Sort numbers by binary 1's count

Question

Goal

Write a function or program sort an array of integers in descending order by the number of 1's present in their binary representation. No secondary sort condition is necessary.

Example sorted list

(using 16-bit integers)

  Dec                Bin        1's
16375   0011111111110111        13
15342   0011101111101110        11
32425   0111111010101001        10
11746   0010110111100010         8
28436   0000110111110100         8
19944   0100110111101000         8
28943   0000011100011111         8
 3944   0000011111101000         7
15752   0011110110001000         7
  825   0000000011111001         6
21826   0101010101000010         6

Input

An array of 32-bit integers.

Output

An array of the same integers sorted as described.

Scoring

This is code golf for the least number of bytes to be selected in one week's time.

Answer 1

J (11)

(\:+/"1@#:)

This is a function that takes a list:

     (\:+/"1@#:) 15342 28943 16375 3944 11746 825 32425 28436 21826 15752 19944
16375 15342 32425 28943 11746 28436 19944 3944 15752 825 21826

If you want to give it a name, it costs one extra character:

     f=:\:+/"1@#:
     f 15342 28943 16375 3944 11746 825 32425 28436 21826 15752 19944
16375 15342 32425 28943 11746 28436 19944 3944 15752 825 21826

Explanation:

• \:: downwards sort on
• +/: sum of
• "1: each row of
• #:: binary representation

Answer 2

JavaScript, 39

Update: Now shorter than Ruby.

x.sort(q=(x,y)=>!x|-!y||q(x&x-1,y&y-1))

40

x.sort(q=(x,y)=>x&&y?q(x&x-1,y&y-1):x-y)

Explanation:

q is a recursive function. If x or y are 0, it returns x-y (a negative number if x is zero or a positive number if y is zero). Otherwise it removes the lowest bit (x&x-1) from x and y and repeats.

Previous version (42)

x.sort(q=(x,y)=>x^y&&!x-!y+q(x&x-1,y&y-1))

Answer 3

Ruby 41

f=->a{a.sort_by{|n|-n.to_s(2).count(?1)}}

Test:

a = [28943, 825, 11746, 16375, 32425, 19944, 21826, 15752, 15342, 3944, 28436];
f[a]
=> [16375, 15342, 32425, 11746, 28436, 28943, 19944, 15752, 3944, 21826, 825]

Answer 4

Python 3 (44):

def f(l):l.sort(lambda n:-bin(n).count('1'))

Answer 5

Common Lisp, 35

logcount returns the number of ‘on’-bits in a number. For a list l, we have:

(sort l '> :key 'logcount)

CL-USER> (sort (list 16375 15342 32425 11746 28436 19944 28943 3944 15752 825 21826) '> :key 'logcount)
;=> (16375 15342 32425 11746 28436 19944 28943 3944 15752 825 21826)

As a standalone function, and what I'll base the byte count on:

(lambda(l)(sort l'> :key'logcount))

Answer 6

Python 3, 90 77 72 67 characters.

Our solution takes an input from the command-line, and prints the number in descending order (67 chars), or ascending (66).

Descending order

print(sorted(input().split(),key=lambda x:-bin(int(x)).count("1"))) # 67

Thanks to @daniero, for the suggestion of using a minus in the 1's count to reverse it, instead of using a slice to reverse the array at the end! This effectively saved 5 characters.

Just for the sake of posting it, the ascending order version (which was the first we made) would take one character less.

Ascending order:

print(sorted(input().split(),key=lambda x:bin(int(x)).count("1"))) # 66

Thanks to @Bakuriu for the key=lambda x… suggestion. ;D

Answer 7

JavaScript [76 bytes]

a.sort(function(x,y){r='..toString(2).split(1).length';return eval(y+r+-x+r)})

where a is an input array of numbers.

Test:

[28943,825,11746,16375,32425,19944,21826,15752,15342,3944,28436].sort(function(x, y) {
    r = '..toString(2).split(1).length';
    return eval(y + r + -x + r);
});

[16375, 15342, 32425, 19944, 11746, 28943, 28436, 15752, 3944, 21826, 825]

Answer 8

Mathematica 30

SortBy[#,-DigitCount[#,2,1]&]&

Usage:

SortBy[#,-DigitCount[#,2,1]&]&@
                           {19944,11746,15342,21826,825,28943,32425,16375,28436,3944,15752}

{16375, 15342, 32425, 11746, 19944, 28436, 28943, 3944, 15752, 825, 21826}

Answer 9

k [15 Chars]

{x@|<+/'0b\:'x}

Example 1

{x@|<+/'0b\:'x}19944, 11746, 15342, 21826, 825, 28943, 32425, 16375, 28436, 3944, 15752

16375 15342 32425 28436 28943 11746 19944 15752 3944 825 21826

Example 2 (all numbers are 2^n -1)

{x@|<{+/0b\:x}'x}3 7 15 31 63 127

127 63 31 15 7 3

Answer 10

Java 8 - 87/113 81/111 60/80 60/74/48 characters

This is not a complete java program, it is just a function (a method, to be exact).

