term vectors

Leveraging term vectors
所谓term vector, 就是对于documents的某一field,如title,body这种文本类型的, 建立词频的多维向量空间.每一个词就是一维, 这维的值就是这个词在这个field中的频率.

如果你要使用term vectors, 就要在indexing的时候对该field打开term vectors的选项:

Field options for term vectors
TermVector.YES – record the unique terms that occurred, and their counts, in each document, but do not store any positions or offsets information.
TermVector.WITH_POSITIONS – record the unique terms and their counts, and also the positions of each occurrence of every term, but no offsets.
TermVector.WITH_OFFSETS – record the unique terms and their counts, with the offsets (start & end character position) of each occurrence of every term, but no positions.
TermVector.WITH_POSITIONS_OFFSETS – store unique terms and their counts, along with positions and offsets.
TermVector.NO – do not store any term vector information.
If Index.NO is specified for a field, then you must also specify TermVector.NO.

这样在index完后, 给定这个document id和field名称, 我们就可以从IndexReader读出这个term vector(前提是你在indexing时创建了terms vector):
TermFreqVector termFreqVector = reader.getTermFreqVector(id, "subject");
你可以遍历这个TermFreqVector去取出每个词和词频, 如果你在index时选择存下offsets和positions信息的话, 你在这边也可以取到.

有了这个term vector我们可以做一些有趣的应用:
1) Books like this
比较两本书是否相似,把书抽象成一个document文件, 具有author, subject fields. 那么现在就通过这两个field来比较两本书的相似度.
author这个field是multiple fields, 就是说可以有多个author, 那么第一步就是比author是否相同,
String[] authors = doc.getValues("author");
BooleanQuery authorQuery = new BooleanQuery(); // #3
for (int i = 0; i < authors.length; i++) { // #3
    String author = authors[i]; // #3
    authorQuery.add(new TermQuery(new Term("author", author)), BooleanClause.Occur.SHOULD); // #3
}
authorQuery.setBoost(2.0f);
最后还可以把这个查询的boost值设高, 表示这个条件很重要, 权重较高, 如果作者相同, 那么就很相似了.
第二步就用到term vector了, 这里用的很简单, 单纯的看subject field的term vector中的term是否相同,
TermFreqVector vector = // #4
reader.getTermFreqVector(id, "subject"); // #4
BooleanQuery subjectQuery = new BooleanQuery(); // #4
for (int j = 0; j < vector.size(); j++) { // #4
    TermQuery tq = new TermQuery(new Term("subject", vector.getTerms()[j]));
    subjectQuery.add(tq, BooleanClause.Occur.SHOULD); // #4
}

2) What category?
这个比上个例子高级一点, 怎么分类了,还是对于document的subject, 我们有了term vector.
所以对于两个document, 我们可以比较这两个文章的term vector在向量空间中的夹角, 夹角越小说明这个两个document越相似.
那么既然是分类就有个训练的过程, 我们必须建立每个类的term vector作为个标准, 来给其它document比较.
这里用map来实现这个term vector, (term, frequency), 用n个这样的map来表示n维. 我们就要为每个category来生成一个term vector, category和term vector也可以用一个map来连接.创建这个category的term vector, 这样做:
遍历这个类中的每个document, 取document的term vector, 把它加到category的term vector上.
private void addTermFreqToMap(Map vectorMap, TermFreqVector termFreqVector) {
    String[] terms = termFreqVector.getTerms();
    int[] freqs = termFreqVector.getTermFrequencies();
    for (int i = 0; i < terms.length; i++) {
        String term = terms[i];
        if (vectorMap.containsKey(term)) {
            Integer value = (Integer) vectorMap.get(term);
            vectorMap.put(term, new Integer(value.intValue() + freqs[i]));
        } else {
            vectorMap.put(term, new Integer(freqs[i]));
        }
   }
}
首先从document的term vector中取出term和frequency的list, 然后从category的term vector中取每一个term, 把document的term frequency加上去.OK了

有了这个每个类的category, 我们就要开始计算document和这个类的向量夹角了
cos = A*B/|A||B|
A*B就是点积, 就是两个向量每一维相乘, 然后全加起来.
这里为了简便计算, 假设document中term frequency只有两种情况, 0或1.就表示出现或不出现
private double computeAngle(String[] words, String category) {
    // assume words are unique and only occur once
    Map vectorMap = (Map) categoryMap.get(category);
    int dotProduct = 0;
    int sumOfSquares = 0;
    for (int i = 0; i < words.length; i++) {
        String word = words[i];
        int categoryWordFreq = 0;
        if (vectorMap.containsKey(word)) {
            categoryWordFreq = ((Integer) vectorMap.get(word)).intValue();
        }
        dotProduct += categoryWordFreq; // optimized because we assume frequency in words is 1
        sumOfSquares += categoryWordFreq * categoryWordFreq;
    }
    double denominator;
    if (sumOfSquares == words.length) {
        // avoid precision issues for special case
        denominator = sumOfSquares; // sqrt x * sqrt x = x
    } else {
        denominator = Math.sqrt(sumOfSquares) *
        Math.sqrt(words.length);
    }
    double ratio = dotProduct / denominator;
    return Math.acos(ratio);
}
这个函数就是实现了上面那个公式还是比较简单的.