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();
}

My answer was wrong, but now I fixed it. I posted a link to a web version running the regex against your test cases. — mbomb007
– mbomb007, Commented Apr 30, 2015 at 19:08
Is your output correct? If yes and you are looking for better way to rewrite your code than it looks like your question should be asked on codereview.stackexchange.com. — Pshemo
– Pshemo, Commented Apr 30, 2015 at 19:13
Then it would be nice to add few cases explaining how your code should react on them. — Pshemo
– Pshemo, Commented Apr 30, 2015 at 19:54

M A · Accepted Answer · 2015-04-30 20:49:42Z

3

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();
}

edited Apr 30, 2015 at 20:49

answered Apr 30, 2015 at 18:48

M A

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3 Comments

Chris Stillwell Over a year ago

Or all on one line return s.replaceAll("\\d", "").replaceAll(" +", " ").trim();

mbomb007 Over a year ago

This doesn't work for his example: "123 stack456overflow" => "stack456overflow"

Bharat Khatri Over a year ago

I've added one more test case: "123! @456#567" => "123! @456#567

mbomb007 · Accepted Answer · 2015-04-30 20:27:19Z

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)

edited Apr 30, 2015 at 20:27

answered Apr 30, 2015 at 18:59

mbomb007

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4 Comments

Bharat Khatri Over a year ago

I've added one more test case: "123! @456#567" => "123! @456#567

mbomb007 Over a year ago

@BharatKhatri I changed it for the added test case and updated the hyperlink to the new example. I hope you find it helpful.

Bharat Khatri Over a year ago

One of your earlier approaches using "( |^)[\\d ]+( |$)" works fine. I've added it in an edit to the question, since you've updated your answer and the earlier approach is no longer visible.

mbomb007 Over a year ago

@BharatKhatri If it worked for you, good. It didn't work for all the test cases when I was testing.

Tomek · Accepted Answer · 2015-04-30 19:04:12Z

0

You can also do it with guava library:

 String text = "stack 123 456 overflow 789 com";
 String theLettersWithLargeSpaces = CharMatcher.JAVA_LETTER.or(CharMatcher.WHITESPACE).retainFrom(text); 
 theLetters = CharMatcher.WHITESPACE.collapseFrom(theLettersWithLargeSpaces , ' ');
 System.out.println(theLetters);

I assumed that not only digits can occur, but also other unwanted characters. The output would be: "stack overflow com"

CharMatcher is very powerful tool. I think that it is much more readable than regexes.

If you want just a function:

public String clearUnwantedChars(String text) {
      return CharMatcher.WHITESPACE.collapseFrom(CharMatcher.JAVA_LETTER.or(CharMatcher.WHITESPACE)
            .retainFrom(text), ' ');
}

edited Apr 30, 2015 at 19:04

answered Apr 30, 2015 at 18:58

Tomek

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