从字符串中去掉标点符号的最佳方法

  • 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