从字符串中提取未加引号的文本

Question

我有一个字符串，可能包含引用和不带引号的文本的随机段。例如，

s = "\"java jobs in delhi\" it software \"pune\" hello"。

我想在python中分离出这个字符串的引用和非引用部分。

所以，基本上我希望输出为：

quoted_string = "\"java jobs in delhi\"" "\"pune\""
unquoted_string = "it software hello"

我相信使用正则表达式是最好的方法。但我对正则表达式不是很好。是否有一些正则表达式可以帮助我解决这个问题？ 或者有更好的解决方案吗？

Answer 1

我不喜欢像这样的正则表达式，为什么不使用像这样的分割呢？

s = "\"java jobs in delhi\" it software \"pune\" hello"

print s.split("\"")[0::2] # Unquoted
print s.split("\"")[1::2] # Quoted

Answer 2

如果你的报价与你的例子一样基本，你可以分开;例如：

for s in (
    '"java jobs in delhi" it software "pune" hello',
    'foo "bar"',
):
    result = s.split('"')
    print 'text between quotes: %s' % (result[1::2],)
    print 'text outside quotes: %s' % (result[::2],)

否则你可以尝试：

import re
pattern = re.compile(
    r'(?<!\\)(?:\\\\)*(?P<quote>["\'])(?P<value>.*?)(?<!\\)(?:\\\\)*(?P=quote)'
)

for s in data:
    print pattern.findall(s)

我解释了正则表达式（我在ihih中使用它）：

(?<!\\)(?:\\\\)*           # find backslash
(?P<quote>["\'])           # any quote character (either " or ')
                           # which is *not* escaped (by a backslash)
(?P<value>.*?)             # text between the quotes
(?<!\\)(?:\\\\)*(?P=quote) # end (matching) quote

Debuggex Demo

Answer 3

使用正则表达式：

re.findall(r'"(.*?)"', s)

将返回

['java jobs in delhi', 'pune']

Answer 4

你应该使用Python的shlex模块，这非常好：

>>> from shlex import shlex
>>> def get_quoted_unquoted(s):
...     lexer = shlex(s)
...     items = list(iter(lexer.get_token, ''))
...     return ([i for i in items if i[0] in "\"'"],
                [i for i in items if i[0] not in "\"'"])
... 
>>> get_quoted_unquoted("\"java jobs in delhi\" it software \"pune\" hello")
(['"java jobs in delhi"', '"pune"'], ['it', 'software', 'hello'])
>>> get_quoted_unquoted("hello 'world' \"foo 'bar' baz\" hi")
(["'world'", '"foo \'bar\' baz"'], ['hello', 'hi'])
>>> get_quoted_unquoted("does 'nested \"quotes\" work' yes")
(['\'nested "quotes" work\''], ['does', 'yes'])
>>> get_quoted_unquoted("what's up with single quotes?")
([], ["what's", 'up', 'with', 'single', 'quotes', '?'])
>>> get_quoted_unquoted("what's up when there's two single quotes")
([], ["what's", 'up', 'when', "there's", 'two', 'single', 'quotes'])

我认为这个解决方案就像任何其他解决方案一样简单（基本上是一个oneliner，如果删除函数声明和分组）并且它可以很好地处理嵌套引号等。