I'm trying to use a Regex pattern (in Java) to find a sequence of 3 digits and only 3 digits in a row. 4 digits doesn't match, 2 digits doesn't match.
The obvious pattern to me was:
"\b(\d{3})\b"
That matches against many source string cases, such as:
">123<"
" 123-"
"123"
But it won't match against a source string of "abc123def" because the c/1 boundary and the 3/d boundary don't count as a "word boundary" match that the \b class is expecting.
I would have expected the solution to be adding a character class that includes both non-Digit (\D) and the word boundary (\b). But that appears to be illegal syntax.
"[\b\D](\d{3})[\b\D]"
Does anybody know what I could use as an expression that would extract "123" for a source string situation like:
"abc123def"
I'd appreciate any help. And yes, I realize that in Java one must double-escape the codes like \b to \b, but that's not my issue and I didn't want to limit this to Java folks.
You should use lookarounds for those cases:
(?<!\d)(\d{3})(?!\d)
This means match 3 digits that are NOT followed and preceded by a digit.
Working Demo
Lookarounds can solve this problem, but I personally try to avoid them because not all regex engines fully support them. Additionally, I wouldn't say this issue is complicated enough to merit the use of lookarounds in the first place.
You could match this: (?:\b|\D)(\d{3})(?:\b|\D)
Then return: \1
Or if you're performing a replacement and need to match the entire string: (?:\b|\D)+(\d{3})(?:\b|\D)+
Then replace with: \1
As a side note, the reason \b wasn't working as part of a character class was because within brackets, [\b] actually has a completely different meaning--it refers to a backspace, not a word boundary.
Here's a Working Demo.
Related
I am trying to create a regular expression that will identify possible abbreviations within a given string in Python. I am kind of new to RegEx and I am having difficulties creating an expression though I beleive it should be somewhat simple. The expression should pick up words that have two or more capitalised letter. The expression should also be able to pick up words where a dash have been used in-between and report the whole word (both before and after the dash). If numbers are also present they should also be reported with the word.
As such, it should pick up:
ABC, AbC, ABc, A-ABC, a-ABC, ABC-a, ABC123, ABC-123, 123-ABC.
I have already made the following expression: r'\b(?:[a-z]*[A-Z\-][a-z\d[^\]*]*){2,}'.
However this does also pick up these wrong words:
A-bc, a-b-c
I believe the problem is that it looks for either multiple capitalised letters or dashes. I wish for it to only give me words that have atleast two or more capitalised letters. I understand that it will also "mistakenly" take words as "Abc-Abc" but I don't believe there is a way to avoid these.
If a lookahead is supported and you don't want to match double -- you might use:
\b(?=(?:[a-z\d-]*[A-Z]){2})[A-Za-z\d]+(?:-[A-Za-z\d]+)*\b
Explanation
\b A word boundary
(?= Positive lookahead, assert that from the current location to the right is
(?:[a-z\d-]*[A-Z]){2} Match 2 times the optionally the allowed characters and an uppercase char A-Z
) Close the lookahead
[A-Za-z\d]+ match 1+ times the allowed characters without the hyphen
(?:-[A-Za-z\d]+)* Optionally repeat - and 1+ times the allowed characters
\b A word boundary
See a regex101 demo.
To also not not match when there are hyphens surrounding the characters you can use negative lookarounds asserting not a hyphen to the left or right.
\b(?<!-)(?=(?:[a-z\d-]*[A-Z]){2})[A-Za-z\d]+(?:-[A-Za-z\d]+)*\b(?!-)
See another regex demo.
I am trying to implement a regex which includes all the strings which have any number of words but cannot be followed by a : and ignore the match if it does. I decided to use a negative look ahead for it.
/([a-zA-Z]+)(?!:)/gm
string: lame:joker
since i am using a character range it is matching one character at a time and only ignoring the last character before the : .
How do i ignore the entire match in this case?
Link to regex101: https://regex101.com/r/DlEmC9/1
The issue is related to backtracking: once your [a-zA-Z]+ comes to a :, the engine steps back from the failing position, re-checks the lookahead match and finds a match whenver there are at least two letters before a colon, returning the one that is not immediately followed by :. See your regex demo: c in c:real is not matched as there is no position to backtrack to, and rea in real:c is matched because a is not immediately followed with :.
Adding implicit requirement to the negative lookahead
Since you only need to match a sequence of letters not followed with a colon, you can explicitly add one more condition that is implied: and not followed with another letter:
[A-Za-z]+(?![A-Za-z]|:)
[A-Za-z]+(?![A-Za-z:])
See the regex demo. Since both [A-Za-z] and : match a single character, it makes sense to put them into a single character class, so, [A-Za-z]+(?![A-Za-z:]) is better.
Preventing backtracking into a word-like pattern by using a word boundary
As #scnerd suggests, word boundaries can also help in these situations, but there is always a catch: word boundary meaning is context dependent (see a number of ifs in the word boundary explanation).
[A-Za-z]+\b(?!:)
is a valid solution here, because the input implies the words end with non-word chars (i.e. end of string, or chars other than letter, digits and underscore). See the regex demo.
When does a word boundary fail?