It assumes that java.util.List and java.lang.Long.bitCount are imported, and has 60 characters:

void s(List<Long>a){a.sort((x,y)->bitCount(x)-bitCount(y));}

If no pre-imported stuff are allowed, here it is with 74 characters:

void s(java.util.List<Long>a){a.sort((x,y)->x.bitCount(x)-x.bitCount(y));}

Add more 7 characters if it would be required that it should be static.

[4 years later] Or if you prefer, it could be a lambda (thanks @KevinCruijssen for the suggestion), with 48 bytes:

a->{a.sort((x,y)->x.bitCount(x)-x.bitCount(y));}

Answer 11

Mathematica 39

IntegerDigits[#,2] converts a base 10 number to list of 1's and 0's.

Tr sums the digits.

f@n_:=SortBy[n,-Tr@IntegerDigits[#,2]&]

Test Case

f[{19944, 11746, 15342, 21826, 825, 28943, 32425, 16375, 28436, 3944, 15752}]

{16375, 15342, 32425, 11746, 19944, 28436, 28943, 3944, 15752, 825, 21826}

Answer 12

Python 2.x - 65 characters (bytes)

print sorted(input(),key=lambda x:-sum(int(d)for d in bin(x)[2:]))

That's actually 66 characters, 65 if we make it a function (then you need something to call it which is lamer to present).

f=lambda a:sorted(a,key=lambda x:-sum(int(d)for d in bin(x)[2:]))

Demo in Bash/CMD:

echo [16, 10, 7, 255, 65536, 5] | python -c "print sorted(input(),key=lambda x:-sum(int(d)for d in bin(x)[2:]))"

Answer 13

Matlab, 34

Input in 'a'

[~,i]=sort(-sum(dec2bin(a)'));a(i)

Works for nonnegative numbers.

Answer 14

C - 85 bytes (108 106 bytes)

Portable version on GCC/Clang/wherever __builtin_popcount is available (106 bytes):

#define p-__builtin_popcount(
c(int*a,int*b){return p*b)-p*a);}
void s(int*n,int l){qsort(n,l,sizeof l,c);}

Ultra-condensed, non-portable, barely functional MSVC-only version (85 bytes):

#define p __popcnt
c(int*a,int*b){return p(*b)-p(*a);}
s(int*n,int l){qsort(n,l,4,c);}         /* or 8 as needed */

---

• First newline included in byte count because of the #define, the others are not necessary.

• Function to call is s(array, length) according to specifications.

• Can hardcode the sizeof in the portable version to save another 7 characters, like a few other C answers did. I'm not sure which one is worth the most in terms of length-usability ratio, you decide.

Answer 15

PowerShell v3, 61 58 53

$args|sort{while($_){if($_-band1){1};$_=$_-shr1}}-des

The ScriptBlock for the Sort-Object cmdlet returns an array of 1's for each 1 in the binary representation of the number. Sort-Object sorts the list based on the length of the array returned for each number.

To execute:

script.ps1 15342 28943 16375 3944 11746 825 32425 28436 21826 15752 19944

Answer 16

R, 132 96 94 88 84 75 73 53 51 bytes

-20 thanks to J.Doe's implementation -2 more thanks to Giuseppe

function(x)x[order(colSums(sapply(x,intToBits)<1))]

My original post:

pryr::f(rev(x[order(sapply(x,function(y)sum(as.double(intToBits(y)))))]))

Try it online!

I tried several different methods before I got down to this result.

Matrix Method: Created a two column matrix, one column with the input vector, one of the sum of the binary representation, then I sorted on the sum of binary.

function(x){m=matrix(c(x,colSums(sapply(x,function(y){as.integer(intToBits(y))}))),nc=2,nr=length(x));m[order(m[,2],decreasing=T),]}

Non-Matrix: Realized I could toss out the matrix function and instead create a vector of binary values, sum them, order them, then use the ordered values to reorder the input vector.

function(x){m=colSums(sapply(x,function(y){as.integer(intToBits(y))}));x[order(m,decreasing=T)]}

Minor Changes

function(x){m=colSums(sapply(x,function(y)as.double(intToBits(y))));x[order(m,decreasing=T)]}

More Minor Changes Converting entire thing to one line of code instead of two separated by a semicolon.

function(x)x[order(colSums(sapply(x,function(y)as.double(intToBits(y)))),decreasing=T)]

Sum Method Instead of adding the columns with colSums of the binary matrix created by sapply, I added the elements in the column before sapply "finished."

function(x)x[order(sapply(x,function(y)sum(as.double(intToBits(y)))),decreasing=T)]

Decreasing to Rev I really wanted to shorten decreasing, but R squawks at me if I try to shorten decreasing in the order function, which was necessary to get the order desired as order defaults to increasing, then I remembered the rev function to reverse a vector. EUREKA!!! The last change in the final solution was function to pryr::f to save 2 more bytes

function(x)rev(x[order(sapply(x,function(y)sum(as.double(intToBits(y)))))])

Answer 17

ECMAScript 6, 61

Assumes z is the input

z.sort((a,b)=>{c=d=e=0;while(++c<32)d+=a>>c&1,e+=b>>c&1},e-d)

Test data

[28943,825,11746,16375,32425,19944,21826,15752,15342,3944,28436].sort(
    (a,b)=>{
        c=d=e=0;
        while(++c<32)
            d+=a>>c&1,e+=b>>c&1
    },e-d
)

[16375, 15342, 32425, 11746, 19944, 28436, 28943, 15752, 3944, 21826, 825]

Thanks, toothbrush for the shorter solution.

