- 问题:
-
似乎应该有比这更简单的方法:
import string
s = "string. With. Punctuation?" # Sample string
out = s.translate(string.maketrans("",""), string.punctuation)有吗?在
- 答案:
-
从效率的角度来看,你不会赢的
s.translate(None, string.punctuation)
对于更高版本的Python,请使用以下代码:
s.translate(str.maketrans('', '', string.punctuation))
它在C语言中用查找表执行原始字符串操作—除了编写自己的C代码之外,没有什么比这更好的了
如果不担心速度,另一个选择是:
exclude = set(string.punctuation)
s = ''.join(ch for ch in s if ch not in exclude)这比s.replace with each char快,但性能不如非纯python方法,如regex或字符串.翻译,从下面的时间安排可以看出。对于这类问题,在尽可能低的水平上做是有回报的
定时代码:
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)结果如下:
sets : 19.8566138744
regex : 6.86155414581
translate : 2.12455511093
replace : 28.4436721802