.end())
            {
                pair< map<string, unsigned int>::iterator, bool > ret = c1_itemtemp.insert(make_pair(*vecit++, 1));
                if (!ret.second)
                {
                    ++ret.first->second;
                }
            }
            ++mapit;
        }
        map<string, unsigned int>::iterator item_it = c1_itemtemp.begin();
        map< vector<string>, unsigned int > c1_item;
        while (item_it != c1_itemtemp.end())//构造第一级频繁项集
        {
            vector<string> temp;
            if (item_it->second >= min_value)
            {
                temp.push_back(item_it->first);
                c1_item.insert(make_pair(temp, item_it->second));
            }
            ++item_it;
        }
        return c1_item;
    }
    else
    {
        cout << endl;
        showAprioriItem(K - 1, K_item);
        map< vector<string>, unsigned int >::iterator ck_item_it1 = K_item.begin(), ck_item_it2;
        map< vector<string>, unsigned int > ck_item;
        while (ck_item_it1 != K_item.end())
        {
            ck_item_it2 = ck_item_it1;
            ++ck_item_it2;
            map< vector<string>, unsigned int >::iterator mit = ck_item_it2;
 
            //取当前第K级频繁项与其后面的第K级频繁项集联合，但要注意联合条件
            //联合条件：连个频繁项的前K-1项完全相同，只是第K项不同，然后两个联合生成第K+1级候选频繁项
            while (mit != K_item.end())
            {
                vector<string> vec, vec1, vec2;
                vec1 = ck_item_it1->first;
                vec2 = mit->first;
                vector<string>::iterator vit1, vit2;
 
                vit1 = vec1.begin();
                vit2 = vec2.begin();
                while (vit1 < vec1.end() && vit2 < vec2.end())
                {
                    string str1 = *vit1;
                    string str2 = *vit2;
                    ++vit1;
                    ++vit2;
                    if (K == 2 || str1 == str2)
                    {
                        if (vit1 != vec1.end() && vit2 != vec2.end())
                        {
                            vec.push_back(str1);
                        }
 
                    }
                    else
                        break;
                }
                if (vit1 == vec1.end() && vit2 == vec2.end())
                {
                    --vit1;
                    --vit2;
                    string str1 = *vit1;
                    string str2 = *vit2;
                    if (str1>str2)
                    {
                        vec.push_back(str2);
                        vec.push_back(str1);
                    }
                    else
                    {
                        vec.push_back(str1);
                        vec.push_back(str2);
                    }
                    map< int, vector<string> >::iterator base_item = item.begin();
                    unsigned int Acount = 0;
                    while (base_item != item.end())//统计该K+1级候选项在原事务集出现次数
                    {
                        unsigned int count = 0, mincount = UINT_MAX;
                        vector<string> vv = base_item->second;
                        vector<string>::iterator vecit, bvit;
                        for (vecit = vec.begin(); vecit < vec.end(); vecit++)
                        {
                            string t = *vecit;
                            count = 0;
                            for (bvit = vv.begin(); bvit < vv.end(); bvit++)
                            {
                                if (t == *bvit)
                                    count++;
                            }
                            mincount = (count < mincount ? count : mincount);
                        }
                        if (mincount >= 1 && mincount != UINT_MAX)
                            Acount += mincount;
                        ++base_item;
                    }
                    if (Acount >= min_value && Acount != 0)
                    {
                        sort(vec.begin(), vec.end());
                        //该第K+1级候选项为频繁项，插入频繁项集
                        pair< map< vector<string>, unsigned int >::iterator, bool> ret = ck_item.insert(make_pair(vec, Acount));
                        if (!ret.second)
                        {
                            ret.first->second += Acount;
                        }
                    }
                }
                ++mit;
            }
            ++ck_item_it1;
        }
        if (ck_item.empty())//该第K+1级频繁项集为空，说明调用结束，把上一级频繁项集返回
            return K_item;
        else
            return ck_item;
    }
}
void Apriori::showAprioriItem(unsigned int K, map< vector<string>, unsigned int > showmap)
{
    map< vector<string>, unsigned int >::iterator showit = showmap.begin();
    if (K != UINT_MAX)
        cout << endl << "第 " << K << " 级频繁项集为：" << endl;
    else
        cout << "最终的频繁项集为：" << endl;
    cout << "项  集" << "  \t  " << "频率" << endl;
    while (showit != showmap.end())
    {
        vector<string> vec = showit->first;
        vector<string>::iterator vecit = vec.begin();
        cout << "{ ";
        while (vecit != vec.end())
        {
            cout << *vecit << "  ";
            ++vecit;
        }
        cout << "}" << "\t";
        cout << showit->second << endl;
        ++showit;
    }
}
 
unsigned int parseNumber(const char * str)//对用户输入的数字进行判断和转换
{
    if (str == NULL)
        return 0;
    else
    {
        unsigned int num = 0;
        size_t len = strlen(str);
        for (size_t i = 0; i<len; i++)
        {
            num *= 10;
            if (str[i] >= '0' && str[i] <= '9')
                num += str[i] - '0';
            else
                return 0;
        }
        return num;
    }
}
 
void main()
{
    //Apriori a;
    unsigned int itemsize = 0;
    unsigned int min;
 
    do
    {
        cout << "请输入事务数：";
        char * str = new char;
        cin >> str;
        itemsize = parseNumber(str);
        if (itemsize == 0)
        {
            cout << "请输入大于0正整数！" << endl;
        }
    } while (itemsize == 0);
 
    do
    {
        cout << "请输入最小阈值：";
        char * str = new char;
        cin >> str;
        min = parseNumber(str);
        if (min == 0)
        {
            cout << "请输入大于0正整数！" << endl;
        }
    } while (min == 0);
 
    Apriori a(itemsize, min);
    a.getItem();
    map< vector<string>, unsigned int> AprioriMap = a.find_freitem();
    //a.showAprioriItem(UINT_MAX,AprioriMap);
    system("pause");
}
试试其它关键字
　Apriopri　
同语言下
可能有用的
欣雨贡献的其它代码(10)