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copied!<p>From an efficiency perspective, you're not going to beat </p> <pre><code>s.translate(None, string.punctuation) </code></pre> <p>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.</p> <p>If speed isn't a worry, another option though is:</p> <pre><code>exclude = set(string.punctuation) s = ''.join(ch for ch in s if ch not in exclude) </code></pre> <p>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.</p> <p>Timing code:</p> <pre><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) </code></pre> <p>This gives the following results:</p> <pre><code>sets : 19.8566138744 regex : 6.86155414581 translate : 2.12455511093 replace : 28.4436721802 </code></pre>

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