Ferret talk - Aug 2007 - scoring problem in acts_as_ferret

Hi,

I am using acts_as_ferret and have a problem with scoring. I would like
to organize it in such way that, if any of the searched terms fits, I
get 1.0 score as a result. I will explain it on the example.

I have in index:

a) "one two three four"
b) "one two three"
c) "one two"
d) "one"

When I search for "one" I would like to get 1.0 score for all of
indexed
elements. When I search for "one two" I get 1.0 score for a),b),c).

Thanks for any help!
-- 
Posted via http://www.ruby-forum.com/.

On 23.08.2007, at 14:32, Mlynco Mlynco wrote:
> I am using acts_as_ferret and have a problem with scoring. I would  
> like
> to organize it in such way that, if any of the searched terms fits, I
> get 1.0 score as a result. I will explain it on the example.
Sounds to me like the wrong approach. You won''t get Ferret to score a  
document with 1.0 if there are more terms in the document than you  
search for.
> I have in index:
>
> a) "one two three four"
> b) "one two three"
> c) "one two"
> d) "one"
>
> When I search for "one" I would like to get 1.0 score for all of
> indexed
> elements. When I search for "one two" I get 1.0 score for
a),b),c).
Question is: What do you actually want to achieve? Why do you want  
the documents to be scored this way? I''m sure there''s a better
way to
do it.

You might want to check out phrase queries (i.e. using quotes) and  
boolean operators and see if a combination of them might work for you.

Cheers,
Andreas

Andreas Korth wrote:
> On 23.08.2007, at 14:32, Mlynco Mlynco wrote:
> 
>> I am using acts_as_ferret and have a problem with scoring. I would  
>> like
>> to organize it in such way that, if any of the searched terms fits, I
>> get 1.0 score as a result. I will explain it on the example.
> 
> Sounds to me like the wrong approach. You won''t get Ferret to
score a
> document with 1.0 if there are more terms in the document than you
> search for.
> 
>> I have in index:
>>
>> a) "one two three four"
>> b) "one two three"
>> c) "one two"
>> d) "one"
>>
>> When I search for "one" I would like to get 1.0 score for all
of
>> indexed
>> elements. When I search for "one two" I get 1.0 score for
a),b),c).
> 
> Question is: What do you actually want to achieve? Why do you want
> the documents to be scored this way? I''m sure there''s a
better way to
> do it.
> 
> You might want to check out phrase queries (i.e. using quotes) and
> boolean operators and see if a combination of them might work for you.
> 
> Cheers,
> Andreas

Andreas,

Probably you are right. What I don''t what to do is to decrease the
score
for elements which have some additional terms.

So, ie I am looking for a recipe. I have indexed some of them with such 
ingredients:

recipe1: "beef"
recipe2: "onion beef chicken"
recipe3: "onion beef chicken tomato"

Looking for a "beef" I wouldn''t like to "punish"
recipe2 and recipe3
because they are richer, I would like to treat them in the same way. 
Actually the score does not have to be 1.0, but I would like it to be 
the same for all recipes mentioned above.

Any suggestions?

Thanks a lot,
         mlynco
-- 
Posted via http://www.ruby-forum.com/.

On Thu, Aug 23, 2007 at 06:22:52PM +0200, Mlynco Mlynco wrote:
[..]
> 
> So, ie I am looking for a recipe. I have indexed some of them with such 
> ingredients:
> 
> recipe1: "beef"
> recipe2: "onion beef chicken"
> recipe3: "onion beef chicken tomato"
> 
> Looking for a "beef" I wouldn''t like to
"punish" recipe2 and recipe3
> because they are richer, I would like to treat them in the same way. 
> Actually the score does not have to be 1.0, but I would like it to be 
> the same for all recipes mentioned above.
> 
> Any suggestions?
have a look at ConstantScoreQuery. It allows to turn a Filter into a
query where all results score equally. 

To get this Filter, you could use the QueryFilter class, which allows
you to turn your original query into a Filter.

query = ConstantScoreQuery.new(QueryFilter.new(TermQuery.new(:ingredients,
''beef''))

looks like there should be an easier way to achieve this, though ;-)

Jens


-- 
Jens Kr?mer
http://www.jkraemer.net/ - Blog
http://www.omdb.org/     - The new free film database

Jens Kraemer wrote:
> have a look at ConstantScoreQuery. It allows to turn a Filter into a
> query where all results score equally.
> 
> To get this Filter, you could use the QueryFilter class, which allows
> you to turn your original query into a Filter.
> 
> query = 
> ConstantScoreQuery.new(QueryFilter.new(TermQuery.new(:ingredients, 
> ''beef''))
> 
> looks like there should be an easier way to achieve this, though ;-)
> 
> Jens

Jens,

Thanks a lot for help. The ConstantScoreQuery with QueryFilter and 
TermQuery/MultiTermQuery work fine. I am able to get constant value for 
all of the ingredients inserted into the query.

The problem appears when I would like to get more precise results with 
multiple ingredients inserted. So, coming back to my example:

recipe1: "beef"
recipe2: "onion beef chicken"
recipe3: "onion beef chicken tomato"
recipe4: "onion chicken"

All results from that query are equal and that''s great:

query = 
ConstantScoreQuery.new(QueryFilter.new(TermQuery.new(:ingredients, 
''beef''))


When I put more than one ingredient into account I would like to 
distinguish different scores for recipes which: include three of them, 
include two of them, recipes which has one of them only.

So I would like create a query which return 100% hits for recipe2 and 
recipe3,
appropriately less for recipe4 and adequately less for recipe1.

I created multi query like that:

multi_term_query = MultiTermQuery.new(:ingredients)
multi_term_query << "onion" << "beef" <<
"chicken"
query = ConstantScoreQuery.new(QueryFilter.new(multi_term_query)

but it doesn''t work at all.

Any ideas?

Thanks in advance,

mlynco

-- 
Posted via http://www.ruby-forum.com/.