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GSP Algorithm: Sequence Mining.

时间:2014-12-19 22:01:13      阅读:241      评论:0      收藏:0      [点我收藏+]

标签:data mining   sequence mining   gsp algorithm   

参考论文: 

Srikant R, Agrawal R.Mining sequential patterns: Generalizations and performance improvements[M].Springer Berlin Heidelberg, 1996.

1. 参考论文描述:

Srikant R, Agrawal R. 提出的算法,在原有aporiori的基础上,引入了3个新的概念来定义频繁模式子序列:

1) 加入时间约束,使得原有的aporiori关注的连续变成了只要满足min_gapmax_gap约束的序列,都算是连续的。

2) 加入time_window_size。使得transaction有新的定义。只要在window_size内的item,都可以认为是在同一个itemset

3) 加入分类标准。

 

本文的GSP算法实现步骤如下:

1) 候选集生成

a) 合并阶段: 2subsequence s1s2能够合并的标准是,去掉s1中的第一个item,去掉s2中的最后一个item,若此时s1s2相同,则可以对s1s2进行合并,即将s2中的最后一个item加入到s1中,其中最后一个item是否为合并在原来s1的最后一个itemset,还是自成一个新的itemset,取决于s2的最后一个item是否原来就是一个单独的itemset

b) 剪枝阶段:不频繁子序列的超集也不频繁。

2) 候选集计算

a) 减少候选集检验是否满足的个数:在这,使用到了哈希树,即将序列中的第pitem映射在第p层上。然后按照候选集进行检查。

b) 检查语料库中是否有满足要求的特定子序列:此处用到了倒排记录表,而倒排记录表的ID可以是item,也可以是transaction-time。简要步骤就是,首先找到第一个itemset,记录时间,然后对于第二个itemset,在第一个itemset的时间上往下继续查找,当找到时,判断2transaction_time的时间是否满足min_gapmax_gap,如果不满足,则在第二个itemset的基础上先往回找最近第一个itemset时间,如果此时仍不满足,则在第二个itemset的时间的基础上往后查找,如此forward backward的步骤。

3) 分类。

 

2. 算法实现细节:

大致流程:

读入数据à 建立倒排索引表(item to time)à 产生1-frequency itemsetà 产生2-frequency itemsetà …

 

源文件Ubuntu下编译: c++ gsp.cpp–o gsp

输入参数:    ./gsp –i data.txt–t minSupport -min min_gap –max max_gap

例如:        ./gsp –i data.txt–t 0.9 -min 2 -max 4

 

(1).  数据读入:

   读入的数据类型为:

         TransactionID  itemNum item1 item2 item3…

         TransactionID  itemNum item1 item2 item3…

   其中,其中itemNum为一个itemset的item个数,比如读入:

         1 2 3 3

         1 1 2

         1 2 3 4

         2 2 1 2

   则此时的sequence分别为 <(3 3) (2) (3 4)> , <(1 2)>,此时根据TransactionID来判断是否在一个sequence中。

 

(2).  采用STL的vector和map来实现,其中主要的数据结构如下:

// An element in thedata sequence.

struct Itemset

{

    std::vector<elemType> item;     // store the single itemset‘s items

    int timestamp;                // record current itemsettimestamp

    Itemset(){

        timestamp = 0;

    }

};

 

// data sequence.

struct Sequence{

    int tid;                        // transation id

    int num;                      // the number of itemset.

    std::vector<Itemset> itemset;     // store the itemsets.

    Sequence() {

        tid = 0;

        num = 0;

    }

};

 

(3) 建立倒排索引(inverted index)

函数:voidCreateInvertedIndex(int seqNum)

实现细节:vector<vector<map<elemType,int>>> elem2time, 用此来存储每个sequence中每个item到timestamp的映射。在计算支持度的时候将用到。

 

(4) 产生1-frequency itemset.

函数: std::map<elemType,int> GenerateOneSequence(int seqNum)

实现细节: 对给定数据扫描,用map来映射每个item,然后对然后累加每个item在某个sequence中出现的次数,计算是否满足min_support.