Answer 18

PowerShell v2 or later, 44

sort{[convert]::ToString($_,2)-replace0}-des

The long version:

Sort-Object -Property {[convert]::ToString($_, 2) -replace '0', ''} -Descending

"sort" is a default alias for the Sort-Object cmdlet.
Sort-Object allows sorting the incoming list by evaluating an expression on the fly, which is the script block inside the curly braces.
The .NET class "Convert" is used to convert the incoming integer ("$_") into its binary string representation.
The "-replace" Operator will then remove all zeros, so that only a string of ones remains (the string "Replace()" method would have worked, too, but would have been longer).
Since these strings all consist of the same character, they'll be sorted by length, ascending by default. The switch argument "Descending" will reverse the order, and since there's another argument 'Debug', the first three characters are required for PowerShell to identify the switch.

The list to sort is expected in the pipeline/stdin, for example:

$l = 15342, 28943, 16375, 3944, 11746, 825, 32425, 28436, 21826, 15752, 19944
$l | sort{[convert]::ToString($_,2)-replace0}-des

Sort-Object
https://docs.microsoft.com/en-us/powershell/module/microsoft.powershell.utility/sort-object?view=powershell-6

Convert Class
https://msdn.microsoft.com/en-us/library/system.convert(v=vs.110).aspx

Answer 19

05AB1E, 4 bytes

ΣbSO

Try it online!

Answer 20

Perl 6, 27 bytes

*.sort(-*.base(2).comb.sum)

Try it online!

Answer 21

Haskell, 123C

import Data.List
import Data.Ord
b 0=[]
b n=mod n 2:b(div n 2)
c n=(n,(sum.b)n)
q x=map fst$sortBy(comparing snd)(map c x)

This is the first way I thought of solving this, but I bet there's a better way to do it. Also, if anyone knows of a way of golfing Haskell imports, I would be very interested to hear it.

Example

*Main> q [4,2,15,5,3]
[4,2,5,3,15]
*Main> q [7,0,2]
[0,2,7]

Ungolfed version (with explanations)

import Data.List
import Data.Ord

-- Converts an integer into a list of its bits
binary 0 = []
binary n = mod n 2 : binary (div n 2)

-- Creates a tuple where the first element is the number and the second element
-- is the sum of its bits.
createTuple n = (n, (sum.binary) n)

-- 1) Turns the list x into tuples
-- 2) Sorts the list of tuples by its second element (bit sum)
-- 3) Pulls the original number out of each tuple
question x = map fst $ sortBy (comparing snd) (map createTuple x)

Answer 22

Julia 1.0, 41 bytes

a->sort(a,by=x->sum((x>>n)&1 for n=0:63))

Try it online!

Answer 23

DFSORT (IBM Mainframe sorting product) 288 (each source line is 72 characters, must have space in position one)

 INREC IFTHEN=(WHEN=INIT,BUILD=(1,2,1,2,TRAN=BIT)), 
       IFTHEN=(WHEN=INIT,FINDREP=(STARTPOS=3,INOUT=(C'0',C'')))
 SORT FIELDS=(3,16,CH,D) 
 OUTREC BUILD=(1,2)

Just for fun, and no mathematics.

Takes a file (could be executed from a program which used an "array") with the integers. Before sorting, it translates the integers to bits (in a 16-character field). Then changes the 0s in the bits to nothing. SORT Descending on the result of the changed bits. Creates the sorted file with just the integers.

Answer 24

Smalltalk (Smalltalk/X), 36 (or maybe 24)

input in a; destructively sorts in a:

a sort:[:a :b|a bitCount>b bitCount]

functional version: returns a new sorted array:

a sorted:[:a :b|a bitCount>b bitCount]

there is even a shorter variant (passing the name or the function as argument) in 24 chars. But (sigh) it will sort highest last. As I understood, this was not asked for, so I don't take that as golf score:

a sortBySelector:#bitCount

Answer 25

CoffeeScript (94)

Readable code (212):

sort_by_ones_count = (numbers) ->
  numbers.sort (a, b) ->
    a1 = a.toString(2).match(/1/g).length
    b1 = b.toString(2).match(/1/g).length
    if a1 == b1
      0
    else if a1 > b1
      1
    else
      -1

console.log sort_by_ones_count [825, 3944, 11746, 15342, 15752, 16375, 19944, 21826, 28436, 28943, 32425]

