Insert commas into number string

Question

Hey there, I'm trying to perform a backwards regular expression search on a string to divide it into groups of 3 digits. As far as i can see from the AS3 documentation, searching backwards is not possible in the reg ex engine.

The point of this exercise is to insert triplet commas into a number like so:

10000000 => 10,000,000

I'm thinking of doing it like so:

string.replace(/(\d{3})/g, ",$1")

But this is not correct due to the search not happening from the back and the replace $1 will only work for the first match.

I'm getting the feeling I would be better off performing this task using a loop.

UPDATE:

Due to AS3 not supporting lookahead this is how I have solved it.

public static function formatNumber(number:Number):String
{
 var numString:String = number.toString()
 var result:String = ''

 while (numString.length > 3)
 {
  var chunk:String = numString.substr(-3)
  numString = numString.substr(0, numString.length - 3)
  result = ',' + chunk + result
 }

 if (numString.length > 0)
 {
  result = numString + result
 }

 return result
}

Answer 1

This really isn't the best use of RegEx... I'm not aware of a number formatting function, but this thread seems to provide a solution.

function commaCoder(yourNum):String {
    //var yourNum:Number = new Number();
    var numtoString:String = new String();
    var numLength:Number = yourNum.toString().length;
    numtoString = "";

    for (i=0; i<numLength; i++) { 
     if ((numLength-i)%3 == 0 && i != 0) {
      numtoString += ",";
     }
     numtoString += yourNum.toString().charAt(i);
     trace(numtoString);
    }
    return numtoString;
}

If you really are insistent on using RegEx, you could just reverse the string, apply the RegEx replace function, then reverse it back.

Answer 2

A sexeger is good for this. In brief, a sexeger is a reversed regex run against a reversed string that you reverse the output of. It is generally more efficient than the alternative. Here is some pseudocode for what you want to do:

string = reverse string
string.replace(/(\d{3})(?!$)/g, "$1,")
string = reverse string

Here is is a Perl implemntation

#!/usr/bin/perl

use strict;
use warnings;

my $s = 13_456_789;

for my $n (1, 12, 123, 1234, 12345, 123456, 1234567) {
    my $s = reverse $n;
    $s =~ s/([0-9]{3})(?!$)/$1,/g;
    $s = reverse $s;
    print "$s\n";
}

Answer 3

If your regex engine has positive lookaheads, you could do something like this:

string.replace(/(\d)(?=(\d\d\d)+$)/, "$1,")

Where the positive lookahead (?=...) means that the regex only matches when the lookahead expression ... matches.

(Note that lookaround-expressions are not always very efficient.)

Answer 4

If your language supports postive lookahead assertions, then I think the following regex will work:

(\d)(?=(\d{3})+$)

Demonstrated in Java:

import static org.junit.Assert.assertEquals;

import org.junit.Test;

public class CommifyTest {

    @Test
    public void testCommify() {
        String num0 = "1";
        String num1 = "123456";
        String num2 = "1234567";
        String num3 = "12345678";
        String num4 = "123456789";

        String regex = "(\\d)(?=(\\d{3})+$)";

        assertEquals("1", num0.replaceAll(regex, "$1,"));
        assertEquals("123,456", num1.replaceAll(regex, "$1,"));
        assertEquals("1,234,567", num2.replaceAll(regex, "$1,"));
        assertEquals("12,345,678", num3.replaceAll(regex, "$1,"));
        assertEquals("123,456,789", num4.replaceAll(regex, "$1,"));    
    }    
}

The following link suggests AS3 does?

Answer 5

This is how I would do it:

public static function formatNumber(number:Number):String
{
    var firstPart:Number = Math.floor(number/1000)
    var lastPart:Number = number%1000
    if (firstPart > 0) {
        return formatNumber(firstPart) + ',' + lastPart.toString()
      } else {
        return lastPart.toString()
    }
}

I think that it should even work for non-integers, e.g. 1234567.34.