 

(5) 产生2-frequncy itemset.

函数:vector<Sequence>GenerateTwoSequence(vector<int> oneSeq)

实现细节: 对于1-frequency itemset, 分别由此得到<(1 2)> 、<(1) (2)>如此的itemset, 然后对得到的所有sequence进行剪枝,剪枝策略是通过计算他们在原来数据中的支持度。

 

(6) 产生3-more frequency itemset.

函数:vector<Sequence> JoinPhase(vector<Sequence> tmpCandidateSet)

实现细节:按照论文的说法,去掉s1中的第一个item和去掉s2中的最后一个item, 合并成新的sequence. 其中还包含判断加入s2的最后一个item是否自成一个itemset,或者合并在已有的itemset中。

 

函数:vector<Sequence>PrunePhase(std::vector<Sequence> duplicateSet)

实现细节:调用CountSupport函数,进行计算支持度,如果支持度大于阈值,则加入candidateSet。

 

函数:intCountSupport(Sequence curSequence)

实现细节:对每个sequence,根据已经建好的倒排记录表(数据集中的items映射到时间),首先找到每个itemset的时间,进行存储在vector中,然后进行dfs查询,判断是否满足min_gap和max_gap,如果满足则是一个满足的k-frequency sequence。

 

函数:boolDfs(vector<vector<int>>storeTime, int len, int cur, int id)

实现细节: 对已经得到的某个sequence每个itemset的对应时间,递归查找是否满足要求min_gap和max_gap,是的话使当前查询的sequence的支持度加1.

 

(7) 输出

函数:voidOutput(std::vector<Sequence> candidateSet, int id)

实现细节:输出满足的frequent sequence。


#include "structure.h"

using namespace std;

// data literals.
vector<Sequence> literals;

// inverted index
vector<vector<map<elemType, int>>> elem2time;

vector<Sequence> infrequencySeq;

// argv[0]='data.txt', argv[1] = min_support
// argv[2]=min_gap, argv[3] = max_gap
int main(/*int argc, char *argv[]*/)
{
    freopen("100.txt", "r", stdin);
    // minimum support
    minSupport = 0.9;
    // the number of sequence.
    seqNum = 0;
    // continues min_gap
    min_gap = 2;
    // continuous max_gap
    max_gap = 4;
    //Sequence seq;
    ReadData(seqNum);
    //count support
    min_support = seqNum * minSupport;
    // create inverted index
    CreateInvertedIndex(seqNum);
    // record start time.
    int tim1 = time(0);
    // Implementation of Gsp algorithm;
    Gsp(seqNum);
    // record the end time.
    int tim2 = time(0);
    OutputParameter(min_gap, max_gap, minSupport, seqNum, tim1, tim2);
    return 0;
}

/*******************************************************************************
*   @Reference:  Mining Sequence Patterns: Generalizations
*           and Performance Improvements  Srikant.R, Agrawal.R,
            VLDB 96;
*   @description: The main strategy by GSP algorithm here is:
*           1. Candidate Generation:
*                   1) Join phase
*                   2) Prune phase.
*           2. Counting Candidates:
*                   1) Reducing the number of candidates that need
*                     to be checked.
*                           a) Adding candidate sequences to the hash-tree
*                           b) Finding the candidates contained in the
*                               data-sequence
*                   2) Checking whether a data-sequence contains a specific
*                      sequence
*                           Contains test.
*                               a) Forward phase
*                               b) Backward phase
*                           Here we simply using DFS to implement it.
*           3. Taxonomies.
*   @CreateTime: 2014-12-7 10:00
*   @LastModifiedTime: 2014-12-9 1:25
*******************************************************************************/
void Gsp(int seqNum)
{
    freopen("output.txt", "w", stdout);
    map<elemType, int> cand;
    // Scan the database to find 1-sequence.
    //-----------------------------first scan-----------------------------------
    cand = GenerateOneSequence(seqNum);
    //--------------------------first scan end----------------------------------

    vector<elemType> oneSeq;
    // Get 1-frequency sequence
    for(map<elemType, int>::iterator iter1 = cand.begin();
        iter1 != cand.end(); iter1 ++)
        oneSeq.push_back(iter1->first);

