Sort odd numbers first

Question

Rearrange a given list such that all the odd numbers appear before all the even numbers. Besides for this requirement, the output list may be in any order.

The input will only contain integers, but they may be negative and there may be duplicates, and they may appear in any order.

Shortest solution wins.

Test cases

[1,2] → [1,2]

[2,1] → [1,2]

[1,0,0] → [1,0,0]

[0,0,-1] → [-1,0,0]

[3,4,3] → [3,3,4]

[-4,3,3] → [3,3,-4]

[2,2,2,3] → [3,2,2,2]

[3,2,2,2,1,2] → [1,3,2,2,2,2] or [3,1,2,2,2,2]

[-2,-2,-2,-1,-2,-3] → [-1,-3,-2,-2,-2,-2,] or [-3,-1,-2,-2,-2,-2,]

[] → []

Answer 1

05AB1E, 2 bytes

ΣÉ

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Answer 2

Pyth, 4 3 bytes

-1 byte thanks to isaacg

iD2

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_{^{crossed out 4 is still regular 4}}

Answer 3

J, 5 bytes

\:2&|

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\: sort descending by

2&| mod-2

Answer 4

R, 30 24 bytes

(x=scan())[order(!x%%2)]

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-6 bytes thanks to JayCe

Answer 5

C++, 79 76 64 bytes

[](auto a,auto b){while(a<--b)*a%2?*++a:(*a^=*b,*b^=*a,*a^=*b);}

This function accepts a pair of iterators (which must be random access iterators), and steadily moves them towards each other. When a points to an odd number, it is advanced. Otherwise, a points to an even number; b is decremented, and iter_swap'ed with a. (We use XOR swap, which saves us having to include <algorithm> - or <utility> for std::swap).

There are unnecessary swaps when b points to an even number, but we're golfing, not squeezing efficiency!

Demo

auto f=[](auto a,auto b){while(a<--b)*a%2?*++a:(*a^=*b,*b^=*a,*a^=*b);};

#include <array>
#include <iostream>
int main()
{
    auto a = std::array{ 3,2,2,5,2,1,2 };

    f(a.begin(),a.end());

    for (auto i: a)
        std::cout << i << " ";
    std::cout << std::endl;
}

---

Non-competitive answer

The natural C++ method is std::partition, but that comes out at 83 bytes:

#include<algorithm>
[](auto a,auto b){std::partition(a,b,[](auto x){return x&1;});}

Answer 6

Japt, 2 bytes

ñv

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Answer 7

Perl 6, 12 bytes

*.sort(*%%2)

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Some Whatever code that sorts the input by parity, with odd numbers first. You can remove a % to get even numbers first instead. Note that 'Whatever' is the name of this sort of anonymous function.

Answer 8

MATL, 6 bytes

tog&)v

Try it on MATL Online

Alternately:

to_2$S

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Answer 9

Python 2, 37 36 bytes

lambda a:sorted(a,key=lambda x:~x%2)

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1 byte tip o' the hat to Mr. Xcoder.

Answer 10

Haskell, 23 22 bytes

f odd<>f even
f=filter

Try it online! This is equivalent to

g x = filter odd x ++ filter even x

-1 byte thanks to Lynn

---

Other approaches:

(<>).($odd)<*>($even)$filter
f x=[i|m<-[0,1],i<-x,odd$m+i]
f x=[i|m<-[1,0],i<-x,mod i 2==m]
f x=id=<<filter<$>[odd,even]<*>[x]

Answer 11

JavaScript (Node.js), 29 bytes

a=>a.sort((a,b)=>(b&1)-(a&1))

Try it online! Save 4 bytes by only supporting positive values using b%2-a%2. If you write this as:

function(a){return a.sort((a,b)=>(b&1)-(a&1))}

then it will work on all sorts of old JavaScript implementations that didn't sort stably.

