How to take out punctuation from string and find a count of words of a certain length?

Question

I am opening trying to create a function that opens a .txt file and counts the words that have the same length as the number specified by the user.
The .txt file is:

This is a random text document. How many words have a length of one? 
How many words have the length three?  We have the power to figure it out! 
Is a function capable of doing this?

I'm able to open and read the file, but I am unable to exclude punctuation and find the length of each word.

def samplePractice(number):
    fin = open('sample.txt', 'r')
    lstLines = fin.readlines()
    fin.close

    count = 0

    for words in lstLines:
        words = words.split()

    for i in words:
        if len(i) == number:
            count += 1
    return count

What do you mean you are unable to do it? What have you tried to remove punctuation? Is there an error you encountered? — razdi, May 05 '19 at 23:25
Initially I tried splitting the lines like split('[.,?!]') but it just gave me an output of 0 — jackrabbit, May 05 '19 at 23:28
Possible duplicate of [How to delete a character from a string using Python](https://stackoverflow.com/questions/3559559/how-to-delete-a-character-from-a-string-using-python) — Boris Verkhovskiy, May 05 '19 at 23:43

score 0 · Answer 1 · answered May 05 '19 at 23:35

0

You can try using the replace() on the string and pass in the desired punctuation and replace it with an empty string("").

It would look something like this:

puncstr = "Hello!"

nopuncstr = puncstr.replace(".", "").replace("?", "").replace("!", "")

answered May 05 '19 at 23:35

PiAreSquared

324
3
10

score 0 · Answer 2 · answered May 05 '19 at 23:37

I have written a sample code to remove punctuations and to count the number of words. Modify according to your requirement.

    import re
    fin = """This is a random text document. How many words have a length of one? How many words have the length three?  We have the power to figure it out! Is a function capable of doing this?"""
    fin = re.sub(r'[^\w\s]','',fin)
    print(len(fin.split()))

The above code prints the number of words. Hope this helps!!

Mahmoud Elshahat · Answer 3 · 2019-05-06T00:12:57.653

instead of cascading replace() just use strip() a one time call

Edit: a cleaner version

pl = '?!."\'' # punctuation list

def samplePractice(number):
    with open('sample.txt', 'r') as fin:
        words = fin.read().split()

    # clean words
    words = [w.strip(pl) for w in words]

    count = 0
    for word in words:
        if len(word) == number:
            print(word, end=', ')
            count += 1

    return count

result = samplePractice(4)
print('\nResult:', result)

output:

This, text, many, have, many, have, have, this, 
Result: 8

your code is almost ok, it just the second for block in wrong position

pl = '?!."\'' # punctuation list

def samplePractice(number):
    fin = open('sample.txt', 'r')
    lstLines = fin.readlines()
    fin.close

    count = 0

    for words in lstLines:
        words = words.split()

        for i in words:
            i = i.strip(pl)  # clean the word by strip
            if len(i) == number:
                count += 1
    return count

result = samplePractice(4)
print(result)

output:

@Mahmoud_Elshahat the above code doesn't seem to give me the right output. In my editor, if I input samplePractice(3) I expect the output of 6 but I only get 5. Any idea why? — jackrabbit, May 06 '19 at 00:19
@jackrabbit just try my last edited answer, it gave me 6 as expected `How, one,How, the, the, out, Result: 6` — Mahmoud Elshahat, May 06 '19 at 00:21

How to take out punctuation from string and find a count of words of a certain length?

3 Answers3