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Regular expression to match a line that doesn't contain a word
(34 answers)
Closed 7 years ago.
Can regular expression be utilized to match any string except a specific string constant (i.e. "ABC")?
Is it possible to exclude just one specific string constant?
You have to use a negative lookahead assertion.
(?!^ABC$)
You could for example use the following.
(?!^ABC$)(^.*$)
If this does not work in your editor, try this. It is tested to work in ruby and javascript:
^((?!ABC).)*$
In .NET you can use grouping to your advantage like this:
http://regexhero.net/tester/?id=65b32601-2326-4ece-912b-6dcefd883f31
You'll notice that:
(ABC)|(.)
Will grab everything except ABC in the 2nd group. Parenthesis surround each group. So (ABC) is group 1 and (.) is group 2.
So you just grab the 2nd group like this in a replace:
$2
Or in .NET look at the Groups collection inside the Regex class for a little more control.
You should be able to do something similar in most other regex implementations as well.
UPDATE: I found a much faster way to do this here:
http://regexhero.net/tester/?id=997ce4a2-878c-41f2-9d28-34e0c5080e03
It still uses grouping (I can't find a way that doesn't use grouping). But this method is over 10X faster than the first.
This isn't easy, unless your regexp engine has special support for it. The easiest way would be to use a negative-match option, for example:
$var !~ /^foo$/
or die "too much foo";
If not, you have to do something evil:
$var =~ /^(($)|([^f].*)|(f[^o].*)|(fo[^o].*)|(foo.+))$/
or die "too much foo";
That one basically says "if it starts with non-f, the rest can be anything; if it starts with f, non-o, the rest can be anything; otherwise, if it starts fo, the next character had better not be another o".
Try this regular expression:
^(.{0,2}|([^A]..|A[^B].|AB[^C])|.{4,})$
It describes three cases:
less than three arbitrary character
exactly three characters, while either
the first is not A, or
the first is A but the second is not B, or
the first is A, the second B but the third is not C
more than three arbitrary characters
You could use negative lookahead, or something like this:
^([^A]|A([^B]|B([^C]|$)|$)|$).*$
Maybe it could be simplified a bit.
Related
This question sounds like a duplicate, but I've looked at a LOT of similar questions, and none fit the bill either because they restrict their question to a very specific example, or to a specific usercase (e.g: single chars only) or because you need substitution for a successful approach, or because you'd need to use a programming language (e.g: C#'s split, or Match().Value).
I want to be able to get the reverse of any arbitrary Regex expression, so that everything is matched EXCEPT the found match.
For example, let's say I want to find the reverse of the Regex "over" in "The cow jumps over the moon", it would match The cow jumps and also match the moon.
That's only a simple example of course. The Regex could be something more messy such as "o.*?m", in which case the matches would be: The c, ps, and oon.
Here is one possible solution I found after ages of hunting. Unfortunately, it requires the use of substitution in the replace field which I was hoping to keep clear. Also, everything else is matched, but only a character by character basis instead of big chunks.
Just to stress again, the answer should be general-purpose for any arbitrary Regex, and not specific to any particular example.
From post: I want to be able to get the reverse of any arbitrary Regex expression, so that everything is matched EXCEPT the found match.
The answer -
A match is Not Discontinuous, it is continuous !!
Each match is a continuous, unbroken substring. So, within each match there
is no skipping anything within that substring. Whatever matched the
regular expression is included in a particular match result.
So within a single Match, there is no inverting (i.e. match not this only) that can extend past
a negative thing.
This is a Tennant of Regular Expressions.
Further, in this case, since you only want all things NOT something, you have
to consume that something in the process.
This is easily done by just capturing what you want.
So, even with multiple matches, its not good enough to say (?:(?!\bover\b).)+
because even though it will match up to (but not) over, on the next match
it will match ver ....
There are ways to avoid this that are tedious, requiring variable length lookbehinds.
But, the easiest way is to match up to over, then over, then the rest.
Several constructs can help. One is \K.
Unfortunately, there is no magical recipe to negate a pattern.
As you mentioned it in your question when you have an efficient pattern you use with a match method, to obtain the complementary, the more easy (and efficient) way is to use a split method with the same pattern.
To do it with the pattern itself, workarounds are:
1. consuming the characters that match the pattern
"other content" is the content until the next pattern or the end of the string.
alternation + capture group:
(pattern)|other content
Then you must check if the capture group exists to know which part of the alternation succeeds.
"other content" can be for example described in this way: .*?(?=pattern|$)
With PCRE and Perl, you can use backtracking control verbs to avoid the capture group, but the idea is the same:
pattern(*SKIP)(*FAIL)|other content
With this variant, you don't need to check anything after, since the first branch is forced to fail.
or without alternation:
((?:pattern)*)(other content)
variant in PCRE, Perl, or Ruby with the \K feature:
(?:pattern)*\Kother content
Where \K removes all on the left from the match result.
2. checking characters of the string one by one
(?:(?!pattern).)*
if this way is very simple to write (if the lookahead is available), it has the inconvenient to be slow since each positions of the string are tested with the lookahead.
