要保留引号，使用这个函数:

def getArgs(s):
    args = []
    cur = ''
    inQuotes = 0
    for char in s.strip():
        if char == ' ' and not inQuotes:
            args.append(cur)
            cur = ''
        elif char == '"' and not inQuotes:
            inQuotes = 1
            cur += char
        elif char == '"' and inQuotes:
            inQuotes = 0
            cur += char
        else:
            cur += char
    args.append(cur)
    return args

你需要从内置的shlex模块中分离。

>>> import shlex
>>> shlex.split('this is "a test"')
['this', 'is', 'a test']

这应该是你想要的效果。

如果你想保留引号，那么你可以传递posix=False kwarg。

>>> shlex.split('this is "a test"', posix=False)
['this', 'is', '"a test"']

看一下shlex模块，特别是shlex.split。

>>> import shlex
>>> shlex.split('This is "a test"')
['This', 'is', 'a test']

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'"]

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