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

From an efficiency perspective, you’re not going to beat

s.translate(None, string.punctuation) 

For higher versions of Python use the following code:

s.translate(str.maketrans('', '', string.punctuation)) 

It’s performing raw string operations in C with a lookup table – there’s not much that will beat that but writing your own C code.

If speed isn’t a worry, another option though is:

exclude = set(string.punctuation) s = ''.join(ch for ch in s if ch not in exclude) 

This is faster than s.replace with each char, but won’t perform as well as non-pure python approaches such as regexes or string.translate, as you can see from the below timings. For this type of problem, doing it at as low a level as possible pays off.

Timing code:

import re, string, timeit  s = 'string. With. Punctuation' exclude = set(string.punctuation) table = string.maketrans('','') regex = re.compile('[%s]' % re.escape(string.punctuation))  def test_set(s):     return ''.join(ch for ch in s if ch not in exclude)  def test_re(s):  # From Vinko's solution, with fix.     return regex.sub('', s)  def test_trans(s):     return s.translate(table, string.punctuation)  def test_repl(s):  # From S.Lott's solution     for c in string.punctuation:         s=s.replace(c,'')     return s  print 'sets      :',timeit.Timer('f(s)', 'from __main__ import s,test_set as f').timeit(1000000) print 'regex     :',timeit.Timer('f(s)', 'from __main__ import s,test_re as f').timeit(1000000) print 'translate :',timeit.Timer('f(s)', 'from __main__ import s,test_trans as f').timeit(1000000) print 'replace   :',timeit.Timer('f(s)', 'from __main__ import s,test_repl as f').timeit(1000000) 

This gives the following results:

sets      : 19.8566138744 regex     : 6.86155414581 translate : 2.12455511093 replace   : 28.4436721802