Code-Golf: Friendly Number Abbreviator

Question

Based on this question: Is there a way to round numbers into a friendly format?

THE CHALLENGE - UPDATED! (removed hundreds abbreviation from spec)

The shortest code by character count that will abbreviate an integer (no decimals).

Code should include the full program.

Relevant range is from 0 - 9,223,372,036,854,775,807 (the upper limit for signed 64 bit integer).

The number of decimal places for abbreviation will be positive. You will not need to calculate the following: 920535 abbreviated -1 place (which would be something like 0.920535M).

Numbers in the tens and hundreds place (0-999) should never be abbreviated (the abbreviation for the number 57 to 1+ decimal places is 5.7dk - it is unneccessary and not friendly).

Remember to round half away from zero (23.5 gets rounded to 24). Banker's rounding is verboten.

Here are the relevant number abbreviations:

h = hundred (102)
k = thousand (103)
M = million (106)
G = billion (109)
T = trillion (1012)
P = quadrillion (1015)
E = quintillion (1018)

SAMPLE INPUTS/OUTPUTS (inputs can be passed as separate arguments):

First argument will be the integer to abbreviate. The second is the number of decimal places.

12 1                  => 12 // tens and hundreds places are never rounded
1500 2                => 1.5k
1500 0                => 2k // look, ma! I round UP at .5
0 2                   => 0
1234 0                => 1k
34567 2               => 34.57k
918395 1              => 918.4k
2134124 2             => 2.13M
47475782130 2         => 47.48G
9223372036854775807 3 => 9.223E
// ect...

---

Original answer from related question (JavaScript, does not follow spec):

function abbrNum(number, decPlaces) {
    // 2 decimal places => 100, 3 => 1000, etc
    decPlaces = Math.pow(10,decPlaces);

    // Enumerate number abbreviations
    var abbrev = [ "k", "m", "b", "t" ];

    // Go through the array backwards, so we do the largest first
    for (var i=abbrev.length-1; i>=0; i--) {

        // Convert array index to "1000", "1000000", etc
        var size = Math.pow(10,(i+1)*3);

        // If the number is bigger or equal do the abbreviation
        if(size <= number) {
             // Here, we multiply by decPlaces, round, and then divide by decPlaces.
             // This gives us nice rounding to a particular decimal place.
             number = Math.round(number*decPlaces/size)/decPlaces;

             // Add the letter for the abbreviation
             number += abbrev[i];

             // We are done... stop
             break;
        }
    }

    return number;
}

Kekkonen

(1) Since your "relevant abbreviations" are k, M, G, T, etc. your sample output should be changed to match that. (2) Should the code include the full program or not? – Pedant

(1) Fixed (2) I updated to include that the code should include the full program. Thanks! – Kekkonen

Two issues to consider if building a full (rather than code-golf) implementation: 1) in some contexts 2^10 makes more sense than 10^3 as the base and 2) should xxx5 round to even (better statistics) rather than always up (simple rule). I'm offering my votes for answers which implement either (or both) of these. – Coercive

@dmckee: I had a hard enough time make the rules as they are. So, I vote for the simplest implementation to keep it as Code-Golfy as possible. – Kekkonen

Whoa, 2^10 vs 10^3 is a pretty big difference, we should settle on one or the other. – Degradation

Use 44.22Gi if you want binary :p – Pedant

hm, whats the problem with my question? I'm going to get another tumbleweed badge if things don't pick up! – Kekkonen

I would vote to remove hundreds as an abbreviation possibility. I have never seen this done in practice, and it complicates the code, because its not a 3x power of 10. – Bumgardner

@David: No chance for tumbleweed, you've already got 5 upvotes :p – Pedant

UPDATED! I've removed hundreds place as a requirement. – Kekkonen

@sr pt. what else should I do? – Kekkonen

@David: Awesome thread, thanks for picking this up! – Ratiocination

now there is codegolf.stackexchange.com – Maible

Answer 1

J, 61 63 65 characters

((j.&(1&{)":({.%&1e3{:));{&' kMGTPE'@{.)(([:<.1e3^.{.),{:,{.)

Output:

((j.&(1&{)":({.%&1e3{:));{&' kMGTPE'@{.)(([:<.1e3^.{.),{:,{.) 1500 0
┌─┬─┐
│2│k│
└─┴─┘

((j.&(1&{)":({.%&1e3{:));{&' kMGTPE'@{.)(([:<.1e3^.{.),{:,{.) 987654321987654321 4
┌────────┬─┐
│987.6543│P│
└────────┴─┘

(The reason the output is "boxed" like that is because J doesn't support a list consisting of varying types)

Explanation (from right to left):

(([:<.1000^.{.),{:,{.)

We make a new 3-element list, using , to join ([:<.1000^.{.) (the floored <. base 1000 log ^. of the first param {.. We join it with the second param {: and then the first param {..

So after the first bit, we've transformed say 12345 2 into 1 2 12345

((j.&(1&{)":({.%&1000{:));{&' kMGTPE'@{.) uses ; to join the two halves of the expression together in a box to produce the final output.

The first half is ((j.&(1&{)":({.%&1000{:)) which divides (%) the last input number ({:) by 1000, the first number of times. Then it sets the precision ": using the second number in the input list (1&{).

