Best way to strip punctuation from a string in Pyt

Question

It seems like there should be a simpler way than:

import string
s = "string. With. Punctuation?" # Sample string 
out = s.translate(string.maketrans("",""), string.punctuation)

Is there?

Answer 1

myString.translate(None, string.punctuation)

Answer 2

Remove stop words from the text file using Python

print('====THIS IS HOW TO REMOVE STOP WORS====')

with open('one.txt','r')as myFile:

    str1=myFile.read()

    stop_words ="not", "is", "it", "By","between","This","By","A","when","And","up","Then","was","by","It","If","can","an","he","This","or","And","a","i","it","am","at","on","in","of","to","is","so","too","my","the","and","but","are","very","here","even","from","them","then","than","this","that","though","be","But","these"

    myList=[]

    myList.extend(str1.split(" "))

    for i in myList:

        if i not in stop_words:

            print ("____________")

            print(i,end='\n')

Answer 3

>>> s = "string. With. Punctuation?"
>>> s = re.sub(r'[^\w\s]','',s)
>>> re.split(r'\s*', s)


['string', 'With', 'Punctuation']

Answer 4

Regular expressions are simple enough, if you know them.

import re
s = "string. With. Punctuation?"
s = re.sub(r'[^\w\s]','',s)

Answer 5

I like to use a function like this:

def scrub(abc):
    while abc[-1] is in list(string.punctuation):
        abc=abc[:-1]
    while abc[0] is in list(string.punctuation):
        abc=abc[1:]
    return abc

Answer 6

I haven't seen this answer yet. Just use a regex; it removes all characters besides word characters (\w) and number characters (\d), followed by a whitespace character (\s):

import re
s = "string. With. Punctuation?" # Sample string 
out = re.sub(ur'[^\w\d\s]+', '', s)