Reputation: 64603

Variable-length lookbehind-assertion alternatives for regular expressions

Is there an implementation of regular expressions in Python/PHP/JavaScript that supports variable-length lookbehind-assertion?

/(?<!foo.*)bar/

How can I write a regular expression that has the same meaning, but uses no lookbehind-assertion?

Is there a chance that this type of assertion will be implemented some day?

Things are much better that I thought.

Update:

(1) There are regular expressions implementation that support variable-length lookbehind-assertion already.

Python module regex (not standard re, but additional regex module) supports such assertions (and has many other cool features).

>>> import regex
>>> m = regex.search('(?<!foo.*)bar', 'f00bar')
>>> print m.group()
bar
>>> m = regex.search('(?<!foo.*)bar', 'foobar')
>>> print m
None

It was a really big surprise for me that there is something in regular expressions that Perl can't do and Python can. Probably, there is "enhanced regular expression" implementation for Perl also?

(Thanks and +1 to MRAB).

(2) There is a cool feature \K in modern regular expressions.

This symbols means that when you make a substitution (and from my point of view the most interesting use case of assertions is the substitution), all characters that were found before \K must not be changed.

s/unchanged-part\Kchanged-part/new-part/x

That is almost like a look-behind assertion, but not so flexible of course.

More about \K:

As far as I understand, you can't use \K twice in the same regular expression. And you can't say till which point you want to "kill" the characters that you've found. That is always till the beginning of the line.

(Thanks and +1 to ikegami).

My additional questions:

Is it possible to say what point must be the final point of \K effect?
What about enhanced regular expressions implementations for Perl/Ruby/JavaScript/PHP? Something like regex for Python.

Upvotes: 53

Answers (5)

ikegami

Reputation: 386396

Most of the time, you can avoid variable length lookbehinds by using \K.

s/(?<=foo.*)bar/moo/s;

would be

s/foo.*\Kbar/moo/s;

Anything up to the last \K encountered is not considered part of the match (e.g. for the purposes of replacement, $&, etc)

Negative lookbehinds are a little trickier.

s/(?<!foo.*)bar/moo/s;

would be

s/^(?:(?!foo).)*\Kbar/moo/s;

because (?:(?!STRING).)* is to STRING as [^CHAR]* is to CHAR.

If you're just matching, you might not even need the \K.

/foo.*bar/s

/^(?:(?!foo).)*bar/s

Upvotes: 54

twasbrillig

Reputation: 18911

foo.*|(bar)

If foo is in the string first, then the regex will match, but there will be no groups.

Otherwise, it will find bar and assign it to a group.

So you can use this regex and look for your results in the groups found:

>>> import re
>>> m = re.search('foo.*|(bar)', 'f00bar')
>>> if m: print(m.group(1))
bar
>>> m = re.search('foo.*|(bar)', 'foobar')
>>> if m: print(m.group(1))
None
>>> m = re.search('foo.*|(bar)', 'fobas')
>>> if m: print(m.group(1))
>>>

Source.

Upvotes: 3

Alex D

Reputation: 30465

The regexp you show will find any instance of bar which is not preceded by foo.

A simple alternative would be to first match foo against the string, and find the index of the first occurrence. Then search for bar, and see if you can find an occurrence which comes before that index.

If you want to find instances of bar which are not directly preceded by foo, I could also provide a regexp for that (without using lookbehind), but it will be very ugly. Basically, invert the sense of /foo/ -- i.e. /[^f]oo|[^o]o|[^o]|$/.

Upvotes: 3