I'm hoping to use a regex to parse strings which begin with an integer n. After a space, there are n characters, after which there may be more text. I'm hoping to capture n and the n characters that follow. There are no constraints on these n characters. In other words, 5 hello world should match with the capture groups 5 and hello.
I tried this regex, but it wouldn't compile because its structure depends on the input: (\d+) .{\1}.
Is there a way to get the regex compiler to do what I want, or do I have to parse this myself?
I'm using Rust's regex crate, if that matters. And if it's not possible with regex, is it possible with another, more sophisticated regex engine?
Thanks!
As #Cary Swoveland said in the comments, this is not possible in regex in one step without hard-coding the various possible lengths.
However, it is not too difficult to take a substring of the matched string with length from the matched digit:
use regex::Regex;
fn main() {
let re = Regex::new(r"(\d+) (.+)").unwrap();
let test_str = "5 hello world";
for cap in re.captures_iter(test_str) {
let length: usize = cap[1].parse().unwrap_or(0);
let short_match: String = cap[2].chars().take(length).collect();
println!("{}", short_match); // hello
}
}
If you know you'll only be dealing with ASCII characters (no Unicode, accent marks, etc.) then you can use the simpler slice syntax let short_match = &cap[2][..length];.
If Perl is your option, would you please try:
perl -e '
$str = "5 abcdefgh";
$str =~ /(\d+) ((??{".{".($^N)."}"}))/;
print "1st capture group = $1\n";
print "2nd capture group = $2\n";
print "whole capture group = $&\n";
'
Output:
1st capture group = 5
2nd capture group = abcde
whole capture group = 5 abcde
[Explanation]
If the (??{...}) block is encountered in a regex, its contents
are expanded as a Perl code on the fly.
The special variable $^N refers to the last captured group
and is expanded as 5 in the case.
Then the code (??{".{".($^N)."}"}) is evaluated as .{5} which
represents a dot followed by a quantifier.
Related
i searching to find some Perl Regular Expression Syntax about some requirements i have in a project.
First i want to exclude strings from a txt file (dictionary).
For example if my file have this strings:
path.../Document.txt |
tree
car
ship
i using Regular Expression
a1testtre -- match
orangesh1 -- match
apleship3 -- not match [contains word from file ]
Also i have one more requirement that i couldnt solve. I have to create a Regex that not allow a String to have over 3 times a char repeat (two chars).
For example :
adminnisstrator21 -- match (have 2 times a repetition of chars)
kkeeykloakk -- not match have over 3 times repetition
stack22ooverflow -- match (have 2 times a repetition of chars)
for this i have try
\b(?:([a-z])(?!\1))+\b
but it works only for the first char-reppeat
Any idea how to solve these two?
To not match a word from a file you might check whether a string contains a substring or use a negative lookahead and an alternation:
^(?!.*(?:tree|car|ship)).*$
^ Assert start of string
(?! negative lookahead, assert what is on the right is not
.*(?:tree|car|ship) Match 0+ times any char except a newline and match either tree car or ship
) Close negative lookahead
.* Match any char except a newline
$ Assert end of string
Regex demo
To not allow a string to have over 3 times a char repeat you could use:
\b(?!(?:\w*(\w)\1){3})\w+\b
\b Word boundary
(?! Negative lookahead, assert what is on the right is not
(?: NOn capturing group
\w*(\w)\1 Match 0+ times a word character followed by capturing a word char in a group followed by a backreference using \1 to that group
){3} Close non capturing group and repeat 3 times
) close negative lookahead
\w+ Match 1+ word characters
\b word boundary
Regex demo
Update
According to this posted answer (which you might add to the question instead) you have 2 patterns that you want to combine but it does not work:
(?=^(?!(?:\w*(.)\1){3}).+$)(?=^(?:(.)(?!(?:.*?\1){4}))*$)
In those 2 patterns you use 2 capturing groups, so the second pattern has to point to the second capturing group \2.
(?=^(?!(?:\w*(.)\1){3}).+$)(?=^(?:(.)(?!(?:.*?\2){4}))*$)
^
Pattern demo
One way to exclude strings that contain words from a given list is to form a pattern with an alternation of the words and use that in a regex, and exclude strings for which it matches.
use warnings;
use strict;
use feature qw(say);
use Path::Tiny;
my $file = shift // die "Usage: $0 file\n"; #/
my #words = split ' ', path($file)->slurp;
my $exclude = join '|', map { quotemeta } #words;
foreach my $string (qw(a1testtre orangesh1 apleship3))
{
if ($string !~ /$exclude/) {
say "OK: $string";
}
}
I use Path::Tiny to read the file into a a string ("slurp"), which is then split by whitespace into words to use for exclusion. The quotemeta escapes non-"word" characters, should any happen in your words, which are then joined by | to form a string with a regex pattern. (With complex patterns use qr.)
