Scoring problem in acts_as_ferret

Ruby Apps Ferret
1

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!

2

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

3

Andreas K. 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

4

Jens K. 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 :wink:

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

5

On Thu, Aug 23, 2007 at 06:22:52PM +0200, Mlynco Mlynco wrote:
[…]

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 :wink:

Jens

–
Jens Krämer
http://www.jkraemer.net/ - Blog
http://www.omdb.org/ - The new free film database