Efficiently count word frequencies in python

user_590847 0
Viewed 36987

I'd like to count frequencies of all words in a text file.

>>> countInFile('test.txt')

should return {'aaa':1, 'bbb': 2, 'ccc':1} if the target text file is like:

# test.txt
aaa bbb ccc
bbb

I've implemented it with pure python following some posts. However, I've found out pure-python ways are insufficient due to huge file size (> 1GB).

I think borrowing sklearn's power is a candidate.

If you let CountVectorizer count frequencies for each line, I guess you will get word frequencies by summing up each column. But, it sounds a bit indirect way.

What is the most efficient and straightforward way to count words in a file with python?

Update

My (very slow) code is here:

from collections import Counter

def get_term_frequency_in_file(source_file_path):
    wordcount = {}
    with open(source_file_path) as f:
        for line in f:
            line = line.lower().translate(None, string.punctuation)
            this_wordcount = Counter(line.split())
            wordcount = add_merge_two_dict(wordcount, this_wordcount)
    return wordcount

def add_merge_two_dict(x, y):
    return { k: x.get(k, 0) + y.get(k, 0) for k in set(x) | set(y) }
frequency-distribution scikit-learn word-count nlp python
8 Answers

Here's some benchmark. It'll look strange but the crudest code wins.

[code]:

from collections import Counter, defaultdict
import io, time

import numpy as np
from sklearn.feature_extraction.text import CountVectorizer

infile = '/path/to/file'

def extract_dictionary_sklearn(file_path):
    with io.open(file_path, 'r', encoding='utf8') as fin:
        ngram_vectorizer = CountVectorizer(analyzer='word')
        X = ngram_vectorizer.fit_transform(fin)
        vocab = ngram_vectorizer.get_feature_names()
        counts = X.sum(axis=0).A1
    return Counter(dict(zip(vocab, counts)))

def extract_dictionary_native(file_path):
    dictionary = Counter()
    with io.open(file_path, 'r', encoding='utf8') as fin:
        for line in fin:
            dictionary.update(line.split())
    return dictionary

def extract_dictionary_paddle(file_path):
    dictionary = defaultdict(int)
    with io.open(file_path, 'r', encoding='utf8') as fin:
        for line in fin:
            for words in line.split():
                dictionary[word] +=1
    return dictionary

start = time.time()
extract_dictionary_sklearn(infile)
print time.time() - start

start = time.time()
extract_dictionary_native(infile)
print time.time() - start

start = time.time()
extract_dictionary_paddle(infile)
print time.time() - start

[out]:

38.306814909
24.8241138458
12.1182529926

Data size (154MB) used in the benchmark above:

$ wc -c /path/to/file
161680851

$ wc -l /path/to/file
2176141

Some things to note:

  • With the sklearn version, there's an overhead of vectorizer creation + numpy manipulation and conversion into a Counter object
  • Then native Counter update version, it seems like Counter.update() is an expensive operation
user_69203630

you can try with sklearn

from sklearn.feature_extraction.text import CountVectorizer
    vectorizer = CountVectorizer()

    data=['i am student','the student suffers a lot']
    transformed_data =vectorizer.fit_transform(data)
    vocab= {a: b for a, b in zip(vectorizer.get_feature_names(), np.ravel(transformed_data.sum(axis=0)))}
    print (vocab)
user_91454300

Combining every ones else's views and some of my own :) Here is what I have for you

from collections import Counter
from nltk.tokenize import RegexpTokenizer
from nltk.corpus import stopwords

text='''Note that if you use RegexpTokenizer option, you lose 
natural language features special to word_tokenize 
like splitting apart contractions. You can naively 
split on the regex \w+ without any need for the NLTK.
'''

# tokenize
raw = ' '.join(word_tokenize(text.lower()))

tokenizer = RegexpTokenizer(r'[A-Za-z]{2,}')
words = tokenizer.tokenize(raw)

# remove stopwords
stop_words = set(stopwords.words('english'))
words = [word for word in words if word not in stop_words]

# count word frequency, sort and return just 20
counter = Counter()
counter.update(words)
most_common = counter.most_common(20)
most_common

Output

(All ones)

[('note', 1),
 ('use', 1),
 ('regexptokenizer', 1),
 ('option', 1),
 ('lose', 1),
 ('natural', 1),
 ('language', 1),
 ('features', 1),
 ('special', 1),
 ('word', 1),
 ('tokenize', 1),
 ('like', 1),
 ('splitting', 1),
 ('apart', 1),
 ('contractions', 1),
 ('naively', 1),
 ('split', 1),
 ('regex', 1),
 ('without', 1),
 ('need', 1)]

One can do better than this in terms of efficiency but if you are not worried about it too much, this code is the best.

user_81273900
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