Answer 12

Stax, 5 bytes

{|eom

Run and debug it

Explanation:

{|eom Full program, implicit input
{  o  Sort by key:
 |e     Is odd?
    m Map over result:
        Implicit output with newline

Answer 13

Attache, 11 bytes

SortBy!Even

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Explanation

Even returns true for even numbers and false otherwise. SortBy ranks false < true (by a numerical cast to 0 < 1), thus placing odd numbers before even ones.

Answer 14

T-SQL, 26 bytes

SELECT*FROM t ORDER BY~i&1

Uses the bitwise AND operator "&" to compare the last digit with 1.

EDIT: Bitwise NOT then shorter than adding 1. EDIT2: Reorder to allow removal of the space.

Answer 15

Jelly, 3 bytes

ḂÞṚ

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One of the more oft-wanted atoms seems to be an is-even one (which would make this 2 bytes), without it we must reverse I believe...

ḂÞṚ - Link: list of integers
 Þ  - sort by:
Ḃ   -   bit (least significant bit - i.e. 1 if odd 0 if even)
  Ṛ - reverse

Answer 16

PHP, 55 bytes

~14 months later and I'm a little bit better at golfing now:

for(;''<$n=$argv[++$i];$s=$n%2?"$n $s":"$s $n");echo$s;

Try it online!

---

PHP (>=5.4), 84 82 bytes

(-2 bytes, thanks to Ismael Miguel)

<?array_shift($a=&$argv);usort($a,function($i){return$i%2==0;});echo join(' ',$a);

To run it:

php -n <filename> <number_1> <number_2> ... <number_n>

Example:

php -n sort_odds_first.php 2 0 1 -1 3 8 29 -666

Or Try it online!

Answer 17

JavaScript, 22 20 bytes

a=>a.sort(a=>!(a%2))

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Answer 18

Red, 42 bytes

func[a][sort/compare a func[n][n % 2 = 1]]

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If we need to account for negative values:

Red, 43 bytes

func[a][sort/compare a func[n][n // 2 = 1]]

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Answer 19

Husk, 4 bytes

↔Ö%2

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Explanation

 Ö     sort input according to the result of the following function
  %2   modulo 2
↔      reverse result to get odd numbers to the front

Answer 20

Scala, 18 bytes

_.sortBy(_.+(1)%2)

Answer 21

Ruby, 23 bytes

->a{a.sort_by{|i|~i%2}}

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Explanation:

sort_by sorts every number as if its value were the result of the block (~i%2)

~x is equivalent to -x-1 and takes precedence over %2

Odd numbers will evaluate to 0 and even numbers will evaluate to 1 so odd numbers will be sorted first.

Barely related: this works on ruby from homebrew 2.5.1p57 (because it's based on a small bug) but only for non-negative integers, 20 bytes

->a{a.sort{|i|i%-2}}

Explanation:

This uses sort which expects a block that takes 2 values and returns -1, 0 or 1 depending on whether the first one is bigger, they're equal or the second one is bigger.

The block given here ignores the second value and returns -1 if the first number is odd or 0 if it's even.

It's not guaranteed to work but it does in some (I think buggy) implementations.

Answer 22

PowerShell, 22 19 bytes

$args|sort{!($_%2)}

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Takes input via splatting, e.g., $a=(3,4,3); .\sort-odd-numbers-first.ps1 @a, which on TIO manifests as separate arguments for each entry.

Like some other answers here, Sort-Object can compare based on an expression. Here the expression is !($_%2), i.e., odds get sorted to $false and evens get sorted to $true. Thanks to how Boolean values are compared, falsey values are sorted first. This moves the odds to the beginning of the output, and the evens to the end. Sort-Object is stable, so the ordering of the individual items in their respective categories doesn't change (as in the TIO example).

-3 bytes thanks to mazzy.

Answer 23

JavaScript (Chrome v67) - 24 19 23 bytes

a=>a.sort(a=>!(a&1)-.5)

The use of &1 rather than Math.abs()%2 was stolen from @Neil. Thanks!

