Regex to replace part of the string with spaces

Question

It seems simple, but I can't get it work.

I have a string which look like 'NNDDDDDAAAA', where 'N' is non digit, 'D' is digit, and 'A' is anything. I need to replace each A with a space character. Number of 'N's, 'D's, and 'A's in an input string is always different.

I know how to do it with two expressions. I can split a string in to two, and then replace everything in second group with spaces. Like this

    Pattern pattern = Pattern.compile("(\\D+\\d+)(.+)");
    Matcher matcher = pattern.matcher(input);
    if (matcher.matches()) {
        return matcher.group(1) + matcher.group(2).replaceAll(".", " ");
    }

But I was wondering if it is possible with a single regex expression.

Answer 1

what do you mean by nondigit vs anything?

[^a-zA-Z0-9]
matches everything that is not a letter or digit.

you would want to replace anything that gets matched by the above regex with a space.

is this what you were talking about?

Answer 2

You want to use positive look behind to match the N's and D's then use a normal match for the A's.

Not sure of the positive look behind grammar in Java, but some article on Java regex with look behind

Answer 3

Given your description, I'm assuming that after the NNDDDDD portion, the first A will actually be a N rather than an A, since otherwise there's no solid boundary between the DDDDD and AAAA portions.

So, your string actually looks like NNDDDDDNAAA, and you want to replace the NAAA portion with spaces. Given this, the regex can be rewritten as such: (\\D+\\d+)(\\D.+)

Positive lookbehind in Java requires a fixed length pattern; You can't use the + or * patterns. You can instead use the curly braces and specify a maximum length. For instance, you can use {1,9} in place of each +, and it will match between 1 and 9 characters: (?<=\\D{1,9}\\d{1,9})(\\D.+)

The only problem here is you're matching the NAAA sequence as a single match, so using "NNNDDDDNAAA".replaceAll("(?<=\\D{1,9}\\d{1,9})(\\D.+)", " ") will result in replacing the entire NAAA sequence with a single space, rather than multiple spaces.

You could take the beginning delimiter of the match, and the string length, and use that to append the correct number of spaces, but I don't see the point. I think you're better off with your original solution; Its simple and easy to follow.

If you're looking for a little extra speed, you could compile your Pattern outside the function, and use StringBuilder or StringBuffer to create your output. If you're building a large String out of all these NNDDDDDAAAAA elements, work entirely in StringBuilder until you're done appending.

class Test {

public static Pattern p = Pattern.compile("(\\D+\\d+)(\\D.+)");

public static StringBuffer replace( String input ) {
    StringBuffer output = new StringBuffer();
    Matcher m = Test.p.matcher(input);
    if( m.matches() )
        output.append( m.group(1) ).append( m.group(2).replaceAll("."," ") );

    return output;
}

public static void main( String[] args ) {
    String input = args[0];
    long startTime;

    StringBuffer tests = new StringBuffer();
    startTime = System.currentTimeMillis();
     for( int i = 0; i < 50; i++)
     {
      tests.append( "Input -> Output: '" );
      tests.append( input );
      tests.append( "' -> '" );
      tests.append( Test.replace( input ) );
      tests.append( "'\n" );
     }
    System.out.println( tests.toString() );
    System.out.println( "\n" + (System.currentTimeMillis()-startTime));
}

}

Update: I wrote a quick iterative solution, and ran some random data through both. The iterative solution is around 4-5x faster.

public static StringBuffer replace( String input )
{
    StringBuffer output = new StringBuffer();
 boolean second = false, third = false;
 for( int i = 0; i < input.length(); i++ )
 {
  if( !second && Character.isDigit(input.charAt(i)) )
   second = true;

  if( second && !third && Character.isLetter(input.charAt(i)) )
   third = true;

  if( second && third )
   output.append( ' ' );
  else
   output.append( input.charAt(i) );

 }

    return output;
}

Answer 4

I know you asked for a regex, but why do you even need a regex for this? How about:

StringBuilder sb = new StringBuilder(inputString);
for (int i = sb.length() - 1; i >= 0; i--) {
    if (Character.isDigit(sb.charAt(i)))
        break;
    sb.setCharAt(i, ' ');
}
String output = sb.toString();

You might find this post interesting. Of course, the above code assumes there will be at least one digit in the string - all characters following the last digit are converted to spaces. If there are no digits, every character is converted to a space.