I am trying to write a regex that finds the first word in each line that contains the character a.
For a string like:
The cat ate the dog
and the mouse
The expression should find cat and
So far, I have:
/\b\w*a\w*\b/g
However this will return every match in each line, not just the first match (cat ate and).
What is the easiest way to only return the first occurrence?
Assuming you are onluy looking for words without numbers and underscores (\w would include those), I'd advise to maybe use:
(?i)^.*?(?<!\S)([b-z]*a[a-z]*)(?!\S)
And use whatever is in the 1st capture group. See an online demo. Or, if supported:
(?i)^.*?\K(?<!\S)[b-z]*a[a-z]*(?!\S)
See an online demo.
Please note that I used lookaround to assert that the word is not inbetween anything other than whitespace characters. You may also use word-boundaries if you please and swap those lookarounds for \b. Also, depending on your application you can probably scratch the inline case-insensitive switch to a 'flag'. For example, if you happen to use JavaScript /^.*?(?<!\S)([b-z]*a[a-z]*)(?!\S)/gmi should probably be your option. See for example:
var myString = "The cat ate the dog\nand the mouse";
var myRegexp = new RegExp("^.*?(?<!\S)([b-z]*a[a-z]*)(?!\S)", "gmi");
m = myRegexp.exec(myString);
while (m != null) {
console.log(m[1])
m = myRegexp.exec(myString);
}
If you want to match a word using \w you might also use a negated character class matching any character except a or a newline.
Then match a word that consists of at least an a char with word boundaries \b
^[^a\n\r]*\b([^\Wa]*a\w*)
The pattern matches:
^ Start of string
[^a\n\r]*\b Optionally match any character except a or a newline
( Capture group 1
[^\Wa]*a\w* Optionally match a word character without a, then match a and optional word characters
) Close group 1
Regex demo
Using whitespace boundaries on the left and right:
^[^a\n\r]*(?<!\S)([^\Wa]*a\w*)(?!\S)
Regex demo
The text could be matched with the regular expression
(?=(\b[a-z]*a[a-z]*\b)).*\r?\n
with the multiline and case-indifferent flags set. For each match capture group 1 contains the first word (comprised only of letters) in a line that contains an "a". There are no matches in lines that do not contain an "a".
Demo
The expression can be broken down as follows.
(?= # begin a positive lookahead
\b # match a word boundary
([a-z]*a[a-z]*) # match a word containing an "a" and save to
# capture group 1
)
.*\r?\n # match the remainder of the line including the
# line terminator
I want to delete all invalid letters from a string which should represent a phone number. Only a '+' prefix and numbers are allowed.
I tried in Kotlin with
"+1234abc567+".replace("[^+0-9]".toRegex(), "")
It works nearly perfect, but it does not replace the last '+'.
How can I modify the regex to only allow the first '+'?
You could do a regex replacement on the following pattern:
(?<=.)\+|[^0-9+]+
Sample script:
String input = "+1234abc567+";
String output = input.replaceAll("(?<=.)\\+|[^0-9+]+", "");
System.out.println(input); // +1234abc567+
System.out.println(output); // +1234567
Here is an explanation of the regex pattern:
(?<=.)\+ match a literal + which is NOT first (i.e. preceded by >= 1 character)
| OR
[^0-9+]+ match one or more non digit characters, excluding +
You can use
^(\+)|\D+
Replace with the backreference to the first group, $1. See the regex demo.
Details:
^(\+) - a + at the start of string captured into Group 1
| - or
\D+ - one or more non-digit chars.
NOTE: a raw string literal delimited with """ allows the use of a single backslash to form regex escapes, such as \D, \d, etc. Using this type of string literals greatly simplifies regex definitions inside code.
See the Kotlin demo:
val s = "+1234abc567+"
val regex = """^(\+)|\D+""".toRegex()
println(s.replace(regex, "$1"))
// => +1234567
I'm trying to use a regular expression to capture substrings delimited by another substring. For example, if I had the sentence
My cat is a cat.
and the delimiter I wanted to use was "cat", the output should be
My
is a
.
I've been unable to find a solution where the delimiter isn't a single character.
Edit: I'm writing this in Java, and the output represents groups returned by Java's Matcher class in a call like "myMatcher.group()". Sorry for the confusion.
What you need is String#split as Tushar pointed out in the comment.
String s = "My cat is a cat.";
String[] res = s.split("cat");
System.out.println(Arrays.toString(res));
This is the only correct way to do it.
Now, you want to know how to match any text other than cat with the Matcher.
DISCLAIMER: do not use it in Java since it is highly impractical and non-performance-wise.
You may match the cat and capture it into a Group, and add another alternative to the pattern that will match any text other than cat.
String s = "My cat is a cat.";
Pattern pattern = Pattern.compile("(?i)(cat)|[^c]*(?:c(?!at)[^c]*)*");
Matcher matcher = pattern.matcher(s);
while (matcher.find()){
if (matcher.group(1) == null) { // Did we match "cat"?
if (!matcher.group(0).isEmpty()) // Is the match text NOT empty? System.out.println(matcher.group(0)); // Great, print it
}
}
See the IDEONE demo
Pattern details:
(?i) - case insensitive inline modifier
(cat) - Group 1 capturing a substring cat
| - or
[^c]*(?:c(?!at)[^c]*)* - a substring that is not a starting point for a cat substring. It is an unrolled (?s)(?:(?!cat).)* tempered greedy token.
[^c]* - 0+ chars other than c or C
(?:c(?!at)[^c]*)* - zero or more sequences of:
c(?!at) - c or C not followed with at, At, AT, aT
[^c]* - 0+ chars other than c or C
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