public void seek(TermEnum terms) throws IOException {
original.seek(terms);
docFreq = terms.docFreq();
pointer = -1;
if (docFreq > postingMaps.length) { // grow postingsMap
PostingMap[] newMap = new PostingMap[docFreq];
System.arraycopy(postingMaps, 0, newMap, 0, postingMaps.length);
for (int i = postingMaps.length; i < docFreq; i++) {
newMap[i] = new PostingMap();
}
postingMaps = newMap;
}
out.reset();
int i = 0;
while (original.next()) {
PostingMap map = postingMaps[i++];
map.newDoc = oldToNew[original.doc()]; // remap the newDoc id
map.offset = out.getFilePointer(); // save pointer to buffer
final int tf = original.freq(); // buffer tf & positions
out.writeVInt(tf);
int prevPosition = 0;
for (int j = tf; j > 0; j--) { // delta encode positions
int p = original.nextPosition();
out.writeVInt(p - prevPosition);
prevPosition = p;
}
}
out.flush();
docFreq = i; // allow for deletions
Arrays.sort(postingMaps, 0, docFreq); // resort by mapped doc ids
// HeapSorter.sort(postingMaps,docFreq); // TODO MC - due to the lack of space
// NOTE: this might be substantially faster if RAMInputStream were public
// and supported a reset() operation.
in = tempDir.openInput(TEMP_FILE);
}
public static void main(String[] args) throws Exception {
// the IndexReader object is the main handle that will give you
// all the documents, terms and inverted index
IndexReader r = IndexReader.open(FSDirectory.open(new File("index")));
// You can figure out the number of documents using the maxDoc() function
System.out.println("The number of documents in this index is: " + r.maxDoc());
int i = 0;
// You can find out all the terms that have been indexed using the terms() function
TermEnum t = r.terms();
while (t.next()) {
// Since there are so many terms, let us try printing only term #100000-#100010
if (i > 100000) System.out.println("[" + i + "] " + t.term().text());
if (++i > 100010) break;
}
// You can create your own query terms by calling the Term constructor, with the field
// 'contents'
// In the following example, the query term is 'brute'
Term te = new Term("contents", "brute");
// You can also quickly find out the number of documents that have term t
System.out.println("Number of documents with the word 'brute' is: " + r.docFreq(te));
// You can use the inverted index to find out all the documents that contain the term 'brute'
// by using the termDocs function
TermDocs td = r.termDocs(te);
while (td.next()) {
System.out.println(
"Document number ["
+ td.doc()
+ "] contains the term 'brute' "
+ td.freq()
+ " time(s).");
}
// You can find the URL of the a specific document number using the document() function
// For example, the URL for document number 14191 is:
Document d = r.document(14191);
String url =
d.getFieldable("path")
.stringValue(); // the 'path' field of the Document object holds the URL
System.out.println(url.replace("%%", "/"));
// -------- Now let us use all of the functions above to make something useful --------
// The following bit of code is a worked out example of how to get a bunch of documents
// in response to a query and show them (without ranking them according to TF/IDF)
Scanner sc = new Scanner(System.in);
String str = "";
System.out.print("query> ");
while (!(str = sc.nextLine()).equals("quit")) {
String[] terms = str.split("\\s+");
for (String word : terms) {
Term term = new Term("contents", word);
TermDocs tdocs = r.termDocs(term);
while (tdocs.next()) {
String d_url =
r.document(tdocs.doc()).getFieldable("path").stringValue().replace("%%", "/");
System.out.println("[" + tdocs.doc() + "] " + d_url);
}
}
System.out.print("query> ");
}
}