Counting bi-gram frequencies

匆匆过客 提交于 2019-12-03 03:56:33

This should get you started:

def bigrams(words):
    wprev = None
    for w in words:
        yield (wprev, w)
        wprev = w

Note that the first bigram is (None, w1) where w1 is the first word, so you have a special bigram that marks start-of-text. If you also want an end-of-text bigram, add yield (wprev, None) after the loop.

Generalized to n-grams with optional padding, also uses defaultdict(int) for frequencies, to work in 2.6:

from collections import defaultdict

def ngrams(words, n=2, padding=False):
    "Compute n-grams with optional padding"
    pad = [] if not padding else [None]*(n-1)
    grams = pad + words + pad
    return (tuple(grams[i:i+n]) for i in range(0, len(grams) - (n - 1)))

# grab n-grams
words = ['the','cat','sat','on','the','dog','on','the','cat']
for size, padding in ((3, 0), (4, 0), (2, 1)):
    print '\n%d-grams padding=%d' % (size, padding)
    print list(ngrams(words, size, padding))

# show frequency
counts = defaultdict(int)
for ng in ngrams(words, 2, False):
    counts[ng] += 1

print '\nfrequencies of bigrams:'
for c, ng in sorted(((c, ng) for ng, c in counts.iteritems()), reverse=True):
    print c, ng

Output:

3-grams padding=0
[('the', 'cat', 'sat'), ('cat', 'sat', 'on'), ('sat', 'on', 'the'), 
 ('on', 'the', 'dog'), ('the', 'dog', 'on'), ('dog', 'on', 'the'), 
 ('on', 'the', 'cat')]

4-grams padding=0
[('the', 'cat', 'sat', 'on'), ('cat', 'sat', 'on', 'the'), 
 ('sat', 'on', 'the', 'dog'), ('on', 'the', 'dog', 'on'), 
 ('the', 'dog', 'on', 'the'), ('dog', 'on', 'the', 'cat')]

2-grams padding=1
[(None, 'the'), ('the', 'cat'), ('cat', 'sat'), ('sat', 'on'), 
 ('on', 'the'), ('the', 'dog'), ('dog', 'on'), ('on', 'the'), 
 ('the', 'cat'), ('cat', None)]

frequencies of bigrams:
2 ('the', 'cat')
2 ('on', 'the')
1 ('the', 'dog')
1 ('sat', 'on')
1 ('dog', 'on')
1 ('cat', 'sat')

I've rewritten the first bit for you, because it's icky. Points to note:

  1. List comprehensions are your friend, use more of them.
  2. collections.Counter is great!

OK, code:

import re
import nltk
import collections

# Quran subset
filename = raw_input('Enter name of file to convert to ARFF with extension, eg. name.txt: ')

# punctuation and numbers to be removed
punctuation = re.compile(r'[-.?!,":;()|0-9]')

# create list of lower case words
word_list = re.split('\s+', open(filename).read().lower())
print 'Words in text:', len(word_list)

words = (punctuation.sub("", word).strip() for word in word_list)
words = (word for word in words if word not in ntlk.corpus.stopwords.words('english'))

# create dictionary of word:frequency pairs
frequencies = collections.Counter(words)

print '-'*30

print "sorted by highest frequency first:"
# create list of (val, key) tuple pairs
print frequencies

# display result as top 10 most frequent words
print frequencies.most_common(10)

[word for word, frequency in frequencies.most_common(10)]

Life is much more easier if you start using NLTK's FreqDist function to do the counting. Also NLTK has bigram feature. Examples for both of them are in the following page.

http://nltk.googlecode.com/svn/trunk/doc/book/ch01.html

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