Optimized (213):

count_ones = (number) -> number.toString(2).match(/1/g).length
sort_by_ones_count = (numbers) -> numbers.sort (a, b) ->
  a1 = count_ones(a)
  b1 = count_ones(b)
  if a1 == b1 then 0 else if a1 > b1 then 1 else -1

Obfuscating (147):

c = (n) -> n.toString(2).match(/1/g).length
s = (n) -> n.sort (a, b) ->
  a1 = c(a)
  b1 = c(b)
  if a1 == b1 then 0 else if a1 > b1 then 1 else -1

Ternary operators are excessively long (129):

c = (n) -> n.toString(2).match(/1/g).length
s = (n) -> n.sort (a, b) ->
  a1 = c(a)
  b1 = c(b)
  (0+(a1!=b1))*(-1)**(0+(a1>=b1))

Too long yet, stop casting (121):

c = (n) -> n.toString(2).match(/1/g).length
s = (n) -> n.sort (a, b) ->
  a1 = c(a)
  b1 = c(b)
  (-1)**(a1>=b1)*(a1!=b1)

Final (94):

c=(n)->n.toString(2).match(/1/g).length
s=(n)->n.sort((a, b)->(-1)**(c(a)>=c(b))*(c(a)!=c(b)))

Answer 26

Scala, 58

def c(l:List[Int])=l.sortBy(-_.toBinaryString.count(_>48))

Answer 27

PHP 5.4+ 131

I don't even know why I bother with PHP, in this case:

<?unset($argv[0]);usort($argv,function($a,$b){return strcmp(strtr(decbin($b),[0=>'']),strtr(decbin($a),[0=>'']));});print_r($argv);

Usage:

> php -f sortbybinaryones.php 15342 28943 16375 3944 11746 825 32425 28436 21826 15752 19944
Array
(
    [0] => 16375
    [1] => 15342
    [2] => 32425
    [3] => 28436
    [4] => 19944
    [5] => 11746
    [6] => 28943
    [7] => 3944
    [8] => 15752
    [9] => 825
    [10] => 21826
)

Answer 28

C

void main()
{
 int a[]={7,6,15,16};
 int b,i,n=0;
 for(i=0;i<4;i++)
 {  for(b=0,n=0;b<=sizeof(int);b++)
      (a[i]&(1<<b))?n++:n;   
    a[i]=n;
 }
 for (i = 1; i < 4; i++) 
  {   int tmp = a[i];
      for (n = i; n >= 1 && tmp < a[n-1]; n--)
         a[n] = a[n-1];
      a[n] = tmp;
  }    
}

Answer 29

C#, 88 89

int[] b(int[] a){return a.OrderBy(i=>-Convert.ToString(i,2).Count(c=>c=='1')).ToArray();}

Edit: descending order adds a character.

Answer 30

ECMAScript 6 (61 characters):

_=_=>_.toString(2).replace(/0/g,'');x.sort((a,b)=>_(b)-_(a))

Expects the input array to be in x.

Answer 31

Javascript 84

Inspired by other javascript answers, but without eval and regex.

var r=(x)=>(+x).toString(2).split('').reduce((p,c)=>p+ +c)
[28943,825,11746,16375,32425,19944,21826,15752,15342,3944,28436].sort((x,y)=>r(x)-r(y));

Answer 32

C - 124 111

Implemented as a method and using the standard library for the sorting. A pointer to the array and the size should be passed as parameters. This will only work on systems with 32-bit pointers. On 64-bit systems, some characters have to be spent specifying pointer definitions.

Indentation for readability

c(int*a,int*b){
    int d,e,i;
    for(d=e=i=0;i-32;){
        d+=*a>>i&1;e+=*b>>i++&1;
    }
    return d>e?-1:d<e;
}
o(r,s){qsort(r,s,4,c);}

Sample call:

main() {
    static int a[] ={1, 2, 3, 4, 5, 6, 7, 8, 9};
    o(a, 9);
}

Answer 33

Javascript (82)

a.sort(function(b,c){q=0;while(b|c){b%2?c%2?0:q++:c%2?q--:0;b>>=1;c>>=1}return q})

Answer 34

Postscript, 126

Because list of values by which we sort is known beforehand and very limited (32), this task can be easily done even if there's no built-in for sorting, by picking matching values for 1..32. (Is it O(32n)? Probably).

Procedure expects array on stack and returns 'sorted' array.

