Code-Golf: Friendly Number Abbreviator

Question

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.

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]

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

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 :)

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.

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.