Code-Golf: Friendly Number Abbreviator

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;
}

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.