Thanks to @Shaggy for showing my hacky 19 byte solution wasn't valid. If anyone wants it:

_{Depends on how the browser handles a hacky return value of 0. Chrome v67, after 100000 iterations of random arrays never sorted it wrong. I'm very sure it works -- and it relies on the specific sort algorithm Chrome uses too, I believe. (It might work in other browsers, that's not the point)}

a=>a.sort(a=>++a&1)

Answer 24

C#, 23 bytes

i=>i.OrderBy(u=>u%2==0)

Quite straigt forward really: This basically converts the numbers to booleans, whilst true means that the number is even and false that it's odd. Because true is higher than false the even numbers appear first.

The formatted version looks like that:

i => i.OrderBy (u => u % 2 == 0)

And you can test it like that:

Console.WriteLine (string.Join (",", new Func <IEnumerable <int>, IEnumerable <int>> (
                                    i => i.OrderBy (u => u % 2 == 0)
                                ).Invoke (new [] {3, 2, 2, 2, 1, 2, 5, 5})));

Which results in the following:

3,1,5,5,2,2,2,2

Answer 25

JavaScript, 23 bytes

6 bytes shorter than @Neil's answer using the same language :D

a=>a.sort(n=>-(n&1)||1)

Explanation:

The function passed to sort only cares about the first parameter. If it is odd it returns -1 (the result of -(n&1)). Otherwise (when -(n&1) yields 0) it returns 1.

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Answer 26

Python, 35 bytes

lambda l:sorted(l,key=(-1).__pow__)

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Sorts by the function x -> (-1)**x, which gives -1 for odd and 1 for even.

Answer 27

JavaScript, 21 bytes

a=>a.sort(n=>(-1)**n)

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Answer 28

6502 machine code routine, 47 bytes

A0 00 84 FE B1 FB 4A 90 07 C8 C4 FD F0 20 D0 F4 2A 85 02 84 FE A4 FD 88 C4 FE
F0 12 B1 FB 4A 90 F6 2A AA A5 02 91 FB A4 FE 8A 91 FB 90 D6 60

Expects a pointer to an array of numbers in $fb/$fc and the length of this array in $fd. Manipulates the array in place to have all odd numbers in front. This is position independent code, so no load address is needed.

As the 6502 is an 8bit chip (so the instructions only deal with 8bit values, optionally signed), the valid number range is [-128 .. 127] and the maximum array size is 256.

Commented disassembly

; function to "partially sort" array, so all odd numbers come before all
; even numbers.
;
; input:
;   $fb/$fc: address of array to sort
;   $fd:     length of array to sort, 0 means 256 (maximum size)
;
; clobbers:
;   $fd/$fe: position from back/front of array
;   $2:      temporary for exchanging two values
;   A, X, Y

 .oddfirst:
A0 00       LDY #$00            ; initialize index from front
84 FE       STY $FE             ; to 0

 .search_front:
B1 FB       LDA ($FB),Y         ; load number from front
4A          LSR A               ; check for even/odd by shifting
90 07       BCC .search_back    ; if odd -> to searching from back
C8          INY                 ; next position from front
C4 FD       CPY $FD             ; same as position searching from back?
F0 20       BEQ .done           ; then we're finished
D0 F4       BNE .search_front   ; else check next from front
 .search_back:
2A          ROL A               ; shift carry back in
85 02       STA $02             ; and save number to temp
84 FE       STY $FE             ; save index from front
A4 FD       LDY $FD             ; load index from back
 .sb_loop:
88          DEY                 ; previous position from back
C4 FE       CPY $FE             ; same as position searching from front?
F0 12       BEQ .done           ; then we're finished
B1 FB       LDA ($FB),Y         ; load number from back
4A          LSR A               ; check for even/odd by shifting
90 F6       BCC .sb_loop        ; if odd -> check previous position
2A          ROL A               ; shift carry back in
AA          TAX                 ; remember in X
A5 02       LDA $02             ; load temporary from front
91 FB       STA ($FB),Y         ; store at current position
A4 FE       LDY $FE             ; load index from front
8A          TXA                 ; load remembered number
91 FB       STA ($FB),Y         ; store at current position
90 D6       BCC .search_front   ; and back to searching from front
 .done:
60          RTS

