计算二元词频率

Question

我编写了一段代码，主要用于计算词频并将它们插入到 ARFF 文件中以与 weka 一起使用。我想改变它，以便它可以计算二元词频率，即单词对而不是单个单词，尽管我的尝试最多被证明是不成功的。

我意识到有很多东西需要注意，但非常感谢对此的任何帮助。 这是我的代码：

    import re
    import nltk

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

    # create list of lower case words
    word_list = re.split('\s+', file(filename).read().lower())
    print 'Words in text:', len(word_list)
    # punctuation and numbers to be removed
    punctuation = re.compile(r'[-.?!,":;()|0-9]')
    word_list = [punctuation.sub("", word) for word in word_list]

    word_list2 = [w.strip() for w in word_list if w.strip() not in nltk.corpus.stopwords.words('english')]



    # create dictionary of word:frequency pairs
    freq_dic = {}


    for word in word_list2:

        # form dictionary
        try: 
            freq_dic[word] += 1
        except: 
            freq_dic[word] = 1


    print '-'*30

    print "sorted by highest frequency first:"
    # create list of (val, key) tuple pairs
    freq_list2 = [(val, key) for key, val in freq_dic.items()]
    # sort by val or frequency
    freq_list2.sort(reverse=True)
    freq_list3 = list(freq_list2)
    # display result as top 10 most frequent words
    freq_list4 =[]
    freq_list4=freq_list3[:10]

    words = []

    for item in freq_list4:
        a = str(item[1])
        a = a.lower()
        words.append(a)



    f = open(filename)

    newlist = []

    for line in f:
        line = punctuation.sub("", line)
        line = line.lower()
        newlist.append(line)

    f2 = open('Lines.txt','w')

    newlist2= []
    for line in newlist:
        line = line.split()
        newlist2.append(line)
        f2.write(str(line))
        f2.write("\n")


    print newlist2

    # ARFF Creation

    arff = open('output.arff','w')
    arff.write('@RELATION wordfrequency\n\n')
    for word in words:
        arff.write('@ATTRIBUTE ')
        arff.write(str(word))
        arff.write(' numeric\n')

    arff.write('@ATTRIBUTE class {endofworld, notendofworld}\n\n')
    arff.write('@DATA\n')
    # Counting word frequencies for each verse
    for line in newlist2:
        word_occurrences = str("")
        for word in words:
            matches = int(0)
            for item in line:
                if str(item) == str(word):
                matches = matches + int(1)
                else:
                continue
            word_occurrences = word_occurrences + str(matches) + ","
        word_occurrences = word_occurrences + "endofworld"
        arff.write(word_occurrences)
        arff.write("\n")

    print words

Answer 1

这应该可以帮助您开始：

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

请注意，第一个二元组是 (None, w1) ，其中 w1 是第一个单词，因此您有一个特殊的二元组来标记开始-文本。如果您还需要文本结尾二元组，请在循环后添加 yield (wprev, None)。

Answer 2

推广到具有可选填充的 n-gram，还使用 ​​defaultdict(int) 作为频率，在 2.6 中工作：

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

输出：

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')

Answer 3

如果您开始使用 NLTK 的 FreqDist 函数进行计数，生活会变得更加轻松。 NLTK 还具有二元组特征。下页中提供了它们的示例。

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

Answer 4

我已经为你重写了第一部分，因为它很恶心。注意事项：

1. 列表推导式是你的朋友，多使用它们。
2. collections.Counter 太棒了！

好的，代码：

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)]