A regex to detect string not enclosed in double quotes

Question

I have a string something like this

"quick" "brown" fox jumps "over" "the" lazy dog

I need a regex to detect words not enclosed in double quotes. After some random tries I found this ("([^"]+)"). This detects a string enclosed in double quotes. But I want the opposite. I really can't come up with it even after trying to reverse the above mentioned regex. I am quite weak in regex. Please help me

you can match all strings, that a quoted with `"` and add `NOT` for this filter :) — gaussblurinc, Jul 04 '12 at 07:58
okay, if you use `Perl`, you can use `$str!~/patternMatchesQuotedStrings/` - `NOT` filter for your word. — gaussblurinc, Jul 04 '12 at 10:02
Related: [How to match something with regex that is not between two special characters?](https://stackoverflow.com/q/11502598/3357935) — Stevoisiak, Sep 17 '18 at 15:29

Igor Chubin · Accepted Answer · 2012-07-04T08:19:03.553

33

Use lookahead/lookbehind assertions:

(?<![\S"])([^"\s]+)(?![\S"])

Example:

>>> import re
>>> a='"quick" "brown" fox jumps "over" "the" lazy dog'
>>> print re.findall('(?<![\S"])([^"\s]+)(?![\S"])',a)
['fox', 'jumps', 'lazy', 'dog']

The main thing here is lookahead/lookbehind assertions. You can say: I want this symbol before the expression but I don't want it to be a part of the match itself. Ok. For that you use assertions:

(?<![\S"])abc

That is a negative lookbehind. That means you want abc but without [\S"] before it, that means there must be no non-space character (beginning of the word) or " before.

That is the same but in the other direction:

abc(?![\S"])

That is a negative lookahead. That means you want abc but without [\S"] after it.

There are four differenet assertions of the type in general:

(?=pattern)
    is a positive look-ahead assertion
(?!pattern)
    is a negative look-ahead assertion
(?<=pattern)
    is a positive look-behind assertion
(?<!pattern)
    is a negative look-behind assertion

edited Jul 04 '12 at 08:19

answered Jul 04 '12 at 08:05

Igor Chubin

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thanks a lot, that worked like magic :) Just one more favor, can you explain it a bit? Looks quite a bit complex – Shades88 Jul 04 '12 at 08:11
1

c'mon...you know their type as `lookahead/lookbehind assertions`. you can definitely explain me how that is working !! – Shades88 Jul 04 '12 at 08:18
1

does Python supports an Negation of regexps like `findAllExcept(/pattern/)`? – gaussblurinc Jul 04 '12 at 10:04
1

@loldop: what do you mean, "find all except"? you can use the same re but get not a list of strings but a list of indexes (something like: (a,b), (c,d) and so on) and then (0,a), (b,c), (d,-1) is what youa re looking for. right? – Igor Chubin Jul 04 '12 at 10:21
1

i mean, if i have simple condition, i will find all words, that DON'T match this condition – gaussblurinc Jul 04 '12 at 10:41
@IgorChubin , Can we improve this to exclude singlequoted strings as well?. I tried (?<![\\S\"[\\S']])([^\"\\s]+)(?![\\S\"[\\S']]) in vain. – NishM Jun 20 '16 at 22:48
@NishM: Why double brackets? Just write ["\'] – Igor Chubin Jun 21 '16 at 10:03
The double quote seems redundant in `[\S"]` since non-space characters include `"`. This seems to do the same `(?<=\s)([^"\s]+)(?=\s)`. – Scratte Sep 16 '21 at 17:01

score 0 · Answer 2 · answered Jul 04 '12 at 08:03

0

use this regex:

\s+(?<myword>([^\"\s]+)*)\s+

this should be work; and get group named myword. else you need to trim your result string.

answered Jul 04 '12 at 08:03

Ria

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that will not work for words with just one space between: `fox jumps` for example – Igor Chubin Jul 04 '12 at 08:06

score -3 · Answer 3 · answered Jul 04 '12 at 08:03

-3

Remove the first quote from the string

answered Jul 04 '12 at 08:03

Vilius Gaidelis

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A regex to detect string not enclosed in double quotes

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