    // Get 2-frequency sequence;
    vector<Sequence> candidateSet;
    candidateSet = GenerateTwoSequence(oneSeq);
    Output(candidateSet, 2);

    for(int k = 3;; k ++)
    {
        candidateSet = JoinPhase(candidateSet);
        if(candidateSet.size() == 0) {
            cout << "No further " << k-1 << "-frequency patterns." << endl;
            break;
        }
        else{
            //candidateSet = PrunePhase(candidateSet);
            Output(candidateSet, k);
        }
    }
}

// First scan the database to get frequency is one itemset.
map<elemType, int> GenerateOneSequence(int seqNum)
{
    map<elemType, int> dup;
    map<elemType, int> cand;
    // count every item in the sequence.
    for(int k = 0; k < literals.size(); k ++)
    {
        map<elemType, int>  item;
        for(int i = 0; i < literals[k].itemset.size(); i ++)
        {
            for(int j = 0; j < literals[k].itemset[i].item.size(); j ++)
            {
                item[literals[k].itemset[i].item[j]] ++;
            }
        }

        map<elemType, int>::iterator iter;
        for(iter = item.begin(); iter != item.end(); iter ++)
        {
            if(iter->second > 0)
                dup[iter->first] ++;
        }
    }
    // add up to become support.
    map<elemType, int>::iterator iter1, iter2;
    for (iter1 = dup.begin(); iter1 != dup.end(); iter1 ++)
        if(iter1->second >= min_support)
        {
            cand[iter1->first] = iter1->second;
        }
    cout << "min_support = " << minSupport * seqNum << endl;
    cout << "1-frequent candidate set" << endl;
    for(iter1 = cand.begin(); iter1 != cand.end(); iter1 ++)
        printf("<%d>:  %d\n", iter1->first, iter1->second);
    return cand;
}

// Generate two frequency sequences.
vector<Sequence> GenerateTwoSequence(vector<int> oneSeq)
{
    vector<Sequence> candidateSet;
    // gather all the two-frequency sequence.
    for(int i = 0; i < oneSeq.size(); i ++)
        for(int j = 0; j < oneSeq.size(); j ++)
        {
            // <aa> <ab>
            Sequence duplicate;
            Itemset tmpItem1, tmpItem2;
            tmpItem1.item.push_back(oneSeq[i]);
            duplicate.itemset.push_back(tmpItem1);
            tmpItem2.item.push_back(oneSeq[j]);
            duplicate.itemset.push_back(tmpItem2);
            int cnt = CountSupport(duplicate);
            if(cnt >= min_support)
                candidateSet.push_back(duplicate);
            else infrequencySeq.push_back(duplicate);
        }
    // Generate <(ab)>. <(ac)>
    for(int i = 0; i < oneSeq.size() - 1; i ++)
        for(int j = i+1; j < oneSeq.size(); j ++)
        {
            Sequence duplicate;
            Itemset tmpItem;
            tmpItem.item.push_back(oneSeq[i]);
            tmpItem.item.push_back(oneSeq[j]);
            duplicate.itemset.push_back(tmpItem);
            int cnt = CountSupport(duplicate);
            duplicate.num = cnt;
            if(cnt >= min_support)
                candidateSet.push_back(duplicate);
            else infrequencySeq.push_back(duplicate);
        }
    //return PrunePhase(candidateSet);
    return candidateSet;
}

