How to format reputation numbers similar to Stack Exchange sites

Question

UPDATE

For JavaScript, see CMS's implementation below. It is much more elegant than the one I provide in the body of this Q.

---

// formats a number similar to the way stack exchange sites // format reputation. e.g. // for numbers< 10000 the output is '9,999' // for numbers > 10000 the output is '10k' with one decimal place when needed function getRepString(rep) { var repString;

    if (rep < 1000)
    {
        repString = rep;
    }
    else if (rep < 10000)
    {
        var mod = rep % 1000;

        repString = ((rep - mod) / 1000)
        + ","
        + ('000' + mod.toString()).slice(-3);
    }
    else
    {
        repString = (rep / 1000).toFixed(1).replace(".0", "") + "k";
    }

    return repString.toString();
}

Output:

• getRepString(999) == '999'
• getRepString(1000) == '1,000'
• getRepString(9999) == '9,999'
• getRepString(10000) == '10k'
• getRepString(10100) == '10.1k'

Post an implementation in the language of your choice.

Answer 1

Here you go, another JavaScript approach, originally posted on SO:

function getRepString (rep) {

  rep = rep+''; // coerce to string

  if (rep < 1000) { // return the same number
    return rep; 
  }

  if (rep < 10000) { // place a comma between

    return rep.charAt(0) + ',' + rep.substring(1);
  } 

  // divide and format
  return (rep/1000).toFixed(rep % 1000 != 0)+'k';

}

Check the output results here.

Answer 2

Objective-C:

This should work on iOS 3.2, iOS 4.0 and Mac OS X 10.4 - 10.6 :

I don't have multiple returns because I like a single exit point in my code. Hence I use my trusty friend returnable.

- (NSString*) stringForReputationFormatted:(NSNumber*)reputation
{
    NSString *returnable = nil;
    long rep = [reputation longValue];
    NSString *returnable = nil;

    if (rep < 1000) {
        returnable = [reputation stringValue];
    }
    else if (rep < 10000) {
        NSNumberFormatter *numberFormatter = [[NSNumberFormatter alloc] init];
        [numberFormatter setNumberStyle:NSNumberFormatterDecimalStyle];
        returnable = [numberFormatter stringForObjectValue:reputation];
        [numberFormatter release];
    }
    else {
        NSString *repStr    = [reputation stringValue];
        NSString *whole     = [repStr substringToIndex:[repStr length]-3];
        NSString *decimal   = [repStr substringWithRange:NSMakeRange([repStr length]-3, 1)];
        if ([decimal intValue] != 0) {
            returnable = [whole stringByAppendingFormat:@".%@K", decimal];
        } else {
            returnable = [whole stringByAppendingFormat:@"K"];
        }
    }
    return returnable;
}

Edit: Apologies for the code being so long. NSDateFormatter isn't working right on iOS atm, so I needed to format the string manually.

Tests to confirm:

Raw:-1 Converted:-1
Raw:0 Converted:0
Raw:1 Converted:1
Raw:2 Converted:2
Raw:10 Converted:10
Raw:100 Converted:100
Raw:101 Converted:101
Raw:999 Converted:999
Raw:1000 Converted:1,000
Raw:1001 Converted:1,001
Raw:3500 Converted:3,500
Raw:6790 Converted:6,790
Raw:8900 Converted:8,900
Raw:9999 Converted:9,999
Raw:10000 Converted:10K
Raw:10099 Converted:10K
Raw:10500 Converted:10.5K
Raw:11600 Converted:11.6K
Raw:10799 Converted:10.7K
Raw:99899 Converted:99.8K
Raw:195999 Converted:195.9K

Answer 3

Here's a Java implementation using NumberFormat.

public static void main(String[] args) {

    int[] values = { 999, 1000, 9999, 10000, 10100 };

    for( int rep : values ) {
        System.out.println( formatRep(rep) );
    }
}

public static String formatRep(int rep) {
    if(rep < 1000) {
        return rep + "";
    }
    if(rep < 10000) {
        NumberFormat formatter = new DecimalFormat("#,###");
        return formatter.format(rep);
    }
    else {
        NumberFormat formatter = new DecimalFormat("#,###.#k");
        double d = rep / 1000.0;
        return formatter.format(d);
    }
}

Output:

999
1,000
9,999
10k
10.1k

I'm not sure about that last format for numbers greater than 999k. I guess we have a few years to wait and see. :)

Answer 4

c#

Rounds up - same same SO.

public string FormatReputation(int value)
{
    var rep = System.Convert.ToDouble(value);
        
    if (rep < 10000)
    {
        return rep.ToString("N0");
    }
    
    return (rep / 1000)
         .ToString(rep % 1000 == 0 ? "" : "F1") + 'k';
}

• 100 => "100"
• 1200 => "1,200"
• 9999 => "9,999"
• 10000 => "10k"
• 10200 => "10.2k"

confirmation as per jjng's comment

• 10999 => "11k"
• 12452 => "12.5k"

Answer 5

Here's a quick and dirty way of how I did mine in .NET

Visual Basic

    Public Shared Function GetReputation(ByVal input As Integer) As String

        Dim _input As String = input.ToString
        Select Case input
            Case Is > 99999 : Return _input.Remove(_input.Length - 3) & "k"
            Case Is > 9999 : Return Math.Round(Double.Parse(input / 1000), 1).ToString & "k"
            Case Is > 999 : Return String.Format("{0:N0}", input)
            Case Else : Return _input
        End Select
    End Function

Answer 6

Just for completeness, here is the C# version I am using ported from the VB version given by @rockinthesixstring:

private string FormatReputation(int reputation)
{
     string s = reputation.ToString();
     if (reputation > 99999)
         return s.Remove(s.Length - 3) + "k";
     else if (reputation > 9999)
         return Math.Round((double)reputation / (double)1000, 1).ToString() + "k";
     else if (reputation > 999)
         return String.Format("{0:N0}", reputation);
     return s;
}

Answer 7

These are great feature ideas - thanks for the tips!

I've more or less converted CMS' answer to Python:

class FormattedReputation(int):
    def format(rep):
        """Formats the reputation score like it is formatted on the sites. Heavily based on CMS' JavaScript implementation at
        http://stackapps.com/questions/1012/how-to-format-reputation-numbers-similar-to-stack-exchange-sites/1019#1019"""
        str_rep = str(rep)

        if rep < 1000:
            return str_rep
        elif rep < 10000:
            return '%s,%s' % (str_rep[0], str_rep[1:])
        elif rep % 1000 == 0:
            return '%dk' % (rep / 1000.0)
        else:
            return '%.1fk' % (rep / 1000.0)

This is used automatically as an alternative to int for reputation fields in Py-StackExchange, so using it on its own is slightly awkward:

lucas@ubuntu:~/projects/py-stackexchange$ python -i se_inter.py
>>> FormattedReputation(100).format()
'100'
>>> FormattedReputation(1000).format()
'1,000'
>>> FormattedReputation(1240).format()
'1,240'
>>> FormattedReputation(12403).format()
'12.4k'
>>> FormattedReputation(100000).format()
'100k'

Answer 8

Good ole' PHP:

// Any number larger than 1000 gets returned like '4.5k'

function Numberify($num)
{
    if($num > 1000)
        return (floor($num / 100) / 10) . 'k';
    return $num
}

Do you like the name of the function? :)