在做英文文本处理时，常常会遇到这样的情况，需要我们提取出里面的词组进行主题抽取，尤其是具有行业特色的，比如金融年报等。其中主要进行的是进行双连词和三连词的抽取，那如何进行双连词和三连词的抽取呢？这是本文将要介绍的具体内容。
1. nltk.bigrams(tokens) 和 nltk.trigrams(tokens)
一般如果只是要求穷举双连词或三连词，则可以直接用nltk中的函数bigrams()或trigrams()， 效果如下面代码：
>>> import nltk
>>> str='you are my sunshine, and all of things are so beautiful just for you.'
>>> tokens=nltk.wordpunct_tokenize(str)
>>> bigram=nltk.bigrams(tokens)
>>> bigram
<generator object bigrams at 0x025C1C10>
>>> list(bigram)
[('you', 'are'), ('are', 'my'), ('my', 'sunshine'), ('sunshine', ','), (',', 'and'), ('and', 'all'), ('all', 'of'), ('of', 'things'), ('things', 'are'), ('are', 'so'), ('so', 'beautiful'), ('beautiful
', 'just'), ('just', 'for'), ('for', 'you'), ('you', '.')]
>>> trigram=nltk.trigrams(tokens)
>>> list(trigram)
[('you', 'are', 'my'), ('are', 'my', 'sunshine'), ('my', 'sunshine', ','), ('sunshine', ',', 'and'), (',', 'and', 'all'), ('and', 'all', 'of'), ('all', 'of', 'things'), ('of', 'things', 'are'), ('thin
gs', 'are', 'so'), ('are', 'so', 'beautiful'), ('so', 'beautiful', 'just'), ('beautiful', 'just', 'for'), ('just', 'for', 'you'), ('for', 'you', '.')]
2. nltk.ngrams(tokens, n)
如果要求穷举四连词甚至更长的多词组，则可以用统一的函数ngrams(tokens, n),其中n表示n词词组， 该函数表达形式较统一，效果如下代码：
>>> nltk.ngrams(tokens, 2)
<generator object ngrams at 0x027AAF30>
>>> list(nltk.ngrams(tokens,2))
[('you', 'are'), ('are', 'my'), ('my', 'sunshine'), ('sunshine', ','), (',', 'and'), ('and', 'all'), ('all', 'of'), ('of', 'things'), ('things', 'are'), ('are', 'so'), ('so', 'beautiful'), ('beautiful
', 'just'), ('just', 'for'), ('for', 'you'), ('you', '.')]
>>> list(nltk.ngrams(tokens,3))
[('you', 'are', 'my'), ('are', 'my', 'sunshine'), ('my', 'sunshine', ','), ('sunshine', ',', 'and'), (',', 'and', 'all'), ('and', 'all', 'of'), ('all', 'of', 'things'), ('of', 'things', 'are'), ('thin
gs', 'are', 'so'), ('are', 'so', 'beautiful'), ('so', 'beautiful', 'just'), ('beautiful', 'just', 'for'), ('just', 'for', 'you'), ('for', 'you', '.')]
>>> list(nltk.ngrams(tokens,4))
[('you', 'are', 'my', 'sunshine'), ('are', 'my', 'sunshine', ','), ('my', 'sunshine', ',', 'and'), ('sunshine', ',', 'and', 'all'), (',', 'and', 'all', 'of'), ('and', 'all', 'of', 'things'), ('all', '
of', 'things', 'are'), ('of', 'things', 'are', 'so'), ('things', 'are', 'so', 'beautiful'), ('are', 'so', 'beautiful', 'just'), ('so', 'beautiful', 'just', 'for'), ('beautiful', 'just', 'for', 'you'),
('just', 'for', 'you', '.')]
3. nltk.collocations下的相关类
nltk.collocations下有三个类：BigramCollocationFinder， QuadgramCollocationFinder， TrigramCollocationFinder
1）BigramCollocationFinder
它是一个发现二元词组并对其进行排序的工具，一般使用函数from_words()去构建一个搜索器，而不是直接生成一个实例。发现器主要调用以下方法：
above_score(self, score_fn, min_score): 返回分数超过min_score的n元词组，并按分数从大到小对其进行排序。这里当然返回的是二元词组，这里的分数有多种定义，后面将做详细介绍。
apply_freq_filter(self, min_freq):过滤掉词组出现频率小于min_freq的词组。
apply_ngram_filter(self, fn): 过滤掉符合条件fn的词组。在判断条件fn时，是将整个词组进行判断是否满足条件fn，如果满足条件，则将该词组过滤掉。
apply_word_filter(self, fn): 过滤掉符合条件fn的词组。在判断条件fn时，是将词组中的词一一判断，如果有一个词满足条件fn，则该词组满足条件，将会被过滤掉。
nbest(self, score_fn, n): 返回分数最高的前n个词组。
score_ngrams(self, score_fn): 返回由词组和对应分数组成的序列，并将其从高到低排列。
>>> finder=nltk.collocations.BigramCollocationFinder.from_words(tokens)
>>> bigram_measures=nltk.collocations.BigramAssocMeasures()
>>> finder.nbest(bigram_measures.pmi, 10)
[(',', 'and'), ('all', 'of'), ('and', 'all'), ('beautiful', 'just'), ('just', 'for'), ('my', 'sunshine'), ('of', 'things'), ('so', 'beautiful'), ('sunshine', ','), ('are', 'my')]
>>> finder.nbest(bigram_measures.pmi, 100)
[(',', 'and'), ('all', 'of'), ('and', 'all'), ('beautiful', 'just'), ('just', 'for'), ('my', 'sunshine'), ('of', 'things'), ('so', 'beautiful'), ('sunshine', ','), ('are', 'my'), ('are', 'so'), ('for'
, 'you'), ('things', 'are'), ('you', '.'), ('you', 'are')]
>>> finder.apply_ngram_filter(lambda w1,w2: w1 in [',', '.'] and w2 in [',', '.'] )
>>> finder.nbest(bigram_measures.pmi, 100)
[(',', 'and'), ('all', 'of'), ('and', 'all'), ('beautiful', 'just'), ('just', 'for'), ('my', 'sunshine'), ('of', 'things'), ('so', 'beautiful'), ('sunshine', ','), ('are', 'my'), ('are', 'so'), ('for'
, 'you'), ('things', 'are'), ('you', '.'), ('you', 'are')]
>>> finder.apply_word_filter(lambda x: x in [',', '.'])
