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.
Related
In a regular expression, I need to know how to match one thing or another, or both (in order). But at least one of the things needs to be there.
For example, the following regular expression
/^([0-9]+|\.[0-9]+)$/
will match
234
and
.56
but not
234.56
While the following regular expression
/^([0-9]+)?(\.[0-9]+)?$/
will match all three of the strings above, but it will also match the empty string, which we do not want.
I need something that will match all three of the strings above, but not the empty string. Is there an easy way to do that?
UPDATE:
Both Andrew's and Justin's below work for the simplified example I provided, but they don't (unless I'm mistaken) work for the actual use case that I was hoping to solve, so I should probably put that in now. Here's the actual regexp I'm using:
/^\s*-?0*(?:[0-9]+|[0-9]{1,3}(?:,[0-9]{3})+)(?:\.[0-9]*)?(\s*|[A-Za-z_]*)*$/
This will match
45
45.988
45,689
34,569,098,233
567,900.90
-9
-34 banana fries
0.56 points
but it WON'T match
.56
and I need it to do this.
The fully general method, given regexes /^A$/ and /^B$/ is:
/^(A|B|AB)$/
i.e.
/^([0-9]+|\.[0-9]+|[0-9]+\.[0-9]+)$/
Note the others have used the structure of your example to make a simplification. Specifically, they (implicitly) factorised it, to pull out the common [0-9]* and [0-9]+ factors on the left and right.
The working for this is:
all the elements of the alternation end in [0-9]+, so pull that out: /^(|\.|[0-9]+\.)[0-9]+$/
Now we have the possibility of the empty string in the alternation, so rewrite it using ? (i.e. use the equivalence (|a|b) = (a|b)?): /^(\.|[0-9]+\.)?[0-9]+$/
Again, an alternation with a common suffix (\. this time): /^((|[0-9]+)\.)?[0-9]+$/
the pattern (|a+) is the same as a*, so, finally: /^([0-9]*\.)?[0-9]+$/
Nice answer by huon (and a bit of brain-twister to follow it along to the end). For anyone looking for a quick and simple answer to the title of this question, 'In a regular expression, match one thing or another, or both', it's worth mentioning that even (A|B|AB) can be simplified to:
A|A?B
Handy if B is a bit more complex.
Now, as c0d3rman's observed, this, in itself, will never match AB. It will only match A and B. (A|B|AB has the same issue.) What I left out was the all-important context of the original question, where the start and end of the string are also being matched. Here it is, written out fully:
^(A|A?B)$
Better still, just switch the order as c0d3rman recommended, and you can use it anywhere:
A?B|A
Yes, you can match all of these with such an expression:
/^[0-9]*\.?[0-9]+$/
Note, it also doesn't match the empty string (your last condition).
Sure. You want the optional quantifier, ?.
/^(?=.)([0-9]+)?(\.[0-9]+)?$/
The above is slightly awkward-looking, but I wanted to show you your exact pattern with some ?s thrown in. In this version, (?=.) makes sure it doesn't accept an empty string, since I've made both clauses optional. A simpler version would be this:
/^\d*\.?\d+$/
This satisfies your requirements, including preventing an empty string.
Note that there are many ways to express this. Some are long and some are very terse, but they become more complex depending on what you're trying to allow/disallow.
Edit:
If you want to match this inside a larger string, I recommend splitting on and testing the results with /^\d*\.?\d+$/. Otherwise, you'll risk either matching stuff like aaa.123.456.bbb or missing matches (trust me, you will. JavaScript's lack of lookbehind support ensures that it will be possible to break any pattern I can think of).
If you know for a fact that you won't get strings like the above, you can use word breaks instead of ^$ anchors, but it will get complicated because there's no word break between . and (a space).
/(\b\d+|\B\.)?\d*\b/g
That ought to do it. It will block stuff like aaa123.456bbb, but it will allow 123, 456, or 123.456. It will allow aaa.123.456.bbb, but as I've said, you'll need two steps if you want to comprehensively handle that.
