Hi Markus, Thank you for your reply.
I have more questions. what I want to do is sort document by "tfidf score + function query score” there are problems to do this : * if I use function query ( https://wiki.apache.org/solr/FunctionQuery <https://wiki.apache.org/solr/FunctionQuery> ), there is no function for tfidf score. * if I use bf, solr product queryNorm * if I use bq, solr product tfidf what can I do? example what I want to do is when query is like ?q=iphone&bf=recip(ms(NOW,mydatefield),3.16e-11,1,1) get score tfidf + bf score = 2.7 + 1.1 when tfidf =2.7 , bf score = 1.1 not 2.7 + 1.1 * 0.12 when queryNorm =0.12 Best, Calvin. > On Jul 7, 2015, at 8:51 PM, Markus Jelsma <markus.jel...@openindex.io> wrote: > > Hello - you can either use a similarity that does not use query > normalization, or you can just ignore it, it is relative anyway. Also, > consider using boost parameter instead of bf, it is multiplicative where bf > is just additive, which offers less control. You may also want to reduce time > resolution by using NOW/HOUR or NOW/DAY. It saves you a lot of bad cache > entries. > > Markus > > -----Original message----- >> From:Lee Chunki <lck7...@coupang.com> >> Sent: Tuesday 7th July 2015 13:29 >> To: solr-user@lucene.apache.org >> Subject: function query result without queryNorm >> >> Hi, >> >> I want to add some value to score. >> >> so, I tried to use “bf” but it returns >> value * queryNorm >> >> for example, when I use >> q=iphone&bf=div(ms(NOW,start_time),3600000) >> >> solr returns >> >> 799.5687 = (MATCH) >> FunctionQuery(div(ms(const(1436268218591),date(start_time)),const(3600000))), >> product of: >> 6988.394 = >> div(ms(const(1436268218591),date(start_time)=2014-09-19T07:00:00Z),const(3600000)) >> 1.0 = boost >> 0.114413805 = queryNorm >> >> >> how can I get only FunctionQuery() result ? >> >> Thanks, >> Calvin. >> >>