\b will not be the right choice when the main consuming pattern is supposed to match even if glued to other word chars. The most common example is matching numbers:
\d+\b(?!:) matches 12 in 12,, but not in 12:, and also 12c and 12_
\d+(?![\d:]) matches 12 in 12, and 12c and 12_, not in 12: only.
Do a word boundary check \b after the + to require it to get to the end of the word.
([a-zA-Z]+\b)(?!:)
Here's an example run.
I really don't use RegEx that much. You could say I am RegEx n00b. I have been working on this issue for a half a day.
I am trying to write a pattern that looks backward from a number character. For example:
1. bob1 => bob
2. cat3 => cat
3. Mary34 => Mary
So far I have this (?![A-Z][a-z]{1,})([A-Za-z_])
It only matches for individual characters, I want all the characters before the number character. I tried to add the ^ and $ into my pattern and using an online simulator. I am unsure where to put the ^ and $.
NOTE: I am using RegEx for the .NET Framework
You may use a regex like
[\p{L}_]+(?=\d)
or
[\w-[\d]]+(?=\d)
See the regex demo
Pattern details
[\p{L}_]+ - any 1 or more letters (both lower- and uppercase) and/or _
OR
[\w-[\d]]+ - 1 or more word chars except digits (the -[] inside a character class is a character class subtraction construct)
(?=\d) - a positive lookahead that requires a digit to appear immediately to the right of the current location
If we break down your RegEx, we see:
(?![A-Z][a-z]{1,}) which says "look ahead to find a string that is NOT one uppercase letter followed one or more lowercase letters" and ([A-Za-z_]) which says "match one letter or underscore". This should end up matching any single lowercase letter.
If I understand what you want to achieve, then you want all of the letters before a number. I would write something like that as:
\b([a-zA-Z]+)[0-9]
This will start at a word boundary \b, match one or more letters, and require a digit right after the matched string.
(The syntax I used seems to match this document about .NET RegEx: https://learn.microsoft.com/en-us/dotnet/standard/base-types/regular-expressions)
In light of Wiktor Stribizew's comment, here is a pure match RegEx:
\b[a-zA-Z_]+(?=[0-9])
This matches the pattern and then looks ahead for the digit. This is better than my first lookahead attempt. (Thank you Wiktor.)
http://www.rexegg.com/regex-lookarounds.html
I need to find all the words in an inputted text that has (?i:val) in it and are no longer that 5 characters.
So far I got: \b([a-zA-Z]*(?i:val)[a-zA-Z]*){1,4}\b
If we take this sample text to look in: In computer science, a value is an expression which cannot be evaluated any further (a normal form). Val is also a match
I get 3 matches (value, evaluated and Val), however evaluated should not match the pattern, as it is too long. What is the right way to get this straight?
Your pattern does not account for the length of the words matched.
Use word boundaries and a lookahead like this:
(?i)\b(?=\w*val)\w{1,5}\b
See regex demo
The regex matches:
\b - a leading word boundary since the next pattern is \w
(?=\w*val) - a lookahead making sure there is a val substring after zero or more word characters
\w{1,5} - matches 1 to 5 word characters
\b - trailing word boundary that stops words of more than 5 characters long from matching
You may use an ASCII JS version of the regex:
/\b(?=[a-z]*val)[a-z]{1,5}\b/i
It's important to understand why the "evaluated" was matched. Note:
[a-zA-Z]* matches the "e"
(?i:val) matches "val"
[a-zA-Z]* matches "uated"
Actually there's not repetition here! The pattern was matched in only one iteration.
You can achieve what you want using lookarounds, but I think that regex is not the best tool for this task. I highly recommend you using other functions depending on what you have.
I am trying to match terms such as the Dutch ge-berg-te. berg is a noun by itself, and ge...te is a circumfix, i.e. geberg does not exist, nor does bergte. gebergte does. What I want is a RegEx that matches berg or gebergte, working with a lookaround. I was thinking this would work
\b(?i)(ge(?=te))?berg(te)?\b
But it doesn't. I am guessing because a lookahead only checks the immediate following characters, and not across characters. Is there any way to match characters with a lookahead withouth the constraint that those characters have to be immediately behind the others?
Valid matches would be:
Berg
berg
Gebergte
gebergte
Invalid matches could be:
Geberg
geberg
Bergte
bergte
ge-/Ge- and -te always have to occur together. Note that I want to try this with a lookahead. I know it can be done simpler, but I want to see if its methodologically possible to do something like this.
Here is one non-lookaround based regex:
\b(berg|gebergte)\b
Use it with i (ignore case) flag. This regex uses alternation and word boundary to search for complete words berg OR gebergte.
RegEx Demo
Lookaround based regex:
(?<=\bge)berg(?=te\b)|\bberg\b
This regex used a lookahead and lookbehind to search for berg preceded by ge and followed by te. Alternatively it matches complete word berg using word boundary asserter \b which is also 0-width asserter like anchors ^ and $.
To generally forbid a sign, you can put the negative lookaround to the beginning of a string and combine it with random number of other signs before the string you want to forbid:
regex: don't match if containing a specific string
^(?!.\*720).*
This will not match, if the string contains 720, but else match everything else.