---

Example C64 assembler program using the routine:

Online demo

Code in ca65 syntax:

.import oddfirst ; link with routine above

.segment "BHDR" ; BASIC header
                .word   $0801           ; load address
                .word   $080b           ; pointer next BASIC line
                .word   2018            ; line number
                .byte   $9e             ; BASIC token "SYS"
                .byte   "2061",$0,$0,$0 ; 2061 ($080d) and terminating 0 bytes

.bss
linebuf:        .res    5               ; maximum length of a valid signed
                                        ; 8-bit number input
convbuf:        .res    3               ; 3 BCD digits for signed 8-bit
                                        ; number conversion
numbers:        .res    $100            ; maximum array size that can be
                                        ; directly handled with indexing
                                        ; instructions

.data
prompt:         .byte   "> ", $0
message:        .byte   $d, $d, "Enter one number per line.", $d
                .byte   "just press enter (empty line) when done.", $0
errmsg:         .byte   "Error parsing number, try again.", $d, $0

.code
                lda     #$17            ; set upper/lower mode
                sta     $d018

                lda     #0
                sta     $2a             ; index for number array
                sta     $52             ; flag that at least one number was
                                        ; entered

                lda     #<message       ; display message
                ldy     #>message
                jsr     $ab1e

inputloop:
                lda     #<prompt        ; display prompt
                ldy     #>prompt
                jsr     $ab1e

                lda     #<linebuf       ; read string into buffer
                ldy     #>linebuf
                ldx     #5
                jsr     readline

                lda     linebuf         ; empty line?
                beq     process         ; -> start processing

                lda     #<linebuf       ; convert input to int8
                ldy     #>linebuf
                jsr     toint8
                bcc     numok           ; successful -> store number
                lda     #<errmsg        ; else show error message and repeat
                ldy     #>errmsg
                jsr     $ab1e
                bcs     inputloop

numok:          ldx     #$ff            ; set flag that we have a number
                stx     $52
                ldx     $2a
                sta     numbers,x
                inc     $2a             ; next index
                bne     inputloop       ; if array not full, next input

process:        lda     $52             ; check we have some numbers
                beq     exit            ; otherwise exit program

                lda     #<numbers       ; address of array to $fb/fc
                sta     $fb
                lda     #>numbers
                sta     $fc
                lda     $2a             ; length of array to $fd
                sta     $fd
                jsr     oddfirst        ; call "sorting" function

                lda     #$0             ; index variable for output loop
                sta     $52
outloop:        ldy     $52             ; load current index
                lda     numbers,y       ; load current number
                jsr     printnum        ; -> output
                inc     $52             ; next index
                lda     $52             ; compare with ...
                cmp     $2a             ; ... array size
                bne     outloop         ; not reached yet -> repeat