Edit 2: Your use case
If you want to allow whitespace at the beginning, negative/positive marks, and words at the end, those are actually fairly strict rules. That's a good thing. You can just add them on to the simplest pattern above:
/^\s*[-+]?\d*\.?\d+[a-z_\s]*$/i
Allowing thousands groups complicates things greatly, and I suggest you take a look at the answer I linked to. Here's the resulting pattern:
/^\s*[-+]?(\d+|\d{1,3}(,\d{3})*)?(\.\d+)?\b(\s[a-z_\s]*)?$/i
The \b ensures that the numeric part ends with a digit, and is followed by at least one whitespace.
Maybe this helps (to give you the general idea):
(?:((?(digits).^|[A-Za-z]+)|(?<digits>\d+))){1,2}
This pattern matches characters, digits, or digits following characters, but not characters following digits.
The pattern matches aa, aa11, and 11, but not 11aa, aa11aa, or the empty string.
Don't be puzzled by the ".^", which means "a character followd by line start", it is intended to prevent any match at all.
Be warned that this does not work with all flavors of regex, your version of regex must support (?(named group)true|false).
For example, in this string with no \s:
abodnpjdcqe
only d should be matched.
But in my case there are thousands of different characters, is it possible to use ONLY regxp to match all characters that appear in the string more than once? It seems that all other problems use other tools.
It is possible to find characters that are present two times in a string as anubhava demonstrates it, and I don't see any other regex pattern to do it.
However, there are problems with an only regex way:
The complexity of this kind of pattern is very high, and you will experience problems (with backtracking limits and execution time) if your string is long and if there are few duplicates.
This way is unable to see if a duplicate character have been already found. For example the string a123a456a789a, the pattern will return a three times instead of one. If your goal is to obtain a list of unique duplicate characters, it can be problematic (but easy to solve programmatically)
So, to answer your question: my answer is no.
a simple way, to do it with code is to loop over the characters of your string and to build an associative array where the keys are the characters and the values the number of occurences. Then, removes each item that has the value 1 and extract the keys.
Note: you can solve the problem of duplicate results (2.) using this pattern:
(.)(?=(?:(?!\1).)*\1(?:(?!\1).)*$)
or if possessive quantifiers are available:
(.)(?=(?:(?!\1).)*+\1(?:(?!\1).)*+$)
but I'm afraid that the complexity may be even more high.
So, using your favorite language stay from far the best way.
You can use this regex:
([a-zA-Z])(?=.*\1)
Explanation:
Regex uses ([a-zA-Z]) to match any letter and captures it as group #1 i.e. \1
A positive lookahead (?=.*\1) then makes sure this match is successful only when it is followed by at least one of the backreference \1 i.e. the character itself.
RegEx Demo
I've just learned about these two concepts in more detail. I've always been good with RegEx, and it seems I've never seen the need for these 2 zero width assertions.
I'm pretty sure I'm wrong, but I do not see why these constructs are needed. Consider this example:
Match a 'q' which is not followed by a 'u'.
2 strings will be the input:
Iraq
quit
With negative lookahead, the regex looks like this:
q(?!u)
Without it, it looks like this:
q[^u]
For the given input, both of these regex give the same results (i.e. matching Iraq but not quit) (tested with perl). The same idea applies to lookbehinds.
Am I missing a crucial feature that makes these assertions more valuable than the classic syntax?
Why your test probably worked (and why it shouldn't)
The reason you were able to match Iraq in your test might be that your string contained a \n at the end (for instance, if you read it from the shell). If you have a string that ends in q, then q[^u] cannot match it as the others said, because [^u] matches a non-u character - but the point is there has to be a character.
What do we actually need lookarounds for?
