t*g
2 楼
有没有现成的api，还是要通过term,用id去比较,自己数呢?
【在 t*g 的大作中提到】
: 一个单词，比如说China, 我怎么能得到这个token在一篇文章中出现过多少次呢?
【在 t*g 的大作中提到】
: 一个单词，比如说China, 我怎么能得到这个token在一篇文章中出现过多少次呢?
t*e
3 楼
try Term Vector Query, and TermFreqVector.getTermFrequencies() method.
Cited from the API doc,
"Array of term frequencies. Locations of the array correspond one to one to the terms in the array obtained from getTerms method. Each location in the array contains the number of times this term occurs in the document or the document field."
Cited from the API doc,
"Array of term frequencies. Locations of the array correspond one to one to the terms in the array obtained from getTerms method. Each location in the array contains the number of times this term occurs in the document or the document field."
t*g
4 楼
Thanks. Will try.
to the terms in the array obtained from getTerms method. Each location in
the array contains the number of times this term occurs in the document or
the document field."
【在 t*******e 的大作中提到】
: try Term Vector Query, and TermFreqVector.getTermFrequencies() method.
: Cited from the API doc,
: "Array of term frequencies. Locations of the array correspond one to one to the terms in the array obtained from getTerms method. Each location in the array contains the number of times this term occurs in the document or the document field."
to the terms in the array obtained from getTerms method. Each location in
the array contains the number of times this term occurs in the document or
the document field."
【在 t*******e 的大作中提到】
: try Term Vector Query, and TermFreqVector.getTermFrequencies() method.
: Cited from the API doc,
: "Array of term frequencies. Locations of the array correspond one to one to the terms in the array obtained from getTerms method. Each location in the array contains the number of times this term occurs in the document or the document field."
t*e
5 楼
Watch out for the slow performance though.
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