CODEWITHSUNDEEP
pythonregexstring
Animesh Sharma
Animesh Sharma

Reputation: 3386

Extract unquoted text from a string

I have a string that may contain random segments of quoted and unquoted texts. For example,

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

I want to separate out the quoted and unquoted parts of this string in python.

So, basically I expect the output to be:

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

I believe using a regex is the best way to do it. But I am not very good with regex. Is there some regex expression that can help me with this? Or is there a better solution available?

Upvotes: 2

Views: 1168

Answers (4)

bufh
bufh

Reputation: 3420

If your quotes are as basic as in your example, you could just split; example:

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],)

Otherwise you could try:

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

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

I explain the regex (I use it in ihih):

(?<!\\)(?:\\\\)*           # 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

Regular expression visualization

Debuggex Demo / Regex101 Demo

Upvotes: 3

Andr&#233; Laszlo
Andr&#233; Laszlo

Reputation: 15537

You should use Python's shlex module, it's very nice:

>>> 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'])

I think this solution is as simple as any other solution (basically a oneliner, if you remove the function declaration and grouping) and it handles nested quotes well etc.

Upvotes: 1

Maroun
Maroun

Reputation: 96016

Use a regex for that:

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

will return 

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

Upvotes: 1

Bas Jansen
Bas Jansen

Reputation: 3343

I dislike regex for something like this, why not just use a split like this?

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

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

Upvotes: 3

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