I think you are asking about this stuff: http://www.google.com/search?q=chuck%20idf%20lucene%20square
Otis --- Yonik Seeley <[EMAIL PROTECTED]> 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 > --------------------------------------------------------------------- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