/sort_by_bit_count {
    [ exch
    32 -1 1 {
        1 index
        {
            dup 2 32 string cvrs
            0 exch
            {48 sub add} forall
            2 index eq 
            {3 1 roll} {pop} ifelse
        } forall
        pop
    } for
    pop ]
} def

Or, ritually stripped of white space and readability:

/s{[exch 32 -1 1{1 index{dup 2 32 string cvrs 0 exch{48 sub add}forall 2 index eq{3 1 roll}{pop}ifelse}forall pop}for pop]}def

Then, if saved to bits.ps it can be used like this:

gs -q -dBATCH bits.ps -c '[(%stdin)(r)file 1000 string readline pop cvx exec] s =='
825 3944 11746 15342 15752 16375 19944 21826 28436 28943 32425
[16375 15342 32425 11746 19944 28436 28943 3944 15752 825 21826]

I think it effectively is the same as this Perl (there's yet no Perl here, too):

sub f{map{$i=$_;grep{$i==(()=(sprintf'%b',$_)=~/1/g)}@_}reverse 1..32}

Though that, unlike Postscript, can be easily golfed:

sub f{sort{j($b)-j($a)}@_}sub j{$_=sprintf'%b',@_;()=/1/g}

Answer 35

Java 8: 144

static void main(String[]a){System.out.print(Stream.of(a).mapToInt(Integer::decode).sorted(Comparable.comparing(Integer::bitCount)).toArray());}

In expanded form:

static void main(String[] args){
    System.out.print(
        Stream.of(args).mapToInt(Integer::decode)
              .sorted(Comparable.comparing(Integer::bitCount))
              .toArray()
        );
}

As you can see, this works by converting the args to a Stream<String>, then converting to a Stream<Integer> with the Integer::decode function reference (shorter than parseInt or valueOf), and then sorting by Integer::bitCount, then putting it in an array, and printing it out.

Streams make everything easier.

Answer 36

Japt, 7 bytes

ñ_¤è1Ãw

Test it

---

Explanation

Implicit input of array U.

ñ_   Ã

Sort (ñ) by passing each element through a function.

¤

Convert to base 2 string.

è1

Count the number of 1s.

w

Reverse and implicitly output the resulting array.

---

Alternatives

ñ_¤o1Ãw
ñ_¤kTÃw

Answer 37

Javascript - Lengthy but result is according to my expectation - sorting numbers in ascending order according to number of binary 1s in them. If binary 1s are equal then sorting done on the base of decimal ascending order.

Thumbs up if anybody can improve or give shorter solution in javascript

  function sortNumberByBinary1s(numbers){
      //sort numbers first
      numbers=numbers.sort();
      console.log(numbers,'numbers');

      //2dimenstional array to store original number and its binary 1s
      var binNumArr = [];
      //loop to count binary 1s in each number
      for(i=0;i<numbers.length;i++){
        var num = numbers[i];
        // var binNum = Number(num.toString(2));

        let binary1 = 0;
        do if(num&1) binary1++;
        while(num>>=1);

        //store original number and its binary 1s in 2dimenstional array
        binNumArr[numbers[i]]={'number':numbers[i],'binary1':binary1};

      }

    //sort binNumArr (22dimenstional array by Number value)
    //filter(() => true) to remove empty values
      var byNumSortedArray=binNumArr.sort(function(a, b){return a.number - b.number}).filter(() => true);

      //sort byNumberSorted array by binary1 value
      var byBinary1SortedArray=byNumSortedArray.sort(function(a, b){return a.binary1 - b.binary1});

      console.log(byBinary1SortedArray,'byBinary1SortedArray');

    //get only sortedNumbers from array and return result
      var sortedNumbers=[]
      for(i=0;i<byBinary1SortedArray.length;i++){
        try{
          var number = byBinary1SortedArray[i].number;
        }catch(e){
          console.log(e,'error');
        }
        console.log(number,'number');
        sortedNumbers[i]=number;

      }
      console.log(sortedNumbers,'sortedNumbers array');
      return sortedNumbers;

    }

Input: [3,4,5,6,7,8]

  sortNumberByBinary1s([3,4,5,6,7,8]);

• {number: 4, binary1: 1}
• {number: 8, binary1: 1}
• {number: 3, binary1: 2}
• {number: 5, binary1: 2}
• {number: 6, binary1: 2}
• {number: 7, binary1: 3}

Output: [4, 8, 3, 5, 6, 7]

Answer 38

Stax, 6 bytes

╖a☻║B⌡

Run and debug it at staxlang.xyz!

This will output the Unicode characters in a string. Add $ at the end (or step through execution) to see decimal.

Unpacked (7 bytes) and Explanation

{:B|+or
{          Begin block.
 :B          Convert to binary array. Ex: 11746 -> [1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0]
   |+        Sum.
     o     Terminate block and use it to sort array (ascending).
      r    Reverse. Implicit print.

Before descending order was specified in the question, there was a 6-byte unpacked solution. This is short enough that packing (into ÇÄì¬♥#) changed nothing:

{:B|+o

Run and debug it at staxlang.xyz!

Answer 39

Haskell, 57 54 bytes

^{-3 bytes thanks to Laikoni! (making the counting function pointfree by using the fact that we only need to support 32bit ints)}

sortOn$sum.(mapM(pure[0,1])[1..32]!!)
import Data.List

Try it online!

