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

Instead of decoding the whole bytes read from the url, I process the binary data. Because bytes.translate expects its second argument to be a byte string, I utf-8 encode punctuation. After removing punctuations, I utf-8 decode the byte string.

The function freq_dist expects an iterable. That's why I've passed data.splitlines().

from urllib2 import urlopen
from collections import Counter
from string import punctuation
from time import time
import sys
from pprint import pprint

url = 'https://raw.githubusercontent.com/Simdiva/DSL-Task/master/data/DSLCC-v2.0/test/test.txt'

data = urlopen(url).read()

def freq_dist(data):
    """
    :param data: file-like object opened in binary mode or
                 sequence of byte strings separated by '\n'
    :type data: an iterable sequence
    """
    #For readability   
    #return Counter(word for line in data
    #    for word in line.translate(
    #    None,bytes(punctuation.encode('utf-8'))).decode('utf-8').split())

    punc = punctuation.encode('utf-8')
    words = (word for line in data for word in line.translate(None, punc).decode('utf-8').split())
    return Counter(words)


start = time()
word_dist = freq_dist(data.splitlines())
print('elapsed: {}'.format(time() - start))
pprint(word_dist.most_common(10))

Output;

elapsed: 0.806480884552

[(u'de', 11106),
 (u'a', 6742),
 (u'que', 5701),
 (u'la', 4319),
 (u'je', 4260),
 (u'se', 3938),
 (u'\u043d\u0430', 3929),
 (u'na', 3623),
 (u'da', 3534),
 (u'i', 3487)]

It seems dict is more efficient than Counter object.

def freq_dist(data):
    """
    :param data: A string with sentences separated by '\n'
    :type data: str
    """
    d = {}
    punc = punctuation.encode('utf-8')
    words = (word for line in data for word in line.translate(None, punc).decode('utf-8').split())
    for word in words:
        d[word] = d.get(word, 0) + 1
    return d

start = time()
word_dist = freq_dist(data.splitlines())
print('elapsed: {}'.format(time() - start))
pprint(sorted(word_dist.items(), key=lambda x: (x[1], x[0]), reverse=True)[:10])

Output;

elapsed: 0.642680168152

[(u'de', 11106),
 (u'a', 6742),
 (u'que', 5701),
 (u'la', 4319),
 (u'je', 4260),
 (u'se', 3938),
 (u'\u043d\u0430', 3929),
 (u'na', 3623),
 (u'da', 3534),
 (u'i', 3487)]

To be more memory efficient when opening huge file, you have to pass just the opened url. But the timing will include file download time too.

data = urlopen(url)
word_dist = freq_dist(data)