>>> finder.nbest(bigram_measures.pmi, 100)
[('all', 'of'), ('and', 'all'), ('beautiful', 'just'), ('just', 'for'), ('my', 'sunshine'), ('of', 'things'), ('so', 'beautiful'), ('are', 'my'), ('are', 'so'), ('for', 'you'), ('things', 'are'), ('yo
u', 'are')]
2)TrigramCollocationFinder 和 QuadgramCollocationFinder
用法同BigramCollocationFinder, 只不过这里生产的是三元词组搜索器, 而QuadgramCollocationFinder产生的是四元词组搜索器。对应函数也同上。
4. 计算词组词频
>>> sorted(finder.ngram_fd.items(), key=lambda t: (-t[1], t[0]))[:10]
[(('all', 'of'), 1), (('and', 'all'), 1), (('are', 'my'), 1), (('are', 'so'), 1), (('beautiful', 'just'), 1), (('for', 'you'), 1), (('just', 'for'), 1), (('my', 'sunshine'), 1), (('of', 'things'), 1),
(('so', 'beautiful'), 1)] ###这里的key是排序依据，就是说先按t[1](词频)排序，-表示从大到小；再按照词组（t[0]）排序，默认从a-z.
5. 判断的分数
在nltk.collocations.ngramAssocMeasures下，有多种分数：
chi_sq(cls, n_ii, n_ix_xi_tuple, n_xx): 使用卡方分布计算出的各个n元词组的分数。
pmi(cls, *marginals): 使用点互信息计算出的各个n元词组的分数。
likelihood_ratio(cls, *marginals): 使用最大似然比计算出的各个n元词组的分数。
student_t(cls, *marginals): 使用针对单元词组的带有独立假设的学生t检验计算各个n元词组的分数
以上是比较常用的几种分数，当然还有很多其他的分数，比如：poisson_stirling, jaccard, fisher, phi_sq等。
>>> bigram_measures=nltk.collocations.BigramAssocMeasures()
>>> bigram_measures.student_t(8, (15828, 4675), 14307668)
0.9999319894802036
>>> bigram_measures.student_t(8, (42, 20), 14307668)
2.828406367705413
>>> bigram_measures.chi_sq(8, (15828, 4675), 14307668)
1.5488692067282201
>>> bigram_measures.chi_sq(59, (67, 65), 571007)
456399.76190356724
>>> bigram_measures.likelihood_ratio(110, (2552, 221), 31777)
270.721876936225
>>> bigram_measures.pmi(110, (2552, 221), 31777)
2.6317398492166078
>>> bigram_measures.pmi
<bound method type.pmi of <class 'nltk.metrics.association.BigramAssocMeasures'>>
>>> bigram_measures.likelihood_ratio
<bound method type.likelihood_ratio of <class 'nltk.metrics.association.BigramAssocMeasures'>>
>>> bigram_measures.chi_sq
<bound method type.chi_sq of <class 'nltk.metrics.association.BigramAssocMeasures'>>
>>> bigram_measures.student_t
<bound method type.student_t of <class 'nltk.metrics.association.BigramAssocMeasures'>>
6. Ranking and correlation
It is useful to consider the results of finding collocations as a ranking, and the rankings output using different association measures can be compared using the Spearman correlation coefficient.
Ranks can be assigned to a sorted list of results trivially by assigning strictly increasing ranks to each result:
>>> from nltk.metrics.spearman import *
>>> results_list = ['item1', 'item2', 'item3', 'item4', 'item5']
>>> print(list(ranks_from_sequence(results_list)))
[('item1', 0), ('item2', 1), ('item3', 2), ('item4', 3), ('item5', 4)]
If scores are available for each result, we may allow sufficiently similar results (differing by no more than rank_gap) to be assigned the same rank:
>>> results_scored = [('item1', 50.0), ('item2', 40.0), ('item3', 38.0),
... ('item4', 35.0), ('item5', 14.0)]
>>> print(list(ranks_from_scores(results_scored, rank_gap=5)))
[('item1', 0), ('item2', 1), ('item3', 1), ('item4', 1), ('item5', 4)]
The Spearman correlation coefficient gives a number from -1.0 to 1.0 comparing two rankings. A coefficient of 1.0 indicates identical rankings; -1.0 indicates exact opposite rankings.
>>> print('%0.1f' % spearman_correlation(
... ranks_from_sequence(results_list),
... ranks_from_sequence(results_list)))
1.0
>>> print('%0.1f' % spearman_correlation(
... ranks_from_sequence(reversed(results_list)),
... ranks_from_sequence(results_list)))
-1.0
>>> results_list2 = ['item2', 'item3', 'item1', 'item5', 'item4']
>>> print('%0.1f' % spearman_correlation(
... ranks_from_sequence(results_list),
... ranks_from_sequence(results_list2)))
0.6
>>> print('%0.1f' % spearman_correlation(
... ranks_from_sequence(reversed(results_list)),
... ranks_from_sequence(results_list2)))
-0.6