The second half {&' kMGTPE'@{. - this uses the first number to select ({) the appropriate character from the 0-indexed list of abbreviations.

Swiftlet

output for 987654321987654321 4 should be 987.6543P. and out put for 1234567 2 should be 1.23M – Kekkonen

@David: I know, I caught that after posting it. Just spent the last half hour fixing that at a cost of 2 characters. :) – Swiftlet

Also, it supports up to 999,999,999,999,999,999,999 - over 100x higher than the requirements. – Swiftlet

Ok, after looking at your code I would like to say that your family is worried about you – Cluff

@M28: I'll take that as a compliment :) – Swiftlet

Answer 2

Python 2.x, 78 chars

a=input()
i=0
while a>=1e3:a/=1e3;i+=1
print"%g"%round(a,input())+" kMGTPE"[i]

This version (75 chars) uses printf which will print extra zeros and follows the round-to-even rule.

a=input()
i=0
while a>=1e3:a/=1e3;i+=1
print"%%.%df"%input()%a+" kMGTPE"[i]

Norfolk

has anyone verified this with some test cases? – Kekkonen

Answer 3

Perl 114 111 104 chars

My first ever code-golf entry!

Arguments provided from standard input: perl fna.pl 918395 1

($n,$d)=@ARGV;
@n=$n=~/./g;
@s=' kMGTPE'=~/./g;
printf"%.".(@n>3?$d:0)."f%s",$n/(10**($#n-$#n%3)),$s[@n/3];

Output:

918.4k

---

De-golfed version (with explanation):

( $number, $dp ) = @ARGV;      # Read in arguments from standard input

@digits = split //, $number;   # Populate array of digits, use this to count
                               # how many digits are present

@suffix = split //, ' kMGTPE'; # Generate suffix array

$number/(10**($#n-$#n%3));     # Divide number by highest multiple of 3

$precision = @n>3 ? $dp : 0;   # Determine number of decimal points to print

sprintf "%.".$precision."f%s", # "%.2f" prints to 2 dp, "%.0f" prints integer
        $number, $suffix[@n/3];# Select appropriate suffix

Advisee

Answer 4

Javascript 114 chars

function m(n,d){p=M.pow
d=p(10,d)
i=7
while(i)(s=p(10,i--*3))<=n&&(n=M.round(n*d/s)/d+"kMGTPE"[i])
return n}

Also 114 - Using spidermonkey - Input on STDIN

[n,d]=readline().split(' '),x=n.length,p=Math.pow,d=p(10,d)
x-=x%3
print(Math.round(n*d/p(10,x))/d+" kMGTPE"[x/3])

104 - Function

function(a,b,c,d){
    c=(''+a).length;
    d=Math.pow;
    b=d(10,b);
    return((a*b/d(10,c-=c%3))+.5|0)/b+' kMGTPE'[c/3]
}

Which also becomes 99 if you replace the (''+a) with a and promise to only pass strings :)

Thaler

I know someone has some clever way of shortening this! Maybe with some 1.6+ features or 1.8's [].reduce? – Kekkonen

Answer 5

Ruby - 79 77 75 83 chars

n,d=ARGV
l=n.to_s.length
printf"%.#{l>3?d:0}f%s",n.to_f/10**(l-l%3)," kMGTPE"[l/3]

Reads from command line arguments.

74 72 80 chars, prints output within double quotes

n,d=ARGV
l=n.to_s.length
p"%.#{l>3?d:0}f%s"%[n.to_f/10**(l-l%3)," kMGTPE"[l/3]]

66 74 chars, prints extra zeroes

n,d=ARGV
l=n.to_s.length
p"%.#{d}f%s"%[n.to_f/10**(l-l%3)," kMGTPE"[l/3]]

Based on this solution, and the sample code.

Nutwood

Answer 6

dc - 75 chars

A7 1:U77 2:U71 3:U84 4:U80 5:U69 6:U[3+r1-r]sJ?sddZd3~d0=Jrsp-Ar^ldk/nlp;UP

Uses Z (number of digits) %3 to find the unit. Most of the code is for setting the units character array, the real code is 39 chars. The J macro adjusts when %3 equals 0, to avoid printing 0.918M in the 7th. test case. It doesn't round properly.

If you speak dc, feel free to improve it.

Clubwoman

Answer 7

Haskell, 126 (without import, it's a function that takes two arguments):

f n p|l>3=showFFloat (Just p) (c n/c 10^(l-w)) [" kMGTPE"!!f]|True=show n where(f,w)=divMod l 3;c=fromIntegral;l=length$show n

Expanded:

import Numeric

doit :: Integer -> Int -> String
doit n p
    | l > 3 = showFFloat (Just p) d [" kMGTPE" !! f]
    | otherwise = show n
    where
    d = (fromIntegral n) / fromIntegral (10^(l-w))
    (f,w) = divMod l 3
    l = length $ show n

Exscind

Answer 8

PHP 57 chars

for($a=num+1;$a>=1;$a=$a/26)$c=chr(--$a%26+65).$c;echo$c;

Dowdell

Answer 9

Perl 94 Chars

($_,$d)=@ARGV;$l=length;@u=' kMGTPE'=~/./g;printf"%.".($l>3?$d:0)."f$u[$l/3]",$_/10**($l-$l%3)

Usage:

perl abbreviator.pl 47475782130 2

Output:

47.48G

Staminody