// generate k >= 3-frequency sequences
vector<Sequence> JoinPhase(vector<Sequence> tmpCandidateSet)
{
    vector<vector<elemType>> seq;
    vector<Sequence> candidateSet;

    for(int k = 0; k < tmpCandidateSet.size(); k ++)
    {
        vector<elemType> items;
        for(int i = 0; i < tmpCandidateSet[k].itemset.size(); i ++)
            for(int j = 0; j < tmpCandidateSet[k].itemset[i].item.size(); j ++)
                items.push_back(tmpCandidateSet[k].itemset[i].item[j]);
        seq.push_back(items);
    }

    for(int i = 0; i < seq.size(); i ++)
        for(int j = 0; j < seq.size(); j ++)
        {
            vector<elemType> s1;
            vector<elemType> s2;
            for(int p = 1; p < seq[i].size(); p ++)
                s1.push_back(seq[i][p]);
            for(int q = 0; q < seq[j].size(); q ++)
                s2.push_back(seq[j][q]);
            bool flag = false;
            for(int k = 0; k < s1.size(); k ++)
                if(s1[k] != s2[k])
                    flag = true;
            if(!flag){
                Sequence tmpSeq;
                for(int p = 0; p < tmpCandidateSet[i].itemset.size(); p ++)
                {
                    Itemset tmpItems;
                    for(int q = 0; q < tmpCandidateSet[i].itemset[p].item.size(); q ++)
                        tmpItems.item.push_back(tmpCandidateSet[i].itemset[p].item[q]);
                    tmpSeq.itemset.push_back(tmpItems);
                }
                int tp = tmpCandidateSet[j].itemset.size()-1;
                int tq = tmpCandidateSet[j].itemset[tp].item.size()-1;
                int sp = tmpCandidateSet[i].itemset.size()-1;
                int sq = tmpCandidateSet[i].itemset[sp].item.size()-1;

                if(tmpCandidateSet[j].itemset[tp].item.size() > 1)
                {
                    tmpSeq.itemset[sp].item.push_back(s2[s2.size()-1]);
                }
                else{
                    Itemset tmpItems;
                    tmpItems.item.push_back(s2[s2.size()-1]);
                    tmpSeq.itemset.push_back(tmpItems);
                }
                int cnt = CountSupport(tmpSeq);
                tmpSeq.num = cnt;
                if(cnt >= min_support)
                    candidateSet.push_back(tmpSeq);
                else infrequencySeq.push_back(tmpSeq);
            }
        }

    return candidateSet;
}

// Prune Phase;
// Strategy here, we will use forward phase, and backward phase.
vector<Sequence> PrunePhase(std::vector<Sequence> duplicateSet)
{
    vector<Sequence> candidateSet;

    for(int k = 0; k < duplicateSet.size(); k ++)
    {
        Sequence duplicate;
        for(int i = 0; i < duplicateSet[k].itemset.size();  i ++)
        {
            Itemset items;
            for(int j = 0; j < duplicateSet[k].itemset[i].item.size(); j ++)
                items.item.push_back(duplicateSet[k].itemset[i].item[j]);
            duplicate.itemset.push_back(items);
        }
        int cnt = CountSupport(duplicate);
        if(cnt >= min_support)
        {
            duplicate.num = cnt;
            candidateSet.push_back(duplicate);
        }
    }
    return candidateSet;
}

// Count candidate set's support
int CountSupport(Sequence curSequence)
{
    int cnt = 0;
    //cout << "CountSupport" << endl;
    vector<vector<int>> seq;
    for(int i = 0; i < curSequence.itemset.size(); i ++)
    {
        vector<int> subSeq;
        for(int j = 0; j < curSequence.itemset[i].item.size(); j ++)
            subSeq.push_back(curSequence.itemset[i].item[j]);

        seq.push_back(subSeq);
    }

    for(int k = 0; k < literals.size(); k ++)
    {
        vector<vector<int>> storeTime;

        for(int i = 0; i < seq.size(); i ++)
        {
            int matchNum;
            int t = -1;
            vector<int> localTime;
            for(int p = 0; p < elem2time[k].size(); p ++)
            {
                matchNum = 0;
                for(int j = 0; j < seq[i].size(); j ++)
                    if(elem2time[k][p][seq[i][j]] > 0)
                    {
                        t = elem2time[k][p][seq[i][j]];
                        matchNum ++;
                    }
                if(matchNum == seq[i].size())
                    localTime.push_back(t);
            }
            if(localTime.size() > 0)
                storeTime.push_back(localTime);
        }

        if(storeTime.size() != seq.size())
            continue;

        for(int i = 0; i < storeTime[0].size(); i ++)
        {
            bool flag = false;
            flag = Dfs(storeTime, storeTime.size(), storeTime[0][i], 1);
            if (flag)
            {
                cnt ++;
                break;
            }
        }
    }
    return cnt;
}

