Regular expression [13579]?[13579]? returns infinite (as http://regexr.com/ says).
Why? I just want to find two jointed odd numbers (two, not more) 😒.
The ? character in RegEx means zero or one of the preceding set. So, your regular expression would match literally everything, as well as two odd numbers in a row.
You'll probably want something like:
[13579]{2}
Debuggex Demo
Which means two and only two of the preceding set.
Related
I am trying to learn to handle Regex Expressions and got some exercises but no solutions to it. One Question is: all lower-case words except 'if'.
Can I do this one like this:
[a-z][a-z]^[if] | [a-z][a-z][a-z]+
I'm expect that a word has at least two characters. So every word with three or more is okay.
Well... the full real solution would be something like that:
\b(?!if\b)\p{Ll}+\b
Demo
But I suppose it's, well, "higher level" regex that you didn't learn yet.
So, let's keep things simple. If you can accept to ignore words of less than 3 characters, you can write this:
\b[a-hj-z][a-eg-z][a-z]+|i[a-z]{2,}
Demo
The first two character classes are just [a-z] without i and f respectively.
If you want to include words of less than 3 characters, this will do:
\b(?:i|if[a-z]+|i[a-eg-z][a-z]*|[a-hj-z][a-z]*)\b
Demo
But it gets complicated at this point...
All sequences of two or more lower-case letters, except "if":
[a-hj-z][a-z]+|i(?:[a-eg-z][a-z]*|f[a-z]+)
With negative look-ahead, you can also do:
(?!if\b)[a-z]{2,}
A simple solution would be to place what you want to ignore on the left side of the alternation operator and place what you want to match in a capturing group on the right side of the alternation operator as you were attempting.
\bif\b|([a-z]{2,})
Note: The caret ^ outside of a character class does not mean negation, it asserts the position at start of the string. And unless you are using the x (free-spacing) modifier, you need to remove the spaces between the alternation.
This has been gone over but I've not found anything that works consistently... or assist me in learning where I've gone awry.
I have file names that start with 3 or more digits and ^\d{3}(.*) works just fine.
I also have strings that start with the word 'account' and ^ACCOUNT(.*) works just fine for these.
The problem I have is all the other strings that DO NOT meet the two previous criteria. I have been using ^[^\d{3}][^ACCOUNT](.*) but occasionally it fails to catch one.
Any insights would be greatly appreciated.
^[^\d{3}][^ACCOUNT](.*)
That's definitely not what you want. Square brackets create character classes: they match one character from the list of characters in brackets. If you put a ^ then the match is inverted and it matches one character that's not listed. The meaning of ^ inside brackets is completely different from its meaning outside.
In short, [] is not at all what you want. What you can do, if your regex implementation supports it, is use a negative lookahead assertion.
^(?!\d{3}|ACCOUNT)(.*)
This negative lookahead assertion doesn't match anything itself. It merely checks that the next part of the string (.*) does not match either \d{3} or ACCOUNT.
Demorgan's law says: !(A v B) = !A ^ !B.
But unfortunately Regex itself does
not support the negation of expressions. (You always could rewrite it, but sometimes, this is a huge task).
Instead, you should look at your Programming Language, where you can negate values without problems:
let the "matching" function be "match" and you are using match("^(?:\d{3}|ACCOUNT)(.)") to determine, whether the string matches one of both conditions. Then you could simple negate the boolean return value of that matching function and you'll receive every string that does NOT match.
I'm trying to parse a document that has reference numbers littered throughout it.
Text text text {4:2} more incredible text {4:3} much later on
{222:115} and yet some more text.
The references will always be wrapped in brackets, and there will always be a colon between the two. I wrote an expression to find them.
{[0-9]:[0-9]}
However, this obviously fails the moment you come across a two or three digit number, and I'm having trouble figuring out what that should be. There won't ever be more than 3 digits {999:999} is the maximum size to deal with.
Anybody have an idea of a proper expression for handling this?
{[0-9]+:[0-9]+}
try adding plus(es)
What regex engine are you using? Most of them will support the following expression:
\{\d+:\d+\}
The \d is actually shorthand for [0-9], but the important part is the addition of + which means "one or more".
Try this:
{[0-9]{1,3}:[0-9]{1,3}}
The {1,3} means "match between 1 and 3 of the preceding characters".
You can specify how many times you want the previous item to match by using {min,max}.
{[0-9]{1,3}:[0-9]{1,3}}
Also, you can use \d for digits instead of [0-9] for most regex flavors:
{\d{1,3}:\d{1,3}}
You may also want to consider escaping the outer { and }, just to make it clear that they are not part of a repetition definition.
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?)