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2014年11月28日 (五) 15:07Lr讨论 | 贡献的版本

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test idf and tf

  • data
 d0 [{如何,怎么}} {办理,办} {户口,户口本} # 到当地派出所办理  # 如何办理户口
 d1 {办理,办} {户口,户口本} [{流程,步骤}] # 到当地派出所办理  # 如何办理户口
 d2 [{如何,怎么}} {办理,办} {身份证,身份} # 到当地派出所办理  # 如何办理身份证
 d3 {办理,办} {身份证} [{流程,步骤}] # 到当地派出所办理  # 如何办理身份证
  • 搜索
query:"如何办理户口"  => question:如何 question:办理户口
  • result
 doc=0 score=0.114656925 shardIndex=-1|0.114656925 = (MATCH) product of:
   0.22931385 = (MATCH) sum of:
     0.22931385 = (MATCH) weight(question:如何 in 0) [DefaultSimilarity], result of:
       0.22931385 = score(doc=0,freq=1.0 = termFreq=1.0
 ), product of:
       0.4748871 = queryWeight, product of:
         1.287682 = idf(docFreq=2, maxDocs=4)
         0.3687922 = queryNorm
       0.48288077 = fieldWeight in 0, product of:
         1.0 = tf(freq=1.0), with freq of:
           1.0 = termFreq=1.0
         1.287682 = idf(docFreq=2, maxDocs=4)
         0.375 = fieldNorm(doc=0)
  0.5 = coord(1/2)
  • 详细计算流score(query,d0)
  • 参考公式:[1]
QQ截图20141128164958.png
  • tf("如何" in d0)=sqrt{frequency}= sqrt{1}=1
  • idf("如何")=<math>1+ln( {numDocs}/{docFreq+1})=1+ln( {4}/{2+1} )
  • 如何".getboost=1
  • coord(如何,d0) : 0.5 = coord(1/2)
   coord(t,d)=overlap  /maxOverlap .
   overlap - the number of query terms matched in the document
   maxOverlap - the total number of terms in the query
  • queryNorm(q)= 1/sqrt(sumOfSquaredWeights)=1/sqrt(sum(idf("如何")*1+idf("办理户口")))=1/sqrt(1*(1.287682*1.287682+1)).
  sumOfSquaredWeights   =   q.getBoost()*q.getBoost()*∑( idf(t) *t.getBoost() )^2