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;
}

(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? — kennytm, Apr 22 '10 at 16:01
(1) Fixed (2) I updated to include that the code should include the full program. Thanks! — David Murdoch, Apr 22 '10 at 16:07
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. — dmckee --- ex-moderator kitten, Apr 22 '10 at 16:21
@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. — David Murdoch, Apr 22 '10 at 16:35
Whoa, 2^10 vs 10^3 is a pretty big difference, we should settle on one or the other. — Potatoswatter, Apr 22 '10 at 16:47
hm, whats the problem with my question? I'm going to get another tumbleweed badge if things don't pick up! — David Murdoch, Apr 22 '10 at 18:17
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. — Jeff B, Apr 22 '10 at 18:18
@David: No chance for tumbleweed, you've already got 5 upvotes :p — kennytm, Apr 22 '10 at 18:25

David · Accepted Answer · 2010-05-26T22:13:55.897

10

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.

edited May 26 '10 at 22:13

answered Apr 23 '10 at 17:08

David

7,011
1
42
38

output for `987654321987654321 4` should be `987.6543P`. and out put for `1234567 2` should be `1.23M` – David Murdoch Apr 23 '10 at 20:58
@David: I know, I caught that after posting it. Just spent the last half hour fixing that at a cost of 2 characters. :) – David Apr 23 '10 at 21:15
5

Also, it supports up to 999,999,999,999,999,999,999 - over 100x higher than the requirements. – David Apr 23 '10 at 21:42
4

Ok, after looking at your code I would like to say that your family is worried about you – May 26 '10 at 22:16
@M28: I'll take that as a compliment :) – David May 26 '10 at 22:18

kennytm · Answer 2 · 2010-04-22T19:30:56.487

7

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]

edited Apr 22 '10 at 19:30

answered Apr 22 '10 at 18:40

kennytm

510,854
105
1,084
1,005

has anyone verified this with some test cases? – David Murdoch Apr 26 '10 at 00:56

score 5 · Answer 3 · edited Oct 21 '13 at 23:27

5

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 :)

edited Oct 21 '13 at 23:27

yckart

32,460
9
122
129

answered Apr 22 '10 at 19:20

David Murdoch

87,823
39
148
191

I know _someone_ has some clever way of shortening this! Maybe with some 1.6+ features or 1.8's `[].reduce`? – David Murdoch Apr 27 '10 at 14:02

Zaid · Answer 4 · 2010-04-23T11:08:37.493

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

score 4 · Answer 5 · edited May 23 '17 at 12:26

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.

score 3 · Answer 6 · edited Jun 20 '20 at 09:12

3

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.

edited Jun 20 '20 at 09:12

Community

1
1

answered Apr 23 '10 at 18:23

Carlos Gutiérrez

13,972
5
37
47

score 1 · Answer 7 · answered Oct 11 '12 at 18:35

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

score 1 · Answer 8 · answered Jan 19 '11 at 09:03

1

PHP 57 chars

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

answered Jan 19 '11 at 09:03

score 0 · Answer 9 · answered Jul 02 '10 at 12:37

0

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

answered Jul 02 '10 at 12:37

Toto

89,455
62
89
125

Code-Golf: Friendly Number Abbreviator

9 Answers9

J, 61 63 65 characters

Python 2.x, 78 chars

Javascript 114 chars

Also 114 - Using spidermonkey - Input on STDIN

104 - Function

Perl 114 111 104 chars

Output:

Ruby - 79 77 75 83 chars

74 72 80 chars, prints output within double quotes

66 74 chars, prints extra zeroes

dc - 75 chars

Linked