This may be possible to tweak and improve, depending on your use cases, for one in regards to the order of of patterns with common parts in alternation.†
The check that successive duplicate characters do not occur more than three times
foreach my $string (qw(adminnisstrator21 kkeeykloakk stack22ooverflow))
{
my #chars_that_repeat = $string =~ /(.)\1+/g;
if (#chars_that_repeat < 3) {
say "OK: $string";
}
}
A long string of repeated chars (aaaa) counts as one instance, due to the + quantifier in regex; if you'd rather count all pairs remove the + and four as will count as two pairs. The same char repeated at various places in the string counts every time, so aaXaa counts as two pairs.
This snippet can be just added to the above program, which is invoked with the name of the file with words to use for exclusion. They both print what is expected from provided samples.
† Consider an example with exclusion-words: so, sole, and solely. If you only need to check whether any one of these matches then you'd want shorter ones first in the alternation
my $exclude = join '|', map { quotemeta } sort { length $a <=> length $b } #words;
#==> so|sole|solely
for a quicker match (so matches all three). This, by all means, appears to be the case here.
But, if you wanted to correctly identify which word matched then you must have longer words first,
solely|sole|so
so that a string solely is correctly matched by its word before it can be "stolen" by so. Then in this case you'd want it the other way round,
sort { length $b <=> length $a }
I hope someone else will come with a better solution, but this seems to do what you want:
\b Match word boundary
(?: Start capture group
(?:([a-z0-9])(?!\1))* Match all characters until it encounters a double
(?:([a-z0-9])\2)+ Match all repeated characters until a different one is reached
){0,2} Match capture group 0 or 2 times
(?:([a-z0-9])(?!\3))+ Match all characters until it encounters a double
\b Match end of word
I changed the [a-z] to also match numbers, since the examples you gave seem to also include numbers. Perl regex also has the \w shorthand, which is equivalent to [A-Za-z0-9_], which could be handy if you want to match any character in a word.
My problem is that i have 2 regex that working:
Not allow over 3 pairs of chars:
(?=^(?!(?:\w*(.)\1){3}).+$)
Not allow over 4 times a char to repeat:
(?=^(?:(.)(?!(?:.*?\1){4}))*$)
Now i want to combine them into one row like:
(?=^(?!(?:\w*(.)\1){3}).+$)(?=^(?:(.)(?!(?:.*?\1){4}))*$)
but its working only the regex that is first and not both of them
As mentioned in comment to #zdim's answer, take it a bit further by making sure that the order in which your words are assembled into the match pattern doesn't trip you. If the words in the file are not very carefully ordered to start, I use a subroutine like this when building the match string:
# Returns a list of alternative match patterns in tight matching order.
# E.g., TRUSTEES before TRUSTEE before TRUST
# TRUSTEES|TRUSTEE|TRUST
sub tight_match_order {
return #_ unless #_ > 1;
my (#alts, #ordered_alts, %alts_seen);
#alts = map { $alts_seen{$_}++ ? () : $_ } #_;
TEST: {
my $alt = shift #alts;
if (grep m#$alt#, #alts) {
push #alts => $alt;
} else {
push #ordered_alts => $alt;
}
redo TEST if #alts;
}
#ordered_alts
}
So following #zdim's answer:
...
my #words = split ' ', path($file)->slurp;
#words = tight_match_order(#words); # add this line
my $exclude = join '|', map { quotemeta } #words;
...
HTH
I have a string test_demo_0.1.1.
I want in PowerShell script to add before the 0.1.1 some text, for example: test_demo_shay_0.1.1.
I succeeded to detect the first number with RegEx and add the text:
$str = "test_demo_0.1.1"
if ($str - match "(?<number>\d)")
{
$newStr = $str.Insert($str.IndexOf($Matches.number) - 1, "_shay")-
}
# $newStr = test_demo_shay_0.1.1
The problem is, sometimes my string includes a number in another location, for example: test_demo2_0.1.1 (and then the insert is not good).
So I want to detect the first number which the character before is _, how can I do it?
I tried "(_<number>\d)" and "([_]<number>\d)" but it doesn't work.
What you ask for is called a positive lookbehind (a construct that checks for the presence of some pattern immediately to the left of thew current location):
"(?<=_)(?<number>\d)"
^^^^^^
However, it seems all you want is to insert _shay before the first digit preceded with _. A replace operation will suit here best:
$str -replace '_(\d.*)', '_shay_$1'
Result: test_demo_shay_0.1.1.
