Count the word frequencies within a given directory

Question

I have built a tiny script count the word frequencies of all text files with a specified directory. I would be glad to receive any suggestions and feedback.

import re, glob,os
from collections import OrderedDict, Counter

class CountWords:
    def __init__(self,dir,ext):
        self.dir = dir
        self.ext = ext

    @property
    def files(self):
        #collect all the files with `.txt` extension
        os.chdir(self.dir)
        return glob.glob(f"**/*.{self.ext}", recursive=True)
    @property
    def contents(self):
        #concatenate files
        cat_content = []
        for file in self.files:
            with open(file, 'r') as f:
                content = f.read()
                cat_content.append(content)
        contents = ",".join(cat_content)
        return contents
    @property
    def words(self):
        words = re.findall(r"[a-zA-z_]+", self.contents)
        words = [word.lower() for word in words]
        return words

    def count(self):
        words_counter = Counter(self.words)
        words_counter = OrderedDict(sorted(words_counter.items(),key=lambda t:t[1], reverse=True))
        return words_counter

The function version:

def count_words(dir,ext):
    os.chdir(dir)
    #collect all the files with `.txt` extension
    files = glob.glob(f"**/*.ext", recursive=True)
    #concatenate files
    cat_content = []
    for file in files:
        with open(file, 'r') as f:
            content = f.read()
            cat_content.append(content)
    contents = ",".join(cat_content)
    #extract the words
    words = re.findall(r"[a-zA-z_]+", contents)
    words = [word.lower() for word in words]
    words_counter = Counter(words)
    words_counter = OrderedDict(sorted(words_counter.items(),key=lambda t:t[1], reverse=True))
    return words_counter

Answer 1

#!/usr/bin/python3
import os
import re
from os.path import join
from collections import Counter, OrderedDict

def count_words(directory, ext): # don't use the name dir, it's a builtin function
    """Counts word frequencies in a directory of files.

    Keyword arguments:
    directory -- count_words will search this directory recursively
    ext -- the extension of files that you wish to count

    Returns an OrderedDict, from most to least frequent.

    (You don't strictly need a docstring, but it's a good habit to get into.
    See PEP 257.)"""

    # Initialize the counter
    word_counter = Counter()

    # Personally I like to break my code into small, simple functions
    # This code could be inline in the loop below,
    # but I think it's a bit clearer this way.
    def update_counter(word_counter, filename):
        '''updates word_counter with all the words in open(filename)'''
        with open(filename, 'r') as f:
            try:
                # Two things:
                # First, I'm updating counter here,
                # instead of creating an enormous string.
                # Each file string only exists long enough for me to search it.

                # Second and less important,
                # I chose to lower the full text first, to make the regex simpler.
                word_counter.update(re.findall('[a-z_]+', f.read().lower()))
            except UnicodeDecodeError:
                # In python3 if your file isn't UTF-8
                # then read() throws an exception.
                # Maybe you want this behavior,
                # but I'd rather just warn and continue.
                print("Warning: couldn't decode", filename)

    # Using os.walk instead of glob
    for root, dirs, files in os.walk(directory):
        for fname in files:
            if fname.endswith(ext):
                update_counter(word_counter, join(root, fname))

    # words_counter.most_common() does exactly the sort you are looking for
    return OrderedDict(word_counter.most_common())