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Lucene example source code file (SloppyPhraseScorer.java)

This example Lucene source code file (SloppyPhraseScorer.java) is included in the DevDaily.com "Java Source Code Warehouse" project. The intent of this project is to help you "Learn Java by Example" TM.

Java - Lucene tags/keywords

hashset, io, ioexception, ioexception, override, phrasepositions, phrasepositions, phrasescorer, pps, pps, similarity, sloppyphrasescorer, sloppyphrasescorer, util

The Lucene SloppyPhraseScorer.java source code

package org.apache.lucene.search;

/**
 * Licensed to the Apache Software Foundation (ASF) under one or more
 * contributor license agreements.  See the NOTICE file distributed with
 * this work for additional information regarding copyright ownership.
 * The ASF licenses this file to You under the Apache License, Version 2.0
 * (the "License"); you may not use this file except in compliance with
 * the License.  You may obtain a copy of the License at
 *
 *     http://www.apache.org/licenses/LICENSE-2.0
 *
 * Unless required by applicable law or agreed to in writing, software
 * distributed under the License is distributed on an "AS IS" BASIS,
 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
 * See the License for the specific language governing permissions and
 * limitations under the License.
 */

import java.io.IOException;
import java.util.HashSet;

final class SloppyPhraseScorer extends PhraseScorer {
    private int slop;
    private PhrasePositions repeats[];
    private PhrasePositions tmpPos[]; // for flipping repeating pps.
    private boolean checkedRepeats;

    SloppyPhraseScorer(Weight weight, PhraseQuery.PostingsAndFreq[] postings, Similarity similarity,
                       int slop, byte[] norms) {
        super(weight, postings, similarity, norms);
        this.slop = slop;
    }

    /**
     * Score a candidate doc for all slop-valid position-combinations (matches) 
     * encountered while traversing/hopping the PhrasePositions.
     * <br> The score contribution of a match depends on the distance: 
     * <br> - highest score for distance=0 (exact match).
     * <br> - score gets lower as distance gets higher.
     * <br>Example: for query "a b"~2, a document "x a b a y" can be scored twice: 
     * once for "a b" (distance=0), and once for "b a" (distance=2).
     * <br>Possibly not all valid combinations are encountered, because for efficiency  
     * we always propagate the least PhrasePosition. This allows to base on 
     * PriorityQueue and move forward faster. 
     * As result, for example, document "a b c b a"
     * would score differently for queries "a b c"~4 and "c b a"~4, although 
     * they really are equivalent. 
     * Similarly, for doc "a b c b a f g", query "c b"~2 
     * would get same score as "g f"~2, although "c b"~2 could be matched twice.
     * We may want to fix this in the future (currently not, for performance reasons).
     */
    @Override
    protected float phraseFreq() throws IOException {
        int end = initPhrasePositions();
        
        float freq = 0.0f;
        boolean done = (end<0);
        while (!done) {
            PhrasePositions pp = pq.pop();
            int start = pp.position;
            int next = pq.top().position;

            boolean tpsDiffer = true;
            for (int pos = start; pos <= next || !tpsDiffer; pos = pp.position) {
                if (pos<=next && tpsDiffer)
                    start = pos;                  // advance pp to min window
                if (!pp.nextPosition()) {
                    done = true;          // ran out of a term -- done
                    break;
                }
                PhrasePositions pp2 = null;
                tpsDiffer = !pp.repeats || (pp2 = termPositionsDiffer(pp))==null;
                if (pp2!=null && pp2!=pp) {
                  pp = flip(pp,pp2); // flip pp to pp2
                }
            }

            int matchLength = end - start;
            if (matchLength <= slop)
                freq += getSimilarity().sloppyFreq(matchLength); // score match

            if (pp.position > end)
                end = pp.position;
            pq.add(pp);               // restore pq
        }

        return freq;
    }
    
    // flip pp2 and pp in the queue: pop until finding pp2, insert back all but pp2, insert pp back.
    // assumes: pp!=pp2, pp2 in pq, pp not in pq.
    // called only when there are repeating pps.
    private PhrasePositions flip(PhrasePositions pp, PhrasePositions pp2) {
      int n=0;
      PhrasePositions pp3;
      //pop until finding pp2
      while ((pp3=pq.pop()) != pp2) {
        tmpPos[n++] = pp3;
      }
      //insert back all but pp2
      for (n--; n>=0; n--) {
        pq.insertWithOverflow(tmpPos[n]);
      }
      //insert pp back
      pq.add(pp);
      return pp2;
    }

