Regular Expression For Duplicate Words

Question

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

Answer 1

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();
  }
  

Answer 2

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.

Answer 3

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.

Answer 4

The below expression should work correctly to find any number of consecutive 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.

Answer 5

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.

Answer 6

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.