Explanation

The function call sortOn f xs returns xs :: [x] sorted by comparing ys = map (f :: x -> y) xs where the type y has to be an instance of Ord, so we just need to feed it a function that counts the number of 1-bits:

(sum . mapM (const [0,1]) [1..32] !!)

The expression mapM (const [0,1]) [1..n] generates all the sequences of [0,1] with length n (n = 32 is sufficient since we only need to handle 32bit ints) in the right order, so we just need to index into these and take it's sum.

Example: mapM (const [0,1]) [1..2] !! 1 ≡ [[0,0],[0,1],[1,0],[1,1]] !! 2 ≡ [0,1]

Answer 40

Kotlin, 57 44 bytes

{it.sortedBy{-it.toString(2).sumBy{it-'0'}}}

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Answer 41

K (ngn/k), 11 10 bytes

{x@>+/2\x}

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Answer 42

MathGolf, 5 bytes

á{âΣ~

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Explanation:

á{âΣ~
á{     Sort by the
    ~  Negative
   Σ   Digit sum
  â    Of the number converted to binary

Answer 43

Powershell, 40 45 bytes

$args|sort{for(;$_;$_=$_-shr1){$s+=$_%2}$s}-d

Explanation:

Sort all arguments from the predefined $args in a descendant order by sum of bits $s.

Note: The script uses -shr1 instead /2 because a Powershell does not provide integer div.

Test script:

$f = {

$args|sort{for(;$_;$_=$_-shr1){$s+=$_%2}$s}-d

}

&$f 0 1 3 4
&$f 16375 15342 32425 11746 28436 19944 28943 3944 15752 825 21826

Output:

3
0
4
1
16375
15342
32425
19944
28943
11746
28436
15752
3944
21826
825

Answer 44

C#, 191 183 172 thanks to Rik

 using System.Linq;class P{static void Main(string[]a){System.Console.WriteLine(string.Join(" ",a.Select(int.Parse).OrderBy(v=>{int c=0;for(;v>0;c++,v&=v-1);return-c;})));}}

Formatted:

using System.Linq;
class P
{
    static void Main(string[] a)
    {
        System.Console.WriteLine(string.Join(" ", a.Select(int.Parse)
            .OrderBy(v => { 
                int c = 0; 
                for (; v > 0; c++, v &= v - 1);
                return -c; 
            })
        ));
    }
}

When run as foo.exe 32425 28943 28436 21826 19944 16375 15752 15342 11746 3944 825 the output is:

16375 15342 32425 28943 28436 19944 11746 15752 3944 21826 825

Including the bitcount:

16375   13
15342   11
32425   10
28943   8
28436   8
19944   8
11746   8
15752   7
3944    7
21826   6
825     5

As function only: 100 89 chars

...Which some seem to regard as correct too:

int[] v(int[]a){return a.OrderBy(v=>{int c=0;for(;v>0;c++,v&=v-1);return-c;}).ToArray();}

To use, call v() with an array of ints to be sorted by binary 1's count.

Answer 45

C++ and QT

template<typename T>

//Compars the number of ones in the binary representation of some numbers 
bool onesCountSort()(Const T* a, const T* b) const
{
   int aCount = getOnesCount(&a);
   int bCount = getOnesCount(&b);
   return aCount < bCount;
}

//Returns the number of ones in a number
int getOnesCount(T value) const
{
    int rVal = 0;
    while(value > 0)
    {
       rVal++;
       value = value >> 1;
    } 
    return rVal;
}    

int main(int argc, char *argv[])
{
   QList<int> listTest;
   listTest << 16375;
   listTest << 15342; 
   listTest << 32425; 
   listTest << 11746; 
   listTest << 28436;
   //...

   qSort(list.begin(); list.end(), onesCountSort<int>());//Sort by binary ones count
   qSort(list.end(); list.begin(), onesCountSort<int>());//Reverse the sort
}

Edit:

template<typename T>

//Compars the number of ones in the binary representation of some numbers 
bool onesCountSort()(Const T* a, const T* b) const
{
   return getOnesCount(a) < getOnesCount(b);
}

int getOnesCount(const T* value, int count = 0) const
{
    return &value ? count : getOnesCount(value >> 1, count + 1);
}        

Answer 46

GolfScript, 14 chars

~{2base 0-,}$`

Example input (sorted in numerical order):

[825 3944 11746 15342 15752 16375 19944 21826 28436 28943 32425]

Example output (sorted by bit count) for the input above:

[825 21826 15752 3944 28436 28943 19944 11746 32425 15342 16375]

Note that the program above sorts the input in ascending order by bit count. If you insist on descending order, that'll cost one extra char:

~{2base 0-,~}$`

Explanation:

• ~ evals the input. Since you didn't specify the input format, I took the liberty of assuming that the input is provided as a GolfScript array literal (i.e. a series of whitespace-separated numbers wrapped in [ ]).

• { }$ applies the code inside the braces to each element of the array, and sorts them according to the resulting sort keys.

