似乎应该有一个比以下更简单的方法:
import string
s = "string. With. Punctuation?" # Sample string
out = s.translate(string.maketrans("",""), string.punctuation)
在那儿?
正则表达式很简单,如果你知道的话。
import re
s = "string. With. Punctuation?"
s = re.sub(r'[^\w\s]','',s)
效率的角度来看,你不会打败
s.translate(None, string.punctuation)
它使用查找表在C中执行原始字符串操作 - 没有太多的东西会打败你,而是编写你自己的C代码。
如果速度不是一个担心,但另一个选项,虽然是:
exclude = set(string.punctuation)
s = ''.join(ch for ch in s if ch not in exclude)
这比使用每个字符的s.replace更快,但是不会像非正式的Python方法(如regexes或者string.translate)那样执行,正如您从下面的时间点可以看到的那样。对于这种类型的问题,在尽可能低的水平上做到这一点是值得的。
时间码:
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)
这给出了以下结果:
sets : 19.8566138744
regex : 6.86155414581
translate : 2.12455511093
replace : 28.4436721802
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