Obviously in the above case, lookaheads are not vital. You could workaround this by using q(?:[^u]|$). So we match only if q is followed by a non-u character or the end of the string. There are much more sophisticated uses for lookaheads though, which become a pain if you do them without lookaheads.
This answer tries to give an overview of some important standard situations which are best solved with lookarounds.
Let's start with looking at quoted strings. The usual way to match them is with something like "[^"]*" (not with ".*?"). After the opening ", we simply repeat as many non-quote characters as possible and then match the closing quote. Again, a negated character class is perfectly fine. But there are cases, where a negated character class doesn't cut it:
Multi-character delimiters
Now what if we don't have double-quotes to delimit our substring of interest, but a multi-character delimiter. For instance, we are looking for ---sometext---, where single and double - are allowed within sometext. Now you can't just use [^-]*, because that would forbid single -. The standard technique is to use a negative lookahead at every position, and only consume the next character, if it is not the beginning of ---. Like so:
---(?:(?!---).)*---
This might look a bit complicated if you haven't seen it before, but it's certainly nicer (and usually more efficient) than the alternatives.
Different delimiters
You get a similar case, where your delimiter is only one character but could be one of two (or more) different characters. For instance, say in our initial example, we want to allow for both single- and double-quoted strings. Of course, you could use '[^']*'|"[^"]*", but it would be nice to treat both cases without an alternative. The surrounding quotes can easily be taken care of with a backreference: (['"])[^'"]*\1. This makes sure that the match ends with the same character it began with. But now we're too restrictive - we'd like to allow " in single-quoted and ' in double-quoted strings. Something like [^\1] doesn't work, because a backreference will in general contain more than one character. So we use the same technique as above:
(['"])(?:(?!\1).)*\1
That is after the opening quote, before consuming each character we make sure that it is not the same as the opening character. We do that as long as possible, and then match the opening character again.
Overlapping matches
This is a (completely different) problem that can usually not be solved at all without lookarounds. If you search for a match globally (or want to regex-replace something globally), you may have noticed that matches can never overlap. I.e. if you search for ... in abcdefghi you get abc, def, ghi and not bcd, cde and so on. This can be problem if you want to make sure that your match is preceded (or surrounded) by something else.
Say you have a CSV file like
aaa,111,bbb,222,333,ccc
and you want to extract only fields that are entirely numerical. For simplicity, I'll assume that there is no leading or trailing whitespace anywhere. Without lookarounds, we might go with capturing and try:
(?:^|,)(\d+)(?:,|$)
So we make sure that we have the start of a field (start of string or ,), then only digits, and then the end of a field (, or end of string). Between that we capture the digits into group 1. Unfortunately, this will not give us 333 in the above example, because the , that precedes it was already part of the match ,222, - and matches cannot overlap. Lookarounds solve the problem:
(?<=^|,)\d+(?=,|$)
Or if you prefer double negation over alternation, this is equivalent to
(?<![^,])\d+(?![^,])
In addition to being able to get all matches, we get rid of the capturing which can generally improve performance. (Thanks to Adrian Pronk for this example.)
Multiple independent conditions
Another very classic example of when to use lookarounds (in particular lookaheads) is when we want to check multiple conditions on an input at the same time. Say we want to write a single regex that makes sure our input contains a digit, a lower case letter, an upper case letter, a character that is none of those, and no whitespace (say, for password security). Without lookarounds you'd have to consider all permutations of digit, lower case/upper case letter, and symbol. Like:
\S*\d\S*[a-z]\S*[A-Z]\S*[^0-9a-zA_Z]\S*|\S*\d\S*[A-Z]\S*[a-z]\S*[^0-9a-zA_Z]\S*|...
Those are only two of the 24 necessary permutations. If you also want to ensure a minimum string length in the same regex, you'd have to distribute those in all possible combinations of the \S* - it simply becomes impossible to do in a single regex.
