Yes Erik I can use ps=0 but, my problem is that I want phrase which have
same sequence and they can be present with in some distance
E.g.
If I have document masitha xyz 12345
I want that to be boosted since the sequence is in order .That's why I have
use ps=5
Thanks,
Aman Deep Singh
On 12-Jun-2017 5:44 PM, "Erik Hatcher" <erik.hatc...@gmail.com> wrote:
Using ps=5 causes the phrase matching to be unordered matching. You’ll
have to set ps=0, if using edismax, to get exact order phrase matches.
Erik
> On Jun 12, 2017, at 1:09 AM, Aman Deep Singh <amandeep.coo...@gmail.com>
wrote:
>
> Hi,
> I'm using a phrase query ,but it was applying the phrase boost to the
query
> where terms are in reverse order also ,which i don't want.Is their any way
> to avoid the phrase boost for reverse order and apply boost only in case
of
> terms are in same sequence
>
> Solr version 6.5.1
>
> e.g.
> http://localhost:8983/solr/l4_collection/select?debugQuery=o
n&defType=edismax&fl=score,nameSearch&indent=on&mm=100%25&
pf=nameSearch&q=12345%20masitha&qf=nameSearch&wt=xml&ps=5
>
>
> while my document has value
>
> in the debug query it is applying boost as
> 23.28365 = sum of:
> 15.112219 = sum of:
> 9.669338 = weight(nameSearch:12345 in 0) [SchemaSimilarity], result of:
> 9.669338 = score(doc=0,freq=1.0 = termFreq=1.0
> ), product of:
> 7.6397386 = idf, computed as log(1 + (docCount - docFreq + 0.5) / (docFreq
> + 0.5)) from:
> 2.0 = docFreq
> 5197.0 = docCount
> 1.2656635 = tfNorm, computed as (freq * (k1 + 1)) / (freq + k1 * (1 - b +
b
> * fieldLength / avgFieldLength)) from:
> 1.0 = termFreq=1.0
> 1.2 = parameter k1
> 0.75 = parameter b
> 5.2576485 = avgFieldLength
> 2.56 = fieldLength
> 5.44288 = weight(nameSearch:masitha in 0) [SchemaSimilarity], result of:
> 5.44288 = score(doc=0,freq=1.0 = termFreq=1.0
> ), product of:
> 4.3004165 = idf, computed as log(1 + (docCount - docFreq + 0.5) / (docFreq
> + 0.5)) from:
> 70.0 = docFreq
> 5197.0 = docCount
> 1.2656635 = tfNorm, computed as (freq * (k1 + 1)) / (freq + k1 * (1 - b +
b
> * fieldLength / avgFieldLength)) from:
> 1.0 = termFreq=1.0
> 1.2 = parameter k1
> 0.75 = parameter b
> 5.2576485 = avgFieldLength
> 2.56 = fieldLength
> 8.171431 = weight(*nameSearch:"12345 masitha"~5 *in 0) [SchemaSimilarity],
> result of:
> 8.171431 = score(doc=0,freq=0.33333334 = phraseFreq=0.33333334
> ), product of:
> 11.940155 = idf(), sum of:
> 7.6397386 = idf, computed as log(1 + (docCount - docFreq + 0.5) / (docFreq
> + 0.5)) from:
> 2.0 = docFreq
> 5197.0 = docCount
> 4.3004165 = idf, computed as log(1 + (docCount - docFreq + 0.5) / (docFreq
> + 0.5)) from:
> 70.0 = docFreq
> 5197.0 = docCount
> 0.6843655 = tfNorm, computed as (freq * (k1 + 1)) / (freq + k1 * (1 - b +
b
> * fieldLength / avgFieldLength)) from:
> 0.33333334 = phraseFreq=0.33333334
> 1.2 = parameter k1
> 0.75 = parameter b
> 5.2576485 = avgFieldLength
> 2.56 = fieldLength
>
> Thanks,
> Aman Deep Singh
