Hi,
Merry Christmas!!
In case of Boolean query like 'sql AND server' .
I am using parser to get correct document containing both sql and server. Inside for loop in below code I get correct documented and to get frequency I need to sum frequency of 'sql' and 'server' individually with the help of termDocs.read().
As I am searching through millions of document. So, to calculate frequency it takes about 160 Second for 80k document.
Is there any way to get frequency of 'Boolean query' directly without manipulation. As it takes lots of time. In case of single term and phrase query, I got frequency for 80k document within 10 Seconds.
QueryParser parser = new QueryParser(Version.LUCENE_CURRENT, field, analyzer);
parser.setDefaultOperator(QueryParser.AND_OPERATOR);
Query query = parser.parse("sql AND server");
TopDocs docs = searcher.search(query, null, n);
TermDocs termDocs = reader.termDocs();
termDocs.seek(new Term(field, query.toString(field).split(" ")[0].toLowerCase()));
int[] ints = new int[docs.totalHits];
int[] ints1 = new int[docs.totalHits];
termDocs.read(ints, ints1);
List lsts = Arrays.asList(ArrayUtils.toObject(ints));
List lsts1 = Arrays.asList(ArrayUtils.toObject(ints1));
termDocs.seek(new Term(field, query.toString(field).split(" ")[1].toLowerCase()));
int[] inta = new int[docs.totalHits];
int[] inta1 = new int[docs.totalHits];
termDocs.read(inta, inta1);
List lsta = Arrays.asList(ArrayUtils.toObject(inta));
List lsta1 = Arrays.asList(ArrayUtils.toObject(inta1));
int totalFreq = 0;
int docId = -1;
String a = null;
int contId = 0;
String path=null;
for (int i = 0; i < docs2.scoreDocs.length; i++) {
docId = docs2.scoreDocs[i].doc;
path=reader.document(docId).get("path");
a = path.substring(path.lastIndexOf("\\") + 1, path.lastIndexOf("."));
try {
if(a.indexOf("e")>-1){
contId = Integer.parseInt(a.substring(0, a.length()-1));
}else{
contId = Integer.parseInt(a);
}
} catch (Exception e) {
// e.printStackTrace();
}
if ((lsts.indexOf(docId) > -1 && lsta.indexOf(docId) > -1) || (lsta.indexOf(docId) > -1) && lsts.indexOf(docId) > -1) {
totalFreq = (Integer) lsts1.get(lsts.indexOf(docId)) + (Integer) lsta1.get(lsta.indexOf(docId));
} else if (lsts.indexOf(docId) > -1) {
totalFreq = (Integer) lsts1.get(lsts.indexOf(docId));
} else if (lsta.indexOf(docId) > -1) {
totalFreq = (Integer) lsta1.get(lsta.indexOf(docId));
}
w.write(contId+"\t"+ID+"\t"+totalFreq+"\t"+reader.document(docId).get("path")+"\n");
}
Thanks & Regards,
Ranjit Kumar
Associate Software Engineer
[cid:[email protected]]
US: +1 408.540.0001
UK: +44 208.099.1660
India: +91 124.474.8100 | +91 124.410.1350
FAX: +1 408.516.9050
http://www.otssolutions.com
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