This was my mistake.
Thank you.
Taisuke
2020年10月23日(金) 15:02 Taisuke Miyazaki :
> Thanks.
>
> I analyzed it as explain=true and this is what I found.
> Why does this behave this way?
>
> fq=foo:1
> bq=foo:(1)^1
> bf=sum(200)
>
> If you do this, the score will be boosted by bq.
> However, if
Thanks.
I analyzed it as explain=true and this is what I found.
Why does this behave this way?
fq=foo:1
bq=foo:(1)^1
bf=sum(200)
If you do this, the score will be boosted by bq.
However, if you remove fq, the score will not be boosted by bq.
However, if you change the boost value of bq to
You’d get a much better idea of what goes on
if you added =true and analyzed the
output. That’d show you exactly what is
calculated when.
Best,
Erick
> On Oct 22, 2020, at 4:05 AM, Taisuke Miyazaki
> wrote:
>
> Hi,
>
> If you use a high value for the score, the values on the smaller scale
Hi,
If you use a high value for the score, the values on the smaller scale are
ignored.
Example :
bq = foo:(1.0)^1.0
bf = sum(200)
When I do this, the additional score for "foo" at 1.0 does not affect the
sort order.
I'm assuming this is an issue with the precision of the score floating
