首先是英文标点：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 solutionfor 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)速度对比：sets : 19.8566138744regex : 6.86155414581translate : 2.12455511093replace : 28.4436721802
然后是中文标点：
re.sub(u”[\uFF00-\uFFEF]+” ,’ ‘,str.decode(‘utf8′))
参考地址：
http://stackoverflow.com/questions/265960/best-way-to-strip-punctuation-from-a-string-in-python
http://www.unicode.org/charts/PDF/UFF00.pdf