    /**
     * Init PhrasePositions in place.
     * There is a one time initialization for this scorer (taking place at the first doc that matches all terms):
     * <br>- Put in repeats[] each pp that has another pp with same position in the doc.
     *       This relies on that the position in PP is computed as (TP.position - offset) and 
     *       so by adding offset we actually compare positions and identify that the two are 
     *       the same term.
     *       An exclusion to this is two distinct terms in the same offset in query and same 
     *       position in doc. This case is detected by comparing just the (query) offsets, 
     *       and two such PPs are not considered "repeating". 
     * <br>- Also mark each such pp by pp.repeats = true.
     * <br>Later can consult with repeats[] in termPositionsDiffer(pp), making that check efficient.
     * In particular, this allows to score queries with no repetitions with no overhead due to this computation.
     * <br>- Example 1 - query with no repetitions: "ho my"~2
     * <br>- Example 2 - query with repetitions: "ho my my"~2
     * <br>- Example 3 - query with repetitions: "my ho my"~2
     * <br>Init per doc w/repeats in query, includes propagating some repeating pp's to avoid false phrase detection.  
     * @return end (max position), or -1 if any term ran out (i.e. done) 
     * @throws IOException 
     */
    private int initPhrasePositions() throws IOException {
        int end = 0;
        
        // no repeats at all (most common case is also the simplest one)
        if (checkedRepeats && repeats==null) {
            // build queue from list
            pq.clear();
            for (PhrasePositions pp = first; pp != null; pp = pp.next) {
                pp.firstPosition();
                if (pp.position > end)
                    end = pp.position;
                pq.add(pp);         // build pq from list
            }
            return end;
        }
        
        // position the pp's
        for (PhrasePositions pp = first; pp != null; pp = pp.next)
            pp.firstPosition();
        
        // one time initializatin for this scorer
        if (!checkedRepeats) {
            checkedRepeats = true;
            // check for repeats
            HashSet<PhrasePositions> m = null;
            for (PhrasePositions pp = first; pp != null; pp = pp.next) {
                int tpPos = pp.position + pp.offset;
                for (PhrasePositions pp2 = pp.next; pp2 != null; pp2 = pp2.next) {
                    if (pp.offset == pp2.offset) {
                      continue; // not a repetition: the two PPs are originally in same offset in the query! 
                    }
                    int tpPos2 = pp2.position + pp2.offset;
                    if (tpPos2 == tpPos) { 
                        if (m == null)
                            m = new HashSet<PhrasePositions>();
                        pp.repeats = true;
                        pp2.repeats = true;
                        m.add(pp);
                        m.add(pp2);
                    }
                }
            }
            if (m!=null)
                repeats = m.toArray(new PhrasePositions[0]);
        }
        
        // with repeats must advance some repeating pp's so they all start with differing tp's       
        if (repeats!=null) {
            for (int i = 0; i < repeats.length; i++) {
                PhrasePositions pp = repeats[i];
                PhrasePositions pp2;
                while ((pp2 = termPositionsDiffer(pp)) != null) {
                  if (!pp2.nextPosition())  // out of pps that do not differ, advance the pp with higher offset 
                      return -1;           // ran out of a term -- done  
                } 
            }
        }
      
        // build queue from list
        pq.clear();
        for (PhrasePositions pp = first; pp != null; pp = pp.next) {
            if (pp.position > end)
                end = pp.position;
            pq.add(pp);         // build pq from list
        }

        if (repeats!=null) {
          tmpPos = new PhrasePositions[pq.size()];
        }
        return end;
    }

    /**
     * We disallow two pp's to have the same TermPosition, thereby verifying multiple occurrences 
     * in the query of the same word would go elsewhere in the matched doc.
     * @return null if differ (i.e. valid) otherwise return the higher offset PhrasePositions
     * out of the first two PPs found to not differ.
     */
    private PhrasePositions termPositionsDiffer(PhrasePositions pp) {
        // efficiency note: a more efficient implementation could keep a map between repeating 
        // pp's, so that if pp1a, pp1b, pp1c are repeats term1, and pp2a, pp2b are repeats 
        // of term2, pp2a would only be checked against pp2b but not against pp1a, pp1b, pp1c. 
        // However this would complicate code, for a rather rare case, so choice is to compromise here.
        int tpPos = pp.position + pp.offset;
        for (int i = 0; i < repeats.length; i++) {
            PhrasePositions pp2 = repeats[i];
            if (pp2 == pp) {
                continue;
            }
            if (pp.offset == pp2.offset) {
              continue; // not a repetition: the two PPs are originally in same offset in the query! 
            }
            int tpPos2 = pp2.position + pp2.offset;
            if (tpPos2 == tpPos) {
                return pp.offset > pp2.offset ? pp : pp2; // do not differ: return the one with higher offset.
            }
        }
        return null; 
    }
}

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