问题
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;
}
回答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.
回答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]
回答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, $suffix[@n/3];# Select appropriate suffix
回答4:
Javascript 114 chars
function m(n,d){p=M.pow
d=p(10,d)
i=7
while(i)(s=p(10,i--*3))<=n&&(n=M.round(n*d/s)/d+"kMGTPE"[i])
return n}
Also 114 - Using spidermonkey - Input on STDIN
[n,d]=readline().split(' '),x=n.length,p=Math.pow,d=p(10,d)
x-=x%3
print(Math.round(n*d/p(10,x))/d+" kMGTPE"[x/3])
104 - Function
function(a,b,c,d){
c=(''+a).length;
d=Math.pow;
b=d(10,b);
return((a*b/d(10,c-=c%3))+.5|0)/b+' kMGTPE'[c/3]
}
Which also becomes 99 if you replace the (''+a) with a and promise to only pass strings :)
回答5:
Ruby - 79 77 75 83 chars
n,d=ARGV
l=n.to_s.length
printf"%.#{l>3?d:0}f%s",n.to_f/10**(l-l%3)," kMGTPE"[l/3]
Reads from command line arguments.
74 72 80 chars, prints output within double quotes
n,d=ARGV
l=n.to_s.length
p"%.#{l>3?d:0}f%s"%[n.to_f/10**(l-l%3)," kMGTPE"[l/3]]
66 74 chars, prints extra zeroes
n,d=ARGV
l=n.to_s.length
p"%.#{d}f%s"%[n.to_f/10**(l-l%3)," kMGTPE"[l/3]]
Based on this solution, and the sample code.
回答6:
dc - 75 chars
A7 1:U77 2:U71 3:U84 4:U80 5:U69 6:U[3+r1-r]sJ?sddZd3~d0=Jrsp-Ar^ldk/nlp;UP
Uses Z (number of digits) %3 to find the unit. Most of the code is for setting the units character array, the real code is 39 chars. The J macro adjusts when %3 equals 0, to avoid printing 0.918M in the 7th. test case. It doesn't round properly.
If you speak dc, feel free to improve it.
回答7:
PHP 57 chars
for($a=num+1;$a>=1;$a=$a/26)$c=chr(--$a%26+65).$c;echo$c;
回答8:
Haskell, 126 (without import, it's a function that takes two arguments):
f n p|l>3=showFFloat (Just p) (c n/c 10^(l-w)) [" kMGTPE"!!f]|True=show n where(f,w)=divMod l 3;c=fromIntegral;l=length$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
回答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
来源：https://stackoverflow.com/questions/2692323/code-golf-friendly-number-abbreviator