Fuzzy Query Score

serkan.ozal · **Joined:** Tue Feb 26, 2013 5:24 am **Posts:** 2

Hi,
I have a problem about Fuzzy Query scoring.

Here is my entity.

Code:

@Indexed
public class PersonInfo {

    ...
    ...
    ...

    @Field
    @Boost(2.0f)
    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    @Field
    @Boost(2.0f)
    public String getLastnameTitle() {
        return lastnameTitle;
    }

    public void setLastnameTitle(String lastnameTitle) {
        this.lastnameTitle = lastnameTitle;
    }

    @Field
    @Boost(1.0f)
    public String getFatherName() {
        return fatherName;
    }

    public void setFatherName(String fatherName) {
        this.fatherName = fatherName;
    }

    @Field
    @Boost(1.0f)
    public String getMotherName() {
        return motherName;
    }

    public void setMotherName(String motherName) {
        this.motherName = motherName;
    }

    @Field
    @Boost(0.5f)
    public String getBirthPlace() {
        return birthPlace;
    }

    public void setBirthPlace(String birthPlace) {
        this.birthPlace = birthPlace;
    }

    @Field
    @Boost(0.5f)
    @DateBridge(resolution = Resolution.DAY)
    public Date getBirthDate() {
        return birthDate;
    }

    public void setBirthDate(Date birthDate) {
        this.birthDate = birthDate;
    }

}

When I search for "John" as name that is exactly matched to indexed entity named "John", I get score as "0.61".
I expeced that score must be "1.00" because they are exactly same.
When I debug search code, in "FuzzyTermEnum" similarity is calculated as "1.00" that is what I expected. But score is calculated as "0.61".

Why score is calculated as "0.61".
Any suggestions?
or
How can I override scoring behaviour?

serkan.ozal · **Joined:** Tue Feb 26, 2013 5:24 am **Posts:** 2

I override scoring behaviour simply

Code:

   public class CustomSimilarity extends Similarity {

      @Override
      public float computeNorm(String field, FieldInvertState state) {
         return state.getBoost();
      }

      @Override
      public float queryNorm(float sumOfSquaredWeights) {
         return 1;
      }

      @Override
      public float sloppyFreq(int distance) {
         return 1;
      }

      @Override
      public float tf(float freq) {
         return 1;
      }

      @Override
      public float idf(int docFreq, int numDocs) {
         return 1;
      }

      @Override
      public float coord(int overlap, int maxOverlap) {
         return 1;
      }
      
   }

Code:

   @Similarity(impl = CustomSimilarity.class)
   public class PersonInfo {
   
      ...
      ...
      ...
   }

Now it returns score what I expected.

sanne.grinovero · **Posted:** Wed Feb 27, 2013 12:06 pm

Hi, your solution is correct, iff you want it to return a score of 1.

That's not what users need usually: the score is normalized across all documents in the index, so some terms have a higher scoring value as they are very peculiar; very frequent terms are less interesting to the search engine.

Make sure you test the general accuracy on larger data sets, overriding the Similarity might make it harder for people to search what they are looking for.

You can find an accurate explanation of scoring and similarity calculations on the book, chapter 12 of Hibernate Search in Action.