python counts the number of words

The upstairs program has many problems, such as not handling punctuation, wrong way to read files, etc.

Is the building mainly case-sensitive? If you want to make a distinction, let's follow the following:

import re

def get_word_frequencies(file_name):

dic = {}

txt = open(filename, 'r').read().splitlines()

The following sentence replaces except'-'outside all punctuation, since'-'May be present in words.

txt = '[^\u4e00-\u94a5\w\d\-]', ' ', txt)

Replace the separate'-'

txt = ' - ', ' ', txt)for line in :

for word in :

If it is not case-sensitive, it will be treated as lowercase, and the following sentence will be changed to 0), 0).

dic[word] += 1

print dic

if __name__ = '__main__':

get_word_frequencies(''If you have any questions, keep asking.

The idea goes something like this:

Read the content of the file into a linked list, use the count() function of the linked list to count the number of spaces n, and then loop n times, mark the spaces before and after each word, take out the word, put it in an empty linked list of word[], and then use the count() function to count the number of occurrences, put it in another number【] empty linked list, after the loop is executed, the result output=dict(zip(word,number)) is the final dictionary output;

Or you can use the re regular expression to directly find the word in the middle of the space, output it to a new linked list, and then directly count each word.

dic = {}

for line in open(filename):

for word in :

dic[word] += 1

dic about this, dic is what you want, just deal with this kind of thing by yourself;

Get a ['-other don't'].

Summary. How to count the total number of words and sentences in an English article in python.

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