private float freq; //prhase frequency in current doc as computed by phraseFreq().
internal PhraseScorer(Weight weight, TermPositions[] tps, int[] offsets, Similarity similarity, byte[] norms):base(similarity)
{
this.norms = norms;
this.weight = weight;
this.value_Renamed = weight.Value;
// convert tps to a list of phrase positions.
// note: phrase-position differs from term-position in that its position
// reflects the phrase offset: pp.pos = tp.pos - offset.
// this allows to easily identify a matching (exact) phrase
// when all PhrasePositions have exactly the same position.
for (int i = 0; i < tps.Length; i++)
{
PhrasePositions pp = new PhrasePositions(tps[i], offsets[i]);
if (last != null)
{
// add next to end of list
last.next = pp;
}
else
{
first = pp;
}
last = pp;
}
pq = new PhraseQueue(tps.Length); // construct empty pq
first.doc = - 1;
}
private float freq; //prhase frequency in current doc as computed by phraseFreq().
internal PhraseScorer(Weight weight, TermPositions[] tps, int[] offsets, Similarity similarity, byte[] norms) : base(similarity)
{
this.norms = norms;
this.weight = weight;
this.value_Renamed = weight.Value;
// convert tps to a list of phrase positions.
// note: phrase-position differs from term-position in that its position
// reflects the phrase offset: pp.pos = tp.pos - offset.
// this allows to easily identify a matching (exact) phrase
// when all PhrasePositions have exactly the same position.
for (int i = 0; i < tps.Length; i++)
{
PhrasePositions pp = new PhrasePositions(tps[i], offsets[i]);
if (last != null)
{
// add next to end of list
last.next = pp;
}
else
{
first = pp;
}
last = pp;
}
pq = new PhraseQueue(tps.Length); // construct empty pq
first.doc = -1;
}
protected internal override float PhraseFreq()
{
// sort list with pq
pq.Clear();
for (PhrasePositions pp = first; pp != null; pp = pp.next)
{
pp.FirstPosition();
pq.Add(pp); // build pq from list
}
PqToList(); // rebuild list from pq
// for counting how many times the exact phrase is found in current document,
// just count how many times all PhrasePosition's have exactly the same position.
int freq = 0;
do
{
// find position w/ all terms
while (first.position < last.position)
{
// scan forward in first
do
{
if (!first.NextPosition())
{
return(freq);
}
}while (first.position < last.position);
FirstToLast();
}
freq++; // all equal: a match
}while (last.NextPosition());
return(freq);
}
protected internal void FirstToLast()
{
last.next = first; // move first to end of list
last = first;
first = first.next;
last.next = null;
}
private void Sort()
{
pq.Clear();
for (PhrasePositions pp = first; pp != null; pp = pp.next)
{
pq.Add(pp);
}
PqToList();
}
private void Init()
{
for (PhrasePositions pp = first; more && pp != null; pp = pp.next)
{
more = pp.Next();
}
if (more)
{
Sort();
}
}
/// <summary> 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).
/// </summary>
protected internal override float PhraseFreq()
{
int end = InitPhrasePositions();
float freq = 0.0f;
bool done = (end < 0);
while (!done)
{
PhrasePositions pp = pq.Pop();
int start = pp.position;
int next = pq.Top().position;
bool 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 += Similarity.SloppyFreq(matchLength); // score match
}
if (pp.position > end)
{
end = pp.position;
}
pq.Add(pp); // restore pq
}
return(freq);
}
public override int Advance(int target)
{
firstTime = false;
for (PhrasePositions pp = first; more && pp != null; pp = pp.next)
{
more = pp.SkipTo(target);
}
if (more)
{
Sort(); // re-sort
}
if (!DoNext())
{
first.doc = NO_MORE_DOCS;
}
return(first.doc);
}
protected internal void PqToList()
{
last = first = null;
while (pq.Top() != null)
{
PhrasePositions pp = pq.Pop();
if (last != null)
{
// add next to end of list
last.next = pp;
}
else
{
first = pp;
}
last = pp;
pp.next = null;
}
}
// 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);
}
/// <summary> 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.
/// </summary>
/// <returns> null if differ (i.e. valid) otherwise return the higher offset PhrasePositions
/// out of the first two PPs found to not differ.
/// </returns>
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;
}
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);
}
protected internal void FirstToLast()
{
last.next = first; // move first to end of list
last = first;
first = first.next;
last.next = null;
}
protected internal void PqToList()
{
last = first = null;
while (pq.Top() != null)
{
PhrasePositions pp = pq.Pop();
if (last != null)
{
// add next to end of list
last.next = pp;
}
else
first = pp;
last = pp;
pp.next = null;
}
}
// 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;
}
/// <summary> 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.
/// </summary>
/// <returns> null if differ (i.e. valid) otherwise return the higher offset PhrasePositions
/// out of the first two PPs found to not differ.
/// </returns>
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;
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;
}
/// <summary> Init PhrasePositions in place.
/// There is a one time initialization for this scorer:
/// <br/>- Put in repeats[] each pp that has another pp with same position in the doc.
/// <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.
/// </summary>
/// <returns> end (max position), or -1 if any term ran out (i.e. done)
/// </returns>
/// <throws> IOException </throws>
private int InitPhrasePositions()
{
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
HashMap <PhrasePositions, object> 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)
{
int tpPos2 = pp2.position + pp2.offset;
if (tpPos2 == tpPos)
{
if (m == null)
{
m = new HashMap <PhrasePositions, object>();
}
pp.repeats = true;
pp2.repeats = true;
m[pp] = null;
m[pp2] = null;
}
}
}
if (m != null)
{
repeats = m.Keys.ToArray();
}
}
// 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);
}