Regex to find numbers in a string

Question

I'm using removeNumbers to remove all numbers in a given string with the regex
"(^| )\\d+($|( \\d+)+($| )| )"

Here's the code:

public class Regex {    
  private static String removeNumbers(String s) {
     s = s.trim();
     s = s.replaceAll(" +", " ");
     s = s.replaceAll("(^| )\\d+($|( \\d+)+($| )| )", " ");
     return s.trim();
  }

  public static void main(String[] args) {
     String[] tests = new String[] {"123", "123 456 stack 789", "123 456 789 101112 131415 161718 192021", "stack 123 456 overflow 789 com", "stack 123 456 overflow 789", "123stack 456", "123 stack456overflow", "123 stack456", "123! @456#567"};
     for (int i = 0; i < tests.length; i++) {
        String test = tests[i];
        System.out.println("\"" + test + "\" => \"" + removeNumbers(test) + "\"");
     }  
  }    
}

Output :

"123" => ""
" 123 " => ""
"123 456 stack 789" => "stack"
"123 456 789 101112 131415 161718 192021" => ""
"stack 123 456 overflow 789 com" => "stack overflow com"
"stack 123 456 overflow 789" => "stack overflow"
"123stack 456" => "123stack"
"123 stack456overflow" => "stack456overflow"
"123 stack456" => "stack456"
"123! @456#567" => "123! @456#567"

Is there any better way to do this?

Edit :

As suggested by @mbomb007 in his previous answer, the regex "( |^)[\\d ]+( |$)" works as well:

private static String removeNumbers(String s) {
   s = s.trim();
   s = s.replaceAll(" +", " ");
   s = s.replaceAll("( |^)[\\d ]+( |$)", " ");
   return s.trim();
}

Answer 1

AFAIU, you can just do:

private static String removeNumbers(String s) {
    return s.replaceAll("\\b\\d+\\b", "").replaceAll(" +", " ").trim();
}

\b\d+\b matches one or more digits that form a word.

EDIT:

Since the pattern must not match numbers in a string like "123! @456#567", a combination of positive lookbehind and lookahead conditions can be used:

private static String removeNumbers(String s) {
    return s.replaceAll("(?<= |^)\\d+(?= |$)", " ").replaceAll(" +", " ").trim();
}

Answer 2

Your regex is a bit redundant (and also doesn't quite fit your test cases). You can use this:

"\\b[ ]*(?<![^\\d\\s])[\\d]+(?![^\\d\\s])[ ]*\\b"

The \b escape character represents a word border (start or end of a word). I also use [ ]* to ensure the spaces between numbers get removed. This regex also allows words to contain numbers without them getting replaced. Just like you want.

EDIT: I added a negative lookbehind and a positive lookahead.

(?<![^\\d\\s]) - This ensures that the characters immediately preceding the digits are only more digits or spaces.

(?![^\\d\\s]) - This ensures that the characters immediately following the digits are only more digits or spaces.

Try it here with your test cases. (Updated the hyperlink for added test case)