python去/替标点总结

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 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)
速度对比:
sets : 19.8566138744
regex : 6.86155414581
translate : 2.12455511093
replace : 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

 

发表评论

电子邮件地址不会被公开。 必填项已用*标注