How to get back the amount of an existing word within a.txt file?

Question

I’m trying to get as a return the amount of low-level words (bad words) in one text, using as a basis another file . txt containing the words (bad words). The code I made only returns 1 (one) occurrence, and there are 3 (three). What could I do to improve?

def read_file():
    with open(r'C:\movie_quotes.txt') as file: # Abertura do arquivo a ser analisado. 
        contents = file.read()
    print(contents)
    file.close()
    check_file(contents)


def check_file(text_check):
    bad_words = open(r'C:\palavroes_bloqueio.txt') # Palavras a serem procuradas.
    contents = list(bad_words)
    # print(contents)
    for name in contents:
        if name in text_check:
            print('Bad words found.')
            print(text_check.count(name))
    bad_words.close()


read_file()

Answer 1

I think the problem is that your second file has one word per line, and then when you search the text for that word, it looks for the word and the line break and it doesn’t find.

For example, file words_block.txt:

foo
bar

When searching the text, it will search for foo n and not only by foo

What you can try is to change this section:

bad_words = open(r'C:\palavroes_bloqueio.txt') # Palavras a serem procuradas.
contents = list(bad_words)

By that stretch:

with open(r'C:\palavroes_bloqueio.txt') as bad_words:
    contents = bad_words.read()

# Divide as palavras usando o delimitador \n
contents = contents.split('\n')

Note: The file is already automatically closed if you use the directive with open('..') as filename, then you don’t need to call filename.close() next