我试图通过Lucene实现显式语义分析（ESA）。
在匹配文档时，如何在查询中考虑术语TFIDF？
例如：
查询应优先于Doc1匹配。
我希望在不影响性能的情况下工作。
我通过查询提升来做到这一点。通过提高相对于其TFIDF的条款。
有更好的方法吗？
答案 0 :(得分：1)
Lucene当然默认支持TF / IDF评分，所以不太清楚我理解你在寻找什么。
实际上听起来有点像你想根据查询本身的TF / IDF来权衡查询词。因此，让我们考虑一下这两个要素：
TF：Lucene总结每个查询词的分数。如果相同的查询字词出现两次，在查询中（如field:(a a b)），加倍的术语会比较（相当于）提升2的重量更重。
IDF：idf是指多文档语料库中的数据。由于只有一个查询，因此不适用。或者，如果您想获得技术，所有术语的idf均为1.
因此，IDF在这种情况下并没有真正意义，TF已经为你完成了。所以，你真的不需要做任何事情。
记住，想一想，还有其他得分元素！ coord因素在这里很重要。
a b a匹配四个查询字词a b a a，但不匹配c d）a b c匹配五个查询字词a b a c a，但不匹配d）因此，该特定评分元素将更强烈地评分第二个文档。
以下是文档explain的{{1}}（请参阅IndexSearcher.explain）输出：
a b a对于文档0.26880693 = (MATCH) product of:
0.40321037 = (MATCH) sum of:
0.10876686 = (MATCH) weight(text:a in 0) [DefaultSimilarity], result of:
0.10876686 = score(doc=0,freq=2.0 = termFreq=2.0
), product of:
0.25872254 = queryWeight, product of:
0.5945349 = idf(docFreq=2, maxDocs=2)
0.435168 = queryNorm
0.42039964 = fieldWeight in 0, product of:
1.4142135 = tf(freq=2.0), with freq of:
2.0 = termFreq=2.0
0.5945349 = idf(docFreq=2, maxDocs=2)
0.5 = fieldNorm(doc=0)
0.07690979 = (MATCH) weight(text:b in 0) [DefaultSimilarity], result of:
0.07690979 = score(doc=0,freq=1.0 = termFreq=1.0
), product of:
0.25872254 = queryWeight, product of:
0.5945349 = idf(docFreq=2, maxDocs=2)
0.435168 = queryNorm
0.29726744 = fieldWeight in 0, product of:
1.0 = tf(freq=1.0), with freq of:
1.0 = termFreq=1.0
0.5945349 = idf(docFreq=2, maxDocs=2)
0.5 = fieldNorm(doc=0)
0.10876686 = (MATCH) weight(text:a in 0) [DefaultSimilarity], result of:
0.10876686 = score(doc=0,freq=2.0 = termFreq=2.0
), product of:
0.25872254 = queryWeight, product of:
0.5945349 = idf(docFreq=2, maxDocs=2)
0.435168 = queryNorm
0.42039964 = fieldWeight in 0, product of:
1.4142135 = tf(freq=2.0), with freq of:
2.0 = termFreq=2.0
0.5945349 = idf(docFreq=2, maxDocs=2)
0.5 = fieldNorm(doc=0)
0.10876686 = (MATCH) weight(text:a in 0) [DefaultSimilarity], result of:
0.10876686 = score(doc=0,freq=2.0 = termFreq=2.0
), product of:
0.25872254 = queryWeight, product of:
0.5945349 = idf(docFreq=2, maxDocs=2)
0.435168 = queryNorm
0.42039964 = fieldWeight in 0, product of:
1.4142135 = tf(freq=2.0), with freq of:
2.0 = termFreq=2.0
0.5945349 = idf(docFreq=2, maxDocs=2)
0.5 = fieldNorm(doc=0)
0.6666667 = coord(4/6)
：
a b c请注意，根据需要，对于术语0.43768594 = (MATCH) product of:
0.52522314 = (MATCH) sum of:
0.07690979 = (MATCH) weight(text:a in 1) [DefaultSimilarity], result of:
0.07690979 = score(doc=1,freq=1.0 = termFreq=1.0
), product of:
0.25872254 = queryWeight, product of:
0.5945349 = idf(docFreq=2, maxDocs=2)
0.435168 = queryNorm
0.29726744 = fieldWeight in 1, product of:
1.0 = tf(freq=1.0), with freq of:
1.0 = termFreq=1.0
0.5945349 = idf(docFreq=2, maxDocs=2)
0.5 = fieldNorm(doc=1)
0.07690979 = (MATCH) weight(text:b in 1) [DefaultSimilarity], result of:
0.07690979 = score(doc=1,freq=1.0 = termFreq=1.0
), product of:
0.25872254 = queryWeight, product of:
0.5945349 = idf(docFreq=2, maxDocs=2)
0.435168 = queryNorm
0.29726744 = fieldWeight in 1, product of:
1.0 = tf(freq=1.0), with freq of:
1.0 = termFreq=1.0
0.5945349 = idf(docFreq=2, maxDocs=2)
0.5 = fieldNorm(doc=1)
0.07690979 = (MATCH) weight(text:a in 1) [DefaultSimilarity], result of:
0.07690979 = score(doc=1,freq=1.0 = termFreq=1.0
), product of:
0.25872254 = queryWeight, product of:
0.5945349 = idf(docFreq=2, maxDocs=2)
0.435168 = queryNorm
0.29726744 = fieldWeight in 1, product of:
1.0 = tf(freq=1.0), with freq of:
1.0 = termFreq=1.0
0.5945349 = idf(docFreq=2, maxDocs=2)
0.5 = fieldNorm(doc=1)
0.217584 = (MATCH) weight(text:c in 1) [DefaultSimilarity], result of:
0.217584 = score(doc=1,freq=1.0 = termFreq=1.0
), product of:
0.435168 = queryWeight, product of:
1.0 = idf(docFreq=1, maxDocs=2)
0.435168 = queryNorm
0.5 = fieldWeight in 1, product of:
1.0 = tf(freq=1.0), with freq of:
1.0 = termFreq=1.0
1.0 = idf(docFreq=1, maxDocs=2)
0.5 = fieldNorm(doc=1)
0.07690979 = (MATCH) weight(text:a in 1) [DefaultSimilarity], result of:
0.07690979 = score(doc=1,freq=1.0 = termFreq=1.0
), product of:
0.25872254 = queryWeight, product of:
0.5945349 = idf(docFreq=2, maxDocs=2)
0.435168 = queryNorm
0.29726744 = fieldWeight in 1, product of:
1.0 = tf(freq=1.0), with freq of:
1.0 = termFreq=1.0
0.5945349 = idf(docFreq=2, maxDocs=2)
0.5 = fieldNorm(doc=1)
0.8333333 = coord(5/6)
的匹配在第一个文档中获得更高的权重，并且您还会看到单独评估的每个独立a并添加到分数中。
另请注意，coord的差异以及第二个doc中术语“c”的idf。这些得分影响正在消除你从增加同一个词的倍数所获得的提升。如果您在查询中添加了足够的a，它们最终会交换位置。 a上的匹配仅被评估为远更有意义的结果。