Re: BUG #16235: ts_rank ignores match and only considers lower weighted vector
От | Tom Lane |
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Тема | Re: BUG #16235: ts_rank ignores match and only considers lower weighted vector |
Дата | |
Msg-id | 23961.1580164498@sss.pgh.pa.us обсуждение исходный текст |
Ответ на | BUG #16235: ts_rank ignores match and only considers lower weighted vector (PG Bug reporting form <noreply@postgresql.org>) |
Ответы |
Re: BUG #16235: ts_rank ignores match and only considers lowerweighted vector
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Список | pgsql-bugs |
PG Bug reporting form <noreply@postgresql.org> writes: > The following query shows the problem: > select ts_rank(doc1, query) as rank_wrong, ts_rank(doc2, query) as > rank_correct > from (select setweight(to_tsvector('simple', 'foo something'), 'A') || > setweight(to_tsvector('simple', 'foobar'), 'C') as doc1, > setweight(to_tsvector('simple', 'foo something'), 'A') as > doc2, > to_tsquery('simple', 'foo:* & something') as > query) as subquery; > ts_rank on doc1 is only half of the rank of doc2. ts_rank seems to only > consider the 'foobar' term with lower weight when calculating the rank. The > foo:1A is only considered in doc2. No, that's not correct. What it actually is doing is taking some sort of average of the weights of the occurrences, as you can see if you play around with a few more examples besides these two. That could be better documented, perhaps, but I don't think it's obviously broken. I can see that there might be a use for taking the max or even the sum of the weights rather than an average --- in many situations it wouldn't be desirable to rank doc1 of your example lower than doc2. But really that'd be a different ranking algorithm, not a bug fix for this one. The manual claims you can write your own ranking algorithm ... but AFAICS you'd have to code it in C, because we aren't exposing anything at SQL level that would let you get at the raw match data :-(. So there's room for improvement there. Also, you might try using ts_rank_cd() instead, as that uses a different algorithm for combining the weights. At least on this example, doc1 gets a higher score than doc2. regards, tom lane
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