Details
_ - an underscore
(\d.*) - Capturing group #1: a digit and then any 0+ chars to the end of the line.
The $1 in the replacement pattern is the contents matched by the capturing group #1.
I would like to extract certain values from the given String
String --> 1ABCDE23
I need only ABCDE from the above string value. Always skip the first value(1) and get next 5 characters(ABCDE) and skip the rest(23)
Appreciate your help
Always skip the first value(1) and get next 5 characters(ABCDE) and skip the rest(23)
This is just extracting a substring from a string! You don't need a regex for this - there has to be a faster function for that in the language you're using, assuming you're using a sane language.
Here are some examples:
Java: String abcde = "1ABCDE23".substring(1, 6);
JavaScript: var abcde = "1ABCDE23".substring(1, 6);
C++: std::string abcde = std::string("1ABCDE23").substr(1, 5);
Python: abcde = "1ABCDE23"[1:6]
PHP: $abcde = substr("1ABCDE23", 1, 5);
C#: string abcde = "1ABCDE23".Substring(1, 5);
Perl: $abcde = substr "1ABCDE23", 1, 5;
Ruby: abcde = "1ABCDE23"[1...6]
If you're using an insane language that has a regex engine with support for capturing groups but not a facility for extracting substrings from strings, you can run this regex (which is suggested by sln) and take the first capturing group:
^.(.{5}).*
^ the match must be at the beginning of the string
. match any character
( ) put what's matched by the parenthesized expression into the 1st capturing group
.{5} match 5 characters; any character goes
.* match 0 or more characters; any character goes
I'm quite terrible at regexes.
I have a string that may have 1 or more words in it (generally 2 or 3), usually a person name, for example:
$str1 = 'John Smith';
$str2 = 'John Doe';
$str3 = 'David X. Cohen';
$str4 = 'Kim Jong Un';
$str5 = 'Bob';
I'd like to convert each as follows:
$str1 = 'John S.';
$str2 = 'John D.';
$str3 = 'David X. C.';
$str4 = 'Kim J. U.';
$str5 = 'Bob';
My guess is that I should first match the first word, like so:
preg_match( "^([\w\-]+)", $str1, $first_word )
then all the words after the first one... but how do I match those? should I use again preg_match and use offset = 1 in the arguments? but that offset is in characters or bytes right?
Anyway after I matched the words following the first, if the exist, should I do for each of them something like:
$second_word = substr( $following_word, 1 ) . '. ';
Or my approach is completely wrong?
Thanks
ps - it would be a boon if the regex could maintain the whole first two words when the string contain three or more words... (e.g. 'Kim Jong U.').
It can be done in single preg_replace using a regex.
You can search using this regex:
^\w+(?:$| +)(*SKIP)(*F)|(\w)\w+
And replace by:
$1.
RegEx Demo
Code:
$name = preg_replace('/^\w+(?:$| +)(*SKIP)(*F)|(\w)\w+/', '$1.', $name);
Explanation:
(*FAIL) behaves like a failing negative assertion and is a synonym for (?!)
(*SKIP) defines a point beyond which the regex engine is not allowed to backtrack when the subpattern fails later
(*SKIP)(*FAIL) together provide a nice alternative of restriction that you cannot have a variable length lookbehind in above regex.
^\w+(?:$| +)(*SKIP)(*F) matches first word in a name and skips it (does nothing)
(\w)\w+ matches all other words and replaces it with first letter and a dot.
You could use a positive lookbehind assertion.
(?<=\h)([A-Z])\w+
OR
Use this regex if you want to turn Bob F to Bob F.
(?<=\h)([A-Z])\w*(?!\.)
Then replace the matched characters with \1.
DEMO
Code would be like,
preg_replace('~(?<=\h)([A-Z])\w+~', '\1.', $string);
DEMO
(?<=\h)([A-Z]) Captures all the uppercase letters which are preceeded by a horizontal space character.
\w+ matches one or more word characters.
Replace the matched chars with the chars inside the group index 1 \1 plus a dot will give you the desired output.
A simple solution with only look-ahead and word boundary check:
preg_replace('~(?!^)\b(\w)\w+~', '$1.', $string);
(\w)\w+ is a word in the name, with the first character captured
(?!^)\b performs a word boundary check \b, and makes sure the match is not at the start of the string (?!^).
Demo
I'm a regular expression newbie and I can't quite figure out how to write a single regular expression that would "match" any duplicate consecutive words such as:
Paris in the the spring.
Not that that is related.
Why are you laughing? Are my my regular expressions THAT bad??
