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From Otis Gospodnetic <otis_gospodne...@yahoo.com>
Subject Re: term scoring (idf) question
Date Thu, 15 Sep 2005 21:14:44 GMT
I think you are asking about this stuff:

http://www.google.com/search?q=chuck%20idf%20lucene%20square

Otis

--- Yonik Seeley <yseeley@gmail.com> wrote:

> I'm trying to figure out why idf is multiplied twice into the score
> of a 
> term query.
> It sort of makes sense if you have just one term... the original
> weight is 
> idf*boost, and
> the normalization factor is 1/(idf*boost), so you multiply in the idf
> again 
> if you want the final score to contain an idf factor (instead of
> being 
> normalized to 1.0).
> 
> But when you have a boolean query of two term queries, the amount of
> score 
> each term contributes will vary with the square of their idfs, not
> their 
> idfs. Is this what's intended? If so, the scoring formula one
> normally sees 
> (tf*idf*boost*lengthNorm), doesn't seem to convey this fact.
> 
> -Yonik
> Now hiring -- http://tinyurl.com/7m67g
> 


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