  • Inside the braces, 2base uses the built-in base conversion operator to turn the number into an array of bits (e.g. 19 → [1 0 0 1 1]). The 0- then removes all the zero bits from the array (the space before it is needed to keep base0 from being parsed as a single token) and the , counts the length of the remaining array.

  • (In the descending order version, the ~ then bitwise negates the count, effectively applying the map x ↦ −(x + 1) and thus inverting the sort order.)

• Finally, the ` un-evals the sorted array, converting it back into the input format for output.

Answer 47

**

C++(Visual Studio 2013), 112

**

#define n int
#define s __popcnt
n* d(n i[], n l){sort(i, i+l, [](n a, n b){return  s(a) > s(b); });return i;}

Unobfuscated:

#define n int
#define s __popcnt
n* d(n i[], n l){
    std::sort(i, i+l, [](n a, n b){return  s(a) > s(b); });
    return i;
}

haven't touched C++ for a while. While not the shortest, it PROBABLY is the fastest, by far, though it depends highly if the __popcnt is implemented as single CPU instruction, or set.

I might fire up mathbrain later and see if it's possible to compare two numbers by bits and see which one has more bits set(without any string conversions or bit counting). I think there is some kind of bit-hack that can be used, but it's probably longer.

Answer 48

C : 112

C(a){int c=a!=0;while(a&=a-1)c++;return c;}
B(int*a,int*b){return C(*b)-C(*a);}
S(int*a,int n){qsort(a,n,4,B);}

works on gcc with Target: x86_64-linux-gnu

usage: S(array, size);

Answer 49

Haskell - 107100 (including sample list and main call) else 100-22=78

import Data.List
b 0=0
b n=mod n 2+b(div n 2)
c q=map snd$ sort$ zip(map b q)q
main=print$ c [1..100]

Result:

[1,2,4,8,16,32,64,3,5,6,9,10,12,17,18,20,24,33,34,36,40,48,65,66,68,72,80,96,7,11,13,14,19,21,22,25,26,28,35,37,38,41,42,44,49,50,52,56,67,69,70,73,74,76,81,82,84,88,97,98,100,15,23,27,29,30,39,43,45,46,51,53,54,57,58,60,71,75,77,78,83,85,86,89,90,92,99,31,47,55,59,61,62,79,87,91,93,94,63,95]

Answer 50

C, 95 chars

Call s(array, size) to sort an array.
The framework is based on AShelly's answer, with a different bit count function.
Only works on 32bit platforms.

c(unsigned a){return a?a%2+c(a/2):0;}
f(int*a,int*b){return c(*b)-c(*a);}
s(a,n){qsort(a,n,4,f);}

The bit count function treats the numbers as unsigned, so division will discard the low bit.

Answer 51

php 5.3, 117 109

Thank to @mniip to point out some more char save. So New One are

usort($a,function($u,$v){return $u==$v?0:(substr_count(decbin($u),'1')<substr_count(decbin($v),'1')?1:-1);});

Older One

usort($a,function($u,$v){if($u==$v)return 0;return substr_count(decbin($u),'1')<substr_count(decbin($v),'1')?1:-1;});

Answer 52

K, 14

{x@>+/'0b\:'x}

.

k){x@>+/'0b\:'x} 28943 16375 15342
16375 15342 28943

Answer 53

Javascript (ECMAScript 6) - 41 Characters

Takes an array a as input:

a.sort(ພ=(ຟ,ຝ)=>ຟ*ຝ?ພ(ຟ&ຟ-1,ຝ&ຝ-1):(ຝ-ຟ))

Testing (with less obfuscated code):

JSFIDDLE

a=[19944, 11746, 15342, 21826, 825, 28943, 32425, 16375, 28436, 3944, 15752];
a.sort(_=(b,c)=>b*c?_(b&b-1,c&c-1):(c-b))
console.log(a.toString());

Gives this output:

16375,15342,32425,19944,11746,28943,28436,3944,15752,21826,825

Answer 54

C# - 98

C# can't really compete in "the smallest" category. Still fun, though :)

x.Select(i=>new{v=i,c=Convert.ToString(i,2).Count(c=>c=='1')}).OrderByDescending(a=>a.c);

Used:

var y = new [] {28943, 825, 11746, 16375, 32425, 19944, 21826, 15752, 15342, 3944, 28436}.Select(i=>new{v=i,c=Convert.ToString(i,2).Count(c=>c=='1')}).OrderByDescending(a=>a.c);
Console.WriteLine(y.Select(a => a.v.ToString()).Aggregate((s1,s2)=>s1+","+s2));

Result:

16375,15342,32425,28943,11746,19944,28436,15752,3944,825,21826

EDIT: .NetFiddle - http://dotnetfiddle.net/ahMMbq

Answer 55

Longwinded C#:

using System;

namespace P
{
    class P
    {
        static int T(int v) { int i = 0; while (v != 0) { i += (v & 1); v >>= 1; } return i; }
        static int[] S(int[] a)
        { Array.Sort<int>(a, (x, y) => { return (T(x) > T(y)) ? -1 : (T(x) < T(y)) ? 0 : 1; }); return a; }
        static void Main(string[] args)
        {
            foreach (var i in S(new int[]{ 28943, 825, 11746, 16375, 32425, 19944, 21826, 15752, 15342, 3944, 28436 }))
                Console.WriteLine("{0}\t{1}\t{2}", i, Convert.ToString(i, 2), T(i));
            Console.ReadKey();
        }
    }
}