Is there a single regular expression that will match ALL of the bold strings above?
Try this regular expression:
\b(\w+)\s+\1\b
Here \b is a word boundary and \1 references the captured match of the first group.
Regex101 example here
I believe this regex handles more situations:
/(\b\S+\b)\s+\b\1\b/
A good selection of test strings can be found here: http://callumacrae.github.com/regex-tuesday/challenge1.html
The below expression should work correctly to find any number of duplicated words. The matching can be case insensitive.
String regex = "\\b(\\w+)(\\s+\\1\\b)+";
Pattern p = Pattern.compile(regex, Pattern.CASE_INSENSITIVE);
Matcher m = p.matcher(input);
// Check for subsequences of input that match the compiled pattern
while (m.find()) {
input = input.replaceAll(m.group(0), m.group(1));
}
Sample Input : Goodbye goodbye GooDbYe
Sample Output : Goodbye
Explanation:
The regex expression:
\b : Start of a word boundary
\w+ : Any number of word characters
(\s+\1\b)* : Any number of space followed by word which matches the previous word and ends the word boundary. Whole thing wrapped in * helps to find more than one repetitions.
Grouping :
m.group(0) : Shall contain the matched group in above case Goodbye goodbye GooDbYe
m.group(1) : Shall contain the first word of the matched pattern in above case Goodbye
Replace method shall replace all consecutive matched words with the first instance of the word.
Try this with below RE
\b start of word word boundary
\W+ any word character
\1 same word matched already
\b end of word
()* Repeating again
public static void main(String[] args) {
String regex = "\\b(\\w+)(\\b\\W+\\b\\1\\b)*";// "/* Write a RegEx matching repeated words here. */";
Pattern p = Pattern.compile(regex, Pattern.CASE_INSENSITIVE/* Insert the correct Pattern flag here.*/);
Scanner in = new Scanner(System.in);
int numSentences = Integer.parseInt(in.nextLine());
while (numSentences-- > 0) {
String input = in.nextLine();
Matcher m = p.matcher(input);
// Check for subsequences of input that match the compiled pattern
while (m.find()) {
input = input.replaceAll(m.group(0),m.group(1));
}
// Prints the modified sentence.
System.out.println(input);
}
in.close();
}
Regex to Strip 2+ duplicate words (consecutive/non-consecutive words)
Try this regex that can catch 2 or more duplicate words and only leave behind one single word. And the duplicate words need not even be consecutive.
/\b(\w+)\b(?=.*?\b\1\b)/ig
Here, \b is used for Word Boundary, ?= is used for positive lookahead, and \1 is used for back-referencing.
Example
Source
The widely-used PCRE library can handle such situations (you won't achieve the the same with POSIX-compliant regex engines, though):
(\b\w+\b)\W+\1
Here is one that catches multiple words multiple times:
(\b\w+\b)(\s+\1)+
No. That is an irregular grammar. There may be engine-/language-specific regular expressions that you can use, but there is no universal regular expression that can do that.
This is the regex I use to remove duplicate phrases in my twitch bot:
(\S+\s*)\1{2,}
(\S+\s*) looks for any string of characters that isn't whitespace, followed whitespace.
\1{2,} then looks for more than 2 instances of that phrase in the string to match. If there are 3 phrases that are identical, it matches.
Since some developers are coming to this page in search of a solution which not only eliminates duplicate consecutive non-whitespace substrings, but triplicates and beyond, I'll show the adapted pattern.
Pattern: /(\b\S+)(?:\s+\1\b)+/ (Pattern Demo)
Replace: $1 (replaces the fullstring match with capture group #1)
This pattern greedily matches a "whole" non-whitespace substring, then requires one or more copies of the matched substring which may be delimited by one or more whitespace characters (space, tab, newline, etc).
Specifically:
\b (word boundary) characters are vital to ensure partial words are not matched.
The second parenthetical is a non-capturing group, because this variable width substring does not need to be captured -- only matched/absorbed.
the + (one or more quantifier) on the non-capturing group is more appropriate than * because * will "bother" the regex engine to capture and replace singleton occurrences -- this is wasteful pattern design.
*note if you are dealing with sentences or input strings with punctuation, then the pattern will need to be further refined.
The example in Javascript: The Good Parts can be adapted to do this:
var doubled_words = /([A-Za-z\u00C0-\u1FFF\u2800-\uFFFD]+)\s+\1(?:\s|$)/gi;
\b uses \w for word boundaries, where \w is equivalent to [0-9A-Z_a-z]. If you don't mind that limitation, the accepted answer is fine.