The amount of lookahead tests can be reduced if you can use the first character of the pattern (lets say "a"):
[^a]*(?:(?!pattern)a[^a]*)*
3. list all that is not the pattern.
using character classes
Lets say your pattern is /hello/:
([^h]|h(([^eh]|$)|e(([^lh]|$)|l(([^lh]|$)|l([^oh]|$))))*
This way becomes quickly fastidious when the number of characters is important, but it can be useful for regex flavors that haven't many features like POSIX regex.
How do I write a regular expression where x is a string whose characters are either a, b, c but no two consecutive characters are the same
For example
abcacb is true
acbaac is false
^(?!.*(.)\1)[abc]+$ works if you follow the original question exactly. However, this does not work/check multiple "words" of characters a/b/c, ie. "abc cba".
The way it works is it asserts that any character is not followed by itself by utilizing a capture group inside a lookahead and that the entire string consists only of characters "a", "b", or "c".
Since the number of chars is limited, you can get away without a back reference in the look ahead:
^(?!.*(aa|bb|cc)[abc]*$
But I like tenub's answer better :)
using negative lookbehind: ^([abc]([abc](?<!(aa|bb|cc)))*)?$ TRY HERE
using negative lookahead: ^(((?!(aa|bb|cc))[abc])*[abc])?$ TRY HERE
Prefer either (both do the same job but differently) if you are going to use this regex as a part of some bigger regex that you might be creating.
In short, this is reusable. Copy & paste and it will do its work without disturbing any regex that is present around it.
In my humble opinion, regexes provided in #tenub and #Bohemian are not reusable which can cause bugs.
Note: empty string ("") will pass these 2 regexes. If you don't want it to, remove ? from regex.
This question already has answers here:
My regex is matching too much. How do I make it stop? [duplicate]
(5 answers)
Closed 4 years ago.
I have this RegEx:
('.+')
It has to match character literals like in C. For example, if I have 'a' b 'a' it should match the a's and the ''s around them.
However, it also matches the b also (it should not), probably because it is, strictly speaking, also between ''s.
Here is a screenshot of how it goes wrong (I use this for syntax highlighting):
I'm fairly new to regular expressions. How can I tell the regex not to match this?
It is being greedy and matching the first apostrophe and the last one and everything in between.
This should match anything that isn't an apostrophe.
('[^']+')
Another alternative is to try non-greedy matches.
('.+?')
Have you tried a non-greedy version, e.g. ('.+?')?
There are usually two modes of matching (or two sets of quantifiers), maximal (greedy) and minimal (non-greedy). The first will result in the longest possible match, the latter in the shortest. You can read about it (although in perl context) in the Perl Cookbook (Section 6.15).
Try:
('[^']+')
The ^ means include every character except the ones in the square brackets. This way, it won't match 'a' b 'a' because there's a ' in between, so instead it'll give both instances of 'a'
You need to escape the qutoes:
\'[^\']+\'
Edit: Hmm, we'll I suppose this answer depends on what lang/system you're using.
The Greedy Option of Regex is really needed?
Lets say I have following texts, I like to extract texts inside [Optionx] and [/Optionx] blocks
[Option1]
Start=1
End=10
[/Option1]
[Option2]
Start=11
End=20
[/Option2]
But with Regex Greedy Option, its give me
Start=1
End=10
[/Option1]
[Option2]
Start=11
End=20
Anybody need like that? If yes, could you let me know?
If I understand correctly, the question is “why (when) do you need greedy matching?”
The answer is – almost always. Consider a regular expression that matches a sequence of arbitrary – but equal – characters, of length at least two. The regular expression would look like this:
(.)\1+
(\1 is a back-reference that matches the same text as the first parenthesized expression).
Now let’s search for repeats in the following string: abbbbbc. What do we find? Well, if we didn’t have greedy matching, we would find bb. Probably not what we want. In fact, in most application s we would be interested in finding the whole substring of bs, bbbbb.
By the way, this is a real-world example: the RLE compression works like that and can be easily implemented using regex.
In fact, if you examine regular expressions all around you will see that a lot of them use quantifiers and expect them to behave greedily. The opposite case is probably a minority. Often, it makes no difference because the searched expression is inside guard clauses (e.g. a quoted string is inside the quote marks) but like in the example above, that’s not always the case.
Regular expressions can potentially match multiple portion of a text.
For example consider the expression (ab)*c+ and the string "abccababccc". There are many portions of the string that can match the regular expressions:
(abc)cababccc
(abcc)ababccc
abcc(ababccc)
abccab(abccc)
ab(c)cababccc
ab(cc)ababccc
abcabab(c)ccc
....
some regular expressions implementation are actually able to return the entire set of matches but it is most common to return a single match.
There are many possible ways to determine the "winning match". The most common one is to take the "longest leftmost match" which results in the greedy behaviour you observed.
This is tipical of search and replace (a la grep) when with a+ you probably mean to match the entire aaaa rather than just a single a.
