Efficiently count word frequencies in python

Question

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

>>> countInFile(\'test.txt\')

should return {\'aaa\':1, \'bbb\':

Answer 1

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)

Answer 2

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