<p>如果您想找到超过2或3克的克,可以使用<a href="http://scikit-learn.org/stable/" rel="noreferrer">scikit package</a>和Freqdist函数来获取这些克的计数。我试着用nltk.collaborations做这个,但是我认为我们不能找到超过3克的分数。所以我决定用克数。我希望这能帮你一点忙。谢谢</p>
<p>这是密码</p>
<pre><code>from sklearn.metrics.pairwise import cosine_similarity
from sklearn.feature_extraction.text import CountVectorizer
from nltk.collocations import *
from nltk.probability import FreqDist
import nltk
query = "This document gives a very short introduction to machine learning problems"
vect = CountVectorizer(ngram_range=(1,4))
analyzer = vect.build_analyzer()
listNgramQuery = analyzer(query)
listNgramQuery.reverse()
print "listNgramQuery=", listNgramQuery
NgramQueryWeights = nltk.FreqDist(listNgramQuery)
print "\nNgramQueryWeights=", NgramQueryWeights
</code></pre>
<p>这将输出为</p>
<pre><code>listNgramQuery= [u'to machine learning problems', u'introduction to machine learning', u'short introduction to machine', u'very short introduction to', u'gives very short introduction', u'document gives very short', u'this document gives very', u'machine learning problems', u'to machine learning', u'introduction to machine', u'short introduction to', u'very short introduction', u'gives very short', u'document gives very', u'this document gives', u'learning problems', u'machine learning', u'to machine', u'introduction to', u'short introduction', u'very short', u'gives very', u'document gives', u'this document', u'problems', u'learning', u'machine', u'to', u'introduction', u'short', u'very', u'gives', u'document', u'this']
NgramQueryWeights= <FreqDist: u'document': 1, u'document gives': 1, u'document gives very': 1, u'document gives very short': 1, u'gives': 1, u'gives very': 1, u'gives very short': 1, u'gives very short introduction': 1, u'introduction': 1, u'introduction to': 1, ...>
</code></pre>