Answer 56

C# 95

int[]o(int[]d){return d.OrderByDescending(v =>Convert.ToString(v,2).Sum(c =>c-'0')).ToArray();}

Answer 57

PHP - 89 bytes

Here is the best I have been able to do so far.

function f($a){usort($a,function($a,$b){$c=gmp_popcount;return$c($a)<$c($b);});return$a;}

If I am allowed to sort the array in-place (which is idiomatic for PHP), then it is only 81:

function f(&$a){usort($a,function($a,$b){$c=gmp_popcount;return$c($a)<$c($b);});}

Answer 58

Perl, 56 bytes

sub x{$_=sprintf'%b',@_;s/1//g}print sort{x($b)<=>x$a}<>

Input is expected in STDIN, one integer per line. Output is in descending order.

sprintf '%b' converts the number into binary representation. s/1// replaces the 1, the return value is the number of replacements. Thus function x returns the number of 1 in the binary representation of the number. The sorting function sorts the number according to their binary 1's count. By exchanging $a and $b the sorting order is reversed to ascending.

Answer 59

Jelly, 5 bytes (non-competing?)

BS$ÞṚ

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Answer 60

PHP, 71

usort($a,function($a,$b){for(;$a&&$b;$a&=$a-1,$b&=$b-1);return$b-$a;});

If we assume there will be no dulicates, 69

foreach($a as$v)$b[$v]=gmp_popcount($v);arsort($b);$a=array_keys($b);

If we accept ascending sort, 68

foreach($a as$v)$b[$v]=gmp_popcount($v);asort($b);$a=array_keys($b);

That is, given $a. So, I'm unclear on whether we can assume the input, as various solutions go to various lengths to get it.

Answer 61

Perl 5, 51 bytes

sub j{(sprintf'%b',@_)=~y/1//}say sort{j($b)-j$a}<>

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Answer 62

Pyth, 8 bytes

_o/.BN\1

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Explanation:

_o/.BN\1Q #Code with implicit variables
 o      Q #  Input list sorted by
  /   \1  #   How many "1"s are in
   .BN    #    The binary string of each number
_         # Reversed

Answer 63

Python 2, 59 bytes

def f(l):l.sort(key=lambda x:`bin(x)`.count('1'),reverse=1)

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Answer 64

Pyt, 19 bytes

ĐĐ↑Ḷ⌊⁺ᴇĐ3ȘĦ*⇹↔+Ş⇹%Ɩ

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Explanation

                         Implicit input (as a list of integers)
ĐĐ                       Triplicate the list
  ↑                      Get the maximum value of the list (X_m)
   Ḷ⌊⁺ᴇĐ                 Push Y=10^(floor(log_10(X_m))+1) twice
        3Ș               Swap the top three elements on the list [puts the input on top]
          Ħ              Get the Hamming weight of each element of the input
           *             Multiply the Hamming weight by Y
            ⇹            Swap the top two elements on the stack
             ↔           Flip the stack [Puts the list on top, followed by Hamming weights multiplied by Y]
              +          Add the two lists element-wise [resulting in Z=[Z_0,Z_1,...]]
               Ş         Sort the resulting list in descending order
                ⇹        Swap the top two elements on the stack
                 %       Take Z mod Y
                  Ɩ      Convert Z mod Y to integers
                         Implicit print

This is as long as it is because Pyt doesn't allow sorting by anything other than the values in the array; this means that a workaround had to be devised to allow the proper sorting

Answer 65

K (oK), 19 bytes

{t@>+/'(99#2)\'t:x}

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Answer 66

Dart, 102 99 84 bytes

F(a)=>'1'.allMatches(a.toRadixString(2)).length;f(List l)=>l.sort((a,b)=>F(a)-F(b));

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-3 bytes by using allMatches instead of replaceAll

-15 bytes by using - instead of compareTo and replacing List< int > by List

Answer 67

Perl 6, 36 bytes

@a.sort({sum .base(2).comb}).reverse

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Answer 68

APL(NARS), 27 chars, 54 bytes

{⍵[⍒{+/(2⍴⍨⌊1+2⍟1+⍵)⊤⍵}¨⍵]}

test

  f←{⍵[⍒{+/(2⍴⍨⌊1+2⍟1+⍵)⊤⍵}¨⍵]}
  f 825 32425 15342 11746 28943 28436 21826 16375 15752 19944 3944
16375 15342 32425 11746 28943 28436 19944 15752 3944 825 21826 

Answer 69

Python-40 bytes

l.sorted(key=lambda x:bin(x).count('1'))

second to K!! Thwarted by K once again.......