由于性能原因，re似乎更快。下面是我的解决方案，使用最小贪婪操作符，保留外部引号:

re.findall("(?:\".*?\"|\S)+", s)

结果:

['this', 'is', '"a test"']

它将像aaa“bla blub”bbb这样的结构放在一起，因为这些标记没有被空格分隔。如果字符串包含转义字符，你可以这样匹配:

>>> a = "She said \"He said, \\\"My name is Mark.\\\"\""
>>> a
'She said "He said, \\"My name is Mark.\\""'
>>> for i in re.findall("(?:\".*?[^\\\\]\"|\S)+", a): print(i)
...
She
said
"He said, \"My name is Mark.\""

请注意，这也通过模式的\S部分来匹配空字符串“”。

如果你不关心子字符串

>>> 'a short sized string with spaces '.split()

性能:

>>> s = " ('a short sized string with spaces '*100).split() "
>>> t = timeit.Timer(stmt=s)
>>> print "%.2f usec/pass" % (1000000 * t.timeit(number=100000)/100000)
171.39 usec/pass

或者字符串模块

>>> from string import split as stringsplit; 
>>> stringsplit('a short sized string with spaces '*100)

性能:String模块的性能似乎比字符串方法更好

>>> s = "stringsplit('a short sized string with spaces '*100)"
>>> t = timeit.Timer(s, "from string import split as stringsplit")
>>> print "%.2f usec/pass" % (1000000 * t.timeit(number=100000)/100000)
154.88 usec/pass

或者你可以使用RE引擎

>>> from re import split as resplit
>>> regex = '\s+'
>>> medstring = 'a short sized string with spaces '*100
>>> resplit(regex, medstring)

性能

>>> s = "resplit(regex, medstring)"
>>> t = timeit.Timer(s, "from re import split as resplit; regex='\s+'; medstring='a short sized string with spaces '*100")
>>> print "%.2f usec/pass" % (1000000 * t.timeit(number=100000)/100000)
540.21 usec/pass

对于非常长的字符串，您不应该将整个字符串加载到内存中，而是将行分开或使用迭代循环

要解决某些Python 2版本中的unicode问题，我建议:

from shlex import split as _split
split = lambda a: [b.decode('utf-8') for b in _split(a.encode('utf-8'))]

我建议:

测试字符串:

s = 'abc "ad" \'fg\' "kk\'rdt\'" zzz"34"zzz "" \'\''

也要捕捉""和":

import re
re.findall(r'"[^"]*"|\'[^\']*\'|[^"\'\s]+',s)

结果:

['abc', '"ad"', "'fg'", '"kk\'rdt\'"', 'zzz', '"34"', 'zzz', '""', "''"]

忽略空的""和":

import re
re.findall(r'"[^"]+"|\'[^\']+\'|[^"\'\s]+',s)

结果:

['abc', '"ad"', "'fg'", '"kk\'rdt\'"', 'zzz', '"34"', 'zzz']

试试这个:

  def adamsplit(s):
    result = []
    inquotes = False
    for substring in s.split('"'):
      if not inquotes:
        result.extend(substring.split())
      else:
        result.append(substring)
      inquotes = not inquotes
    return result

一些测试字符串:

'This is "a test"' -> ['This', 'is', 'a test']
'"This is \'a test\'"' -> ["This is 'a test'"]

嗯，好像找不到“回复”按钮了…不管怎样，这个答案是基于Kate的方法，但正确地分割了包含转义引号的子字符串，并删除了子字符串的开始和结束引号:

  [i.strip('"').strip("'") for i in re.split(r'(\s+|(?<!\\)".*?(?<!\\)"|(?<!\\)\'.*?(?<!\\)\')', string) if i.strip()]

这适用于'这是' a \\\"test\\\"\\\'s子字符串' '这样的字符串(不幸的是，为了防止Python删除转义，这个疯狂的标记是必要的)。

如果返回列表中的字符串不需要转义，您可以使用稍微修改过的函数版本:

[i.strip('"').strip("'").decode('string_escape') for i in re.split(r'(\s+|(?<!\\)".*?(?<!\\)"|(?<!\\)\'.*?(?<!\\)\')', string) if i.strip()]