CODEWITHSUNDEEP
pythonnltk
marie
marie

Reputation: 417

NLTK Create bigrams with sentence boundaries

I'm trying to create bigrams using nltk which don't cross sentence boundaries. I tried using from_documents, however, it isn't working as I had hoped.

import nltk
from nltk.collocations import *
bigram_measures = nltk.collocations.BigramAssocMeasures()

finder = BigramCollocationFinder.from_documents([['This', 'is', 'sentence', 'one'], ['A', 'second', 'sentence']])
print finder.nbest(bigram_measures.pmi, 10)

>> [(u'A', u'second'), (u'This', u'is'), (u'one', u'A'), (u'is', u'sentence'), (u'second', u'sentence'), (u'sentence', u'one')]

This includes (u'one', u'A'), which I'm trying to avoid.

Upvotes: 3

Views: 486

Answers (1)

marie
marie

Reputation: 417

I ended up ditching nltk and doing the processing by hand:

To create ngrams, I found this handy function on http://locallyoptimal.com/blog/2013/01/20/elegant-n-gram-generation-in-python/

def find_ngrams(input_list, n):
    return zip(*[input_list[i:] for i in range(n)])

From there, I computed bigram probabilities doing the following:

First I created the bigrams 

all_bigrams = [find_ngrams(sentence, 2) for sentence in text]

Then I grouped them by first word

first_words = {}
for bigram in all_bigrams:
    if bigram[0] in first_words.keys():
        first_words[bigram[0]].append(bigram)
    else:
        first_words[bigram[0]] = [bigram]

I then computed the probabilities for each bigram

bi_probabilites = {}
for bigram in (set(all_bigrams)):
    bigram_count = 0
    first_word_list = first_words[bigram[0]]
    for item in first_word_list:
        if item == bigram:
            bigram_count += 1
    bi_probabilites[bigram] = {
        'count': bigram_count, 
        'length': len(first_word_list), 
        'prob': float(bigram_count)/len(first_word_list)
    }

Not the most elegant, but it gets the job done.

Upvotes: 2

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