Python - Finding word frequencies of list of words in text file

Question

I am trying to speed up my project to count word frequencies. I have 360+ text files, and I need to get the total number of words and the number of times each word from anot

Answer 1

A simple functional code to count word frequencies in a text file:

{
import string

def process_file(filename):
hist = dict()
f = open(filename,'rb')
for line in f:
    process_line(line,hist)
return hist

def process_line(line,hist):

line = line.replace('-','.')

for word in line.split():
    word = word.strip(string.punctuation + string.whitespace)
    word.lower()

    hist[word] = hist.get(word,0)+1

hist = process_file(filename)
print hist
}

Answer 2

import re, os, sys, codecs, fnmatch
import decimal
import zipfile
import glob
import csv

path= 'C:\\Users\\user\\Desktop\\sentiment2020\\POSITIVE'

files=[]
for r,d,f in os.walk(path):
    for file in f:
        if'.txt' in  file:
            files.append(os.path.join(r,file))

for f in files:
    print(f)
    file1= codecs.open(f,'r','utf8',errors='ignore')
    content=file1.read()

words=content.split()
for x in words:
    print (x)

dicts=[]
if __name__=="__main__":  
    str =words
    str2 = [] 
    for i in str:              
        if i not in str2: 
              str2.append(i)  
    for i in range(0, len(str2)):
        a= {str2[i]:str.count(str2[i])}
        dicts.append(a)
for i in dicts:        
    print(dicts)



#  for i in range(len(files)):
  #    with codecs.open('C:\\Users\\user\\Desktop\\sentiment2020\\NEGETIVE1\\sad1%s.txt' % i, 'w',"utf8") as filehandle:
  #         filehandle.write('%s\n' % dicts) 

Answer 3

collections.Counter() has this covered if I understand your problem.

The example from the docs would seem to match your problem.

# Tally occurrences of words in a list
cnt = Counter()
for word in ['red', 'blue', 'red', 'green', 'blue', 'blue']:
    cnt[word] += 1
print cnt


# Find the ten most common words in Hamlet
import re
words = re.findall('\w+', open('hamlet.txt').read().lower())
Counter(words).most_common(10)

From the example above you should be able to do:

import re
import collections
words = re.findall('\w+', open('1976.03.txt').read().lower())
print collections.Counter(words)

EDIT naive approach to show one way.

wanted = "fish chips steak"
cnt = Counter()
words = re.findall('\w+', open('1976.03.txt').read().lower())
for word in words:
    if word in wanted:
        cnt[word] += 1
print cnt

Answer 4

One possible implementation (using Counter)...

Instead of printing the output, I think it would be simpler to write to a csv file and import that into Excel. Look at http://docs.python.org/2/library/csv.html and replace print_summary.

import os
from collections import Counter
import glob

def word_frequency(fileobj, words):
    """Build a Counter of specified words in fileobj"""
    # initialise the counter to 0 for each word
    ct = Counter(dict((w, 0) for w in words))
    file_words = (word for line in fileobj for word in line.split())
    filtered_words = (word for word in file_words if word in words)
    return Counter(filtered_words)


def count_words_in_dir(dirpath, words, action=None):
    """For each .txt file in a dir, count the specified words"""
    for filepath in glob.iglob(os.path.join(dirpath, '*.txt')):
        with open(filepath) as f:
            ct = word_frequency(f, words)
            if action:
                action(filepath, ct)


def print_summary(filepath, ct):
    words = sorted(ct.keys())
    counts = [str(ct[k]) for k in words]
    print('{0}\n{1}\n{2}\n\n'.format(
        filepath,
        ', '.join(words),
        ', '.join(counts)))


words = set(['inflation', 'jobs', 'output'])
count_words_in_dir('./', words, action=print_summary)