This expression (inspired from Mike, above) seems to catch all duplicates, triplicates, etc, including the ones at the end of the string, which most of the others don't:
/(^|\s+)(\S+)(($|\s+)\2)+/g, "$1$2")
I know the question asked to match duplicates only, but a triplicate is just 2 duplicates next to each other :)
First, I put (^|\s+) to make sure it starts with a full word, otherwise "child's steak" would go to "child'steak" (the "s"'s would match). Then, it matches all full words ((\b\S+\b)), followed by an end of string ($) or a number of spaces (\s+), the whole repeated more than once.
I tried it like this and it worked well:
var s = "here here here here is ahi-ahi ahi-ahi ahi-ahi joe's joe's joe's joe's joe's the result result result";
print( s.replace( /(\b\S+\b)(($|\s+)\1)+/g, "$1"))
--> here is ahi-ahi joe's the result
Try this regular expression it fits for all repeated words cases:
\b(\w+)\s+\1(?:\s+\1)*\b
I think another solution would be to use named capture groups and backreferences like this:
.* (?<mytoken>\w+)\s+\k<mytoken> .*/
OR
.*(?<mytoken>\w{3,}).+\k<mytoken>.*/
Kotlin:
val regex = Regex(""".* (?<myToken>\w+)\s+\k<myToken> .*""")
val input = "This is a test test data"
val result = regex.find(input)
println(result!!.groups["myToken"]!!.value)
Java:
var pattern = Pattern.compile(".* (?<myToken>\\w+)\\s+\\k<myToken> .*");
var matcher = pattern.matcher("This is a test test data");
var isFound = matcher.find();
var result = matcher.group("myToken");
System.out.println(result);
JavaScript:
const regex = /.* (?<myToken>\w+)\s+\k<myToken> .*/;
const input = "This is a test test data";
const result = regex.exec(input);
console.log(result.groups.myToken);
// OR
const regex = /.* (?<myToken>\w+)\s+\k<myToken> .*/g;
const input = "This is a test test data";
const result = [...input.matchAll(regex)];
console.log(result[0].groups.myToken);
All the above detect the test as the duplicate word.
Tested with Kotlin 1.7.0-Beta, Java 11, Chrome and Firefox 100.
You can use this pattern:
\b(\w+)(?:\W+\1\b)+
This pattern can be used to match all duplicated word groups in sentences. :)
Here is a sample util function written in java 17, which replaces all duplications with the first occurrence:
public String removeDuplicates(String input) {
var regex = "\\b(\\w+)(?:\\W+\\1\\b)+";
var pattern = Pattern.compile(regex, Pattern.CASE_INSENSITIVE);
var matcher = pattern.matcher(input);
while (matcher.find()) {
input = input.replaceAll(matcher.group(), matcher.group(1));
}
return input;
}
As far as I can see, none of these would match:
London in the
the winter (with the winter on a new line )
Although matching duplicates on the same line is fairly straightforward,
I haven't been able to come up with a solution for the situation in which they
stretch over two lines. ( with Perl )
To find duplicate words that have no leading or trailing non whitespace character(s) other than a word character(s), you can use whitespace boundaries on the left and on the right making use of lookarounds.
The pattern will have a match in:
Paris in the the spring.
Not that that is related.
The pattern will not have a match in:
This is $word word
(?<!\S)(\w+)\s+\1(?!\S)
Explanation
(?<!\S) Negative lookbehind, assert not a non whitespace char to the left of the current location
(\w+) Capture group 1, match 1 or more word characters
\s+ Match 1 or more whitespace characters (note that this can also match a newline)
\1 Backreference to match the same as in group 1
(?!\S) Negative lookahead, assert not a non whitespace char to the right of the current location
See a regex101 demo.
To find 2 or more duplicate words:
(?<!\S)(\w+)(?:\s+\1)+(?!\S)
This part of the pattern (?:\s+\1)+ uses a non capture group to repeat 1 or more times matching 1 or more whitespace characters followed by the backreference to match the same as in group 1.
See a regex101 demo.
Alternatives without using lookarounds
You could also make use of a leading and trailing alternation matching either a whitespace char or assert the start/end of the string.
Then use a capture group 1 for the value that you want to get, and use a second capture group with a backreference \2 to match the repeated word.
Matching 2 duplicate words:
(?:\s|^)((\w+)\s+\2)(?:\s|$)
See a regex101 demo.
Matching 2 or more duplicate words:
(?:\s|^)((\w+)(?:\s+\2)+)(?:\s|$)
See a regex101 demo.
Use this in case you want case-insensitive checking for duplicate words.
(?i)\\b(\\w+)\\s+\\1\\b