Lookahead to the rescue! We can simply use several lookaheads at the beginning of the string to check all of these conditions:
^(?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[^0-9a-zA-Z])(?!.*\s)
Because the lookaheads don't actually consume anything, after checking each condition the engine resets to the beginning of the string and can start looking at the next one. If we wanted to add a minimum string length (say 8), we could simply append (?=.{8}). Much simpler, much more readable, much more maintainable.
Important note: This is not the best general approach to check these conditions in any real setting. If you are making the check programmatically, it's usually better to have one regex for each condition, and check them separately - this let's you return a much more useful error message. However, the above is sometimes necessary, if you have some fixed framework that lets you do validation only by supplying a single regex. In addition, it's worth knowing the general technique, if you ever have independent criteria for a string to match.
I hope these examples give you a better idea of why people would like to use lookarounds. There are a lot more applications (another classic is inserting commas into numbers), but it's important that you realise that there is a difference between (?!u) and [^u] and that there are cases where negated character classes are not powerful enough at all.
q[^u] will not match "Iraq" because it will look for another symbol.
q(?!u) however, will match "Iraq":
regex = /q[^u]/
/q[^u]/
regex.test("Iraq")
false
regex.test("Iraqf")
true
regex = /q(?!u)/
/q(?!u)/
regex.test("Iraq")
true
Well, another thing along with what others mentioned with the negative lookahead, you can match consecutive characters (e.g. you can negate ui while with [^...], you cannot negate ui but either u or i and if you try [^ui]{2}, you will also negate uu, ii and iu.
The whole point is to not "consume" the next character(s), so that it can be e.g. captured by another expression that comes afterwards.
If they're the last expression in the regex, then what you've shown are equivalent.
But e.g. q(?!u)([a-z]) would let the non-u character be part of the next group.
I need to define a regex for a string with the following requirements:
Maximum 20 characters
Must be in the form Name,Surname
No numbers and special characters allowed (again, it's a name&surname)
I already tried something like ^[^1-9\?\*\.\?\$\^\_]{1,20}[,][^1-9\?\*\.\?\$\^\_\-]{1,20}$ but as you can find, it also matches a 40 chars long string.
How can I check for the whole string's maximum length and at the same time impose 1 comma inside of it and obviously not at the borders?
Thank you
Try the regex:
^(?=[^,]+,[^,]+$)[a-zA-Z,]{1,20}$
Rubular Link
Explanation:
^ : Start anchor
(?=[^,]+,[^,]+$) : Positive lookahead to ensure string has exactly one comma
surrounded by at least one non-comma character on both sides.
[a-zA-Z,]{1,20} : Ensure entire string is of length max 20 and has only
letters and comma
$ : End anchor
You can do this using forward negative assertions:
^(?!.{21})[A-Za-z]+,[A-Za-z]+$
The regex contains two parts now, the actual definition, and a statement at the start, saying that from that point, there will not be 21 characters.
So for the definition as stated above, the regex becomes
^(?!.{21})[^1-9\?*\.\?\$\^_\,]+,[^1-9\?*\.\?\$\^_\,]+$
The obvious answer would be: Don't ask for name and surname in the same input field.
If you still want to do it: There's no easy way that I know of, but here is a possibility. To see the principle think your [^1-9\?\*\.\?\$\^\_\,] instead of X (I added he \, since it's kind of important :-)).
^(X{1},X{19})|(X{2},X{18})|...|(X{19},X{1})$
Quite ugly, but should work.
On a different note: You don't capture nearly all special characters with your exclusive range. But it's probably still better than an inclusive range.
As I say, I think stated the way you have it, it's not matchable by a regular expression -- it's a pushdown language.
However, you could always split on ',' and match each substring, then total.
I have you tried your example, but removing the
{1,20}
in the middle, leaving to try this:
^[[^1-9\?\*\.\?\$\^\_],[^1-9\?\*\.\?\$\^\_\-]]{1,20}$
Use:
[[a-zA-Z],[a-zA-Z]]{1,20}
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