有人可以给我提示在lucene中应用伪反馈。我在谷歌上找不到太多帮助。我正在使用相似性课程。 lucene中是否有任何类可以扩展以实现反馈? 感谢。
答案 0 :(得分:2)
假设您指的是this relevance feedback method,一旦您获得了原始查询的TopDocs
,请重复多少次(假设我们需要前25个文档中的前25个术语)原始查询)您想要的记录,并致电IndexReader.getTermVectors(int)
,它将获取您需要的信息。遍历每个。将术语频率存储在哈希映射中将是我立即发生的实现。
类似的东西:
//Get the original results
TopDocs docs = indexsearcher.search(query,25);
HashMap<String,ScorePair> map = new HashMap<String,ScorePair>();
for (int i = 0; i < docs.scoreDocs.length; i++) {
//Iterate fields for each result
FieldsEnum fields = indexreader.getTermVectors(docs.scoreDocs[i].doc).iterator();
String fieldname;
while (fieldname = fields.next()) {
//For each field, iterate it's terms
TermsEnum terms = fields.terms().iterator();
while (terms.next()) {
//and store it
putTermInMap(fieldname, terms.term(), terms.docFreq(), map);
}
}
}
List<ScorePair> byScore = new ArrayList<ScorePair>(map.values());
Collections.sort(byScore);
BooleanQuery bq = new BooleanQuery();
//Perhaps we want to give the original query a bit of a boost
query.setBoost(5);
bq.add(query,BooleanClause.Occur.SHOULD);
for (int i = 0; i < 25; i++) {
//Add all our found terms to the final query
ScorePair pair = byScore.get(i);
bq.add(new TermQuery(new Term(pair.field,pair.term)),BooleanClause.Occur.SHOULD);
}
}
//Say, we want to score based on tf/idf
void putTermInMap(String field, String term, int freq, Map<String,ScorePair> map) {
String key = field + ":" + term;
if (map.containsKey(key))
map.get(key).increment();
else
map.put(key,new ScorePair(freq,field,term));
}
private class ScorePair implements Comparable{
int count = 0;
double idf;
String field;
String term;
ScorePair(int docfreq, String field, String term) {
count++;
//Standard Lucene idf calculation. This is calculated once per field:term
idf = (1 + Math.log(indexreader.numDocs()/((double)docfreq + 1))) ^ 2;
this.field = field;
this.term = term;
}
void increment() { count++; }
double score() {
return Math.sqrt(count) * idf;
}
//Standard Lucene TF/IDF calculation, if I'm not mistaken about it.
int compareTo(ScorePair pair) {
if (this.score() < pair.score()) return -1;
else return 1;
}
}
(我没有声称这是功能代码,在它的当前状态。)