Choosing the "shortest non-empty leftmost" match is the usual non-greedy behaviour. It is the most useful when you have delimiters like your case.
It all depends on what you need, sometimes greedy is ok, some other times, like the case you showed, a non-greedy behaviour would be more meaningful. It's good that modern implementations of regular expressions allow us to do both.
If you're looking for text between the optionx blocks, instead of searching for .+, search for anything that's not "[\".
This is really rough, but works:
\[[^\]]+]([^(\[/)]+)
The first bit searches for anything in square brackets, then the second bit searches for anything that isn't "[\". That way you don't have to care about greediness, just tell it what you don't want to see.
One other consideration: In many cases, greedy and non-greedy quantifiers result in the same match, but differ in performance:
With a non-greedy quantifier, the regex engine needs to backtrack after every single character that was matched until it finally has matched as much as it needs to. With a greedy quantifier, on the other hand, it will match as much as possible "in one go" and only then backtrack as much as necessary to match any following tokens.
Let's say you apply a.*c to
abbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbc. This finds a match in 5 steps of the regex engine. Now apply a.*?c to the same string. The match is identical, but the regex engine needs 101 steps to arrive at this conclusion.
On the other hand, if you apply a.*c to abcbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb, it takes 101 steps whereas a.*?c only takes 5.
So if you know your data, you can tailor your regex to match it as efficiently as possible.
just use this algorithm which you can use in your fav language. No need regex.
flag=0
open file for reading
for each line in file :
if check "[/Option" in line:
flag=0
if check "[Option" in line:
flag=1
continue
if flag:
print line.strip()
# you can store the values of each option in this part
Is it possible to write a regex that returns the converse of a desired result? Regexes are usually inclusive - finding matches. I want to be able to transform a regex into its opposite - asserting that there are no matches. Is this possible? If so, how?
http://zijab.blogspot.com/2008/09/finding-opposite-of-regular-expression.html states that you should bracket your regex with
/^((?!^ MYREGEX ).)*$/
, but this doesn't seem to work. If I have regex
/[a|b]./
, the string "abc" returns false with both my regex and the converse suggested by zijab,
/^((?!^[a|b].).)*$/
. Is it possible to write a regex's converse, or am I thinking incorrectly?
Couldn't you just check to see if there are no matches? I don't know what language you are using, but how about this pseudocode?
if (!'Some String'.match(someRegularExpression))
// do something...
If you can only change the regex, then the one you got from your link should work:
/^((?!REGULAR_EXPRESSION_HERE).)*$/
The reason your inverted regex isn't working is because of the '^' inside the negative lookahead:
/^((?!^[ab].).)*$/
^ # WRONG
Maybe it's different in vim, but in every regex flavor I'm familiar with, the caret matches the beginning of the string (or the beginning of a line in multiline mode). But I think that was just a typo in the blog entry.
You also need to take into account the semantics of the regex tool you're using. For example, in Perl, this is true:
"abc" =~ /[ab]./
But in Java, this isn't:
"abc".matches("[ab].")
That's because the regex passed to the matches() method is implicitly anchored at both ends (i.e., /^[ab].$/).
Taking the more common, Perl semantics, /[ab]./ means the target string contains a sequence consisting of an 'a' or 'b' followed by at least one (non-line separator) character. In other words, at ANY point, the condition is TRUE. The inverse of that statement is, at EVERY point the condition is FALSE. That means, before you consume each character, you perform a negative lookahead to confirm that the character isn't the beginning of a matching sequence:
(?![ab].).
And you have to examine every character, so the regex has to be anchored at both ends:
/^(?:(?![ab].).)*$/
That's the general idea, but I don't think it's possible to invert every regex--not when the original regexes can include positive and negative lookarounds, reluctant and possessive quantifiers, and who-knows-what.
You can invert the character set by writing a ^ at the start ([^…]). So the opposite expression of [ab] (match either a or b) is [^ab] (match neither a nor b).
But the more complex your expression gets, the more complex is the complementary expression too. An example:
You want to match the literal foo. An expression, that does match anything else but a string that contains foo would have to match either
any string that’s shorter than foo (^.{0,2}$), or
any three characters long string that’s not foo (^([^f]..|f[^o].|fo[^o])$), or
any longer string that does not contain foo.
All together this may work:
^[^fo]*(f+($|[^o]|o($|[^fo]*)))*$
But note: This does only apply to foo.
You can also do this (in python) by using re.split, and splitting based on your regular expression, thus returning all the parts that don't match the regex, how to find the converse of a regex
In perl you can anti-match with $string !~ /regex/;.
With grep, you can use --invert-match or -v.
Java Regexps have an interesting way of doing this (can test here) where you can create a greedy optional match for the string you want, and then match data after it. If the greedy match fails, it's optional so it doesn't matter, if it succeeds, it needs some extra data to match the second expression and so fails.
It looks counter-intuitive, but works.
Eg (foo)?+.+ matches bar, foox and xfoo but won't match foo (or an empty string).
It might be possible in other dialects, but couldn't get it to work myself (they seem more willing to backtrack if the second match fails?)