在Python中，用空格分隔字符串——保留带引号的子字符串

我有一个这样的字符串:

this is "a test"

我试图在Python中写一些东西，通过空格分割它，同时忽略引号中的空格。我想要的结果是:

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

PS，我知道你会问“如果引号中有引号会发生什么，在我的应用程序中，这永远不会发生。

当前回答

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']

2015-09-09 13:19:59

其他回答

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()]

2010-01-29 01:36:23

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

2017-03-26 23:08:09

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'"]

2008-09-17 04:31:38

不同答案的速度测试:

import re
import shlex
import csv

line = 'this is "a test"'

%timeit [p for p in re.split("( |\\\".*?\\\"|'.*?')", line) if p.strip()]
100000 loops, best of 3: 5.17 µs per loop

%timeit re.findall(r'[^"\s]\S*|".+?"', line)
100000 loops, best of 3: 2.88 µs per loop

%timeit list(csv.reader([line], delimiter=" "))
The slowest run took 9.62 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 2.4 µs per loop

%timeit shlex.split(line)
10000 loops, best of 3: 50.2 µs per loop

2018-04-12 08:28:50