// Dfs to find a sequence contain the min_gap and max_gap conditions
bool Dfs(vector<vector<int>>storeTime, int len, int cur, int id)
{

    if(id == len)
        return true;

    for(int i = 0; i < storeTime[id].size(); i ++){
        if(storeTime[id][i] - cur >= min_gap &&
                storeTime[id][i] - cur <= max_gap)
        {
            return Dfs(storeTime, storeTime.size(), storeTime[id][i], id + 1);
        }
    }
    return false;
}

// Output k-th candidateSet.
void Output(std::vector<Sequence> candidateSet, int id)
{
    int len = candidateSet.size();
    if(len != 0)
        cout << id << "-frequency candidate set" << endl;
    for(int k = 0; k < len; k ++)
    {
        cout << "<";
        for(int i = 0; i < candidateSet[k].itemset.size(); i ++)
        {
            cout << "(";
            for(int j = 0; j < candidateSet[k].itemset[i].item.size(); j ++)
            {
                cout << candidateSet[k].itemset[i].item[j] << " ";

            }
            cout << ")";
        }
        cout << ">:  ";
        cout << candidateSet[k].num << endl;
    }
    cout << endl;
}

// Read data from file
void ReadData(/*Sequence *seq, */int &seqNum)
{
    int id, num, tmp, t;
    int cnt = 1;
    map<int, int> transactionID;

    while(cin >> id)
    {
        Sequence seq;
        if(transactionID[id] == 0)
        {
            Itemset items;
            cin >> num;
            t = 0;
            for(int i = 0; i < num; i ++)
            {
                cin >> tmp;
                items.item.push_back(tmp);
            }
            items.timestamp = t;
            t ++;
            seq.itemset.push_back(items);
            seq.tid = id;
            literals.push_back(seq);
            transactionID[id] = cnt ++;
        }
        else{
            Itemset items;
            cin >> num;
            for(int i = 0; i < num; i ++)
            {
                cin >> tmp;
                items.item.push_back(tmp);
            }
            items.timestamp = t;
            t ++;
            literals[transactionID[id]-1].itemset.push_back(items);
        }
    }
    seqNum = literals.size();
}

inline int min(int a, int b)
{
    return a > b ? b : a;
}

// Create Inverted Index
void CreateInvertedIndex(int seqNum)
{
    for(int k = 0; k < seqNum; k ++)
    {
        vector<map<elemType, int>> invertedIndex;
        for(int i = 0; i < literals[k].itemset.size(); i ++)
        {
            map<elemType, int> tmp;
            for(int j = 0; j < literals[k].itemset[i].item.size(); j ++)
                tmp[literals[k].itemset[i].item[j]] = i+1;
            invertedIndex.push_back(tmp);
        }
        elem2time.push_back(invertedIndex);
    }
}

// output parameters
void OutputParameter(int min_gap, int max_gap, double min_support, int seqNum,
                     int tim1, int tim2)
{
    cout << "---------------------------------" << endl;
    cout << "\nmin_gap = " << min_gap << endl;
    cout << "max_gap = " << max_gap << endl;
    cout << "min_support = " << min_support << endl;
    cout << "the number of sequence = " << seqNum << endl;
    // output time
    cout << "\nbegin time: " << tim1 << '\n';
    cout << "end time  : " << tim2 << '\n';
    cout << "The execution time is:\n" << tim2-tim1;
    cout << "\n\nEnd the program" << endl;
    cout << "---------------------------------" << endl;
}


GSP Algorithm: Sequence Mining.

标签:data mining   sequence mining   gsp algorithm   

原文地址:http://blog.csdn.net/zone_programming/article/details/42032309

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