exit:           rts                     ; done, exit program

; read a line of input from keyboard, terminate it with 0
; expects pointer to input buffer in A/Y, buffer length in X
.proc readline
                dex
                stx     $fb
                sta     $fc
                sty     $fd
                ldy     #$0
                sty     $cc             ; enable cursor blinking
                sty     $fe             ; temporary for loop variable
getkey:         jsr     $f142           ; get character from keyboard
                beq     getkey
                sta     $2              ; save to temporary
                and     #$7f
                cmp     #$20            ; check for control character
                bcs     checkout        ; no -> check buffer size
                cmp     #$d             ; was it enter/return?
                beq     prepout         ; -> normal flow
                cmp     #$14            ; was it backspace/delete?
                bne     getkey          ; if not, get next char
                lda     $fe             ; check current index
                beq     getkey          ; zero -> backspace not possible
                bne     prepout         ; skip checking buffer size for bs
checkout:       lda     $fe             ; buffer index
                cmp     $fb             ; check against buffer size
                beq     getkey          ; if it would overflow, loop again
prepout:        sei                     ; no interrupts
                ldy     $d3             ; get current screen column
                lda     ($d1),y         ; and clear 
                and     #$7f            ;   cursor in
                sta     ($d1),y         ;   current row
output:         lda     $2              ; load character
                jsr     $e716           ;   and output
                ldx     $cf             ; check cursor phase
                beq     store           ; invisible -> to store
                ldy     $d3             ; get current screen column
                lda     ($d1),y         ; and show
                ora     #$80            ;   cursor in
                sta     ($d1),y         ;   current row
                lda     $2              ; load character
store:          cli                     ; enable interrupts
                cmp     #$14            ; was it backspace/delete?
                beq     backspace       ; to backspace handling code
                cmp     #$d             ; was it enter/return?
                beq     done            ; then we're done.
                ldy     $fe             ; load buffer index
                sta     ($fc),y         ; store character in buffer
                iny                     ; advance buffer index
                sty     $fe
                bne     getkey          ; not zero -> ok
done:           lda     #$0             ; terminate string in buffer with zero
                ldy     $fe             ; get buffer index
                sta     ($fc),y         ; store terminator in buffer
                sei                     ; no interrupts
                ldy     $d3             ; get current screen column
                lda     ($d1),y         ; and clear 
                and     #$7f            ;   cursor in
                sta     ($d1),y         ;   current row
                inc     $cc             ; disable cursor blinking
                cli                     ; enable interrupts
                rts                     ; return
backspace:      dec     $fe             ; decrement buffer index
                bcs     getkey          ; and get next key
.endproc

; print an int8 number to the screen
; input:
;   A - the number to print
; clobbers:
;   X, Y
.proc printnum
                bpl     doprint         ; positive? -> direct number output
                eor     #$ff            ; else invert, 
                sta     $2              ; ...
                inc     $2              ; add one,
                lda     #'-'            ; output a minus sign
                jsr     $e716
                lda     $2
doprint:        tax                     ; number to X reg
                lda     #$0             ; set A to 0
                jsr     $bdcd           ; routine for uint16 in X/A output
                lda     #' '            
                jmp     $e716           ; and print a space
.endproc

; parse / convert int8 number using a BCD representation and double-dabble,
; handle negative numbers.
.proc toint8
                sta     $fb
                sty     $fc
                ldy     #$0
                sty     $fd
                sty     $fe
                sty     convbuf
                sty     convbuf+1
                sty     convbuf+2
scanloop:       lda     ($fb),y
                beq     copy
                iny
                cmp     #$20
                beq     scanloop
                cmp     #$2d
                beq     minus
                cmp     #$30
                bcc     error
                cmp     #$3a
                bcs     error
                inc     $fd
                bcc     scanloop
minus:          lda     $fd
                bne     error
                lda     $fe
                bne     error
                inc     $fe
                bne     scanloop
error:          sec
                rts
copy:           dey
                bmi     error
                ldx     #$2
copyloop:       lda     ($fb),y
                cmp     #$30
                bcc     copynext
                cmp     #$3a
                bcs     copynext
                sec
                sbc     #$30
                sta     convbuf,x
                dex
copynext:       dey
                bpl     copyloop
                lda     #$0
                sta     $fb
                ldx     #$8
loop:           lsr     convbuf
                lda     convbuf+1
                bcc     skipbit1
                ora     #$10
skipbit1:       lsr     a
                sta     convbuf+1
                lda     convbuf+2
                bcc     skipbit2
                ora     #$10
skipbit2:       lsr     a
                sta     convbuf+2
                ror     $fb
                dex
                beq     done
                lda     convbuf
                cmp     #$8
                bmi     nosub1
                sbc     #$3
                sta     convbuf
nosub1:         lda     convbuf+1
                cmp     #$8
                bmi     nosub2
                sbc     #$3
                sta     convbuf+1
nosub2:         lda     convbuf+2
                cmp     #$8
                bmi     loop
                sbc     #$3
                sta     convbuf+2
                bcs     loop
done:           lda     $fe
                beq     positive
                lda     #$ff
                eor     $fb
                sta     $fb
                inc     $fb
positive:       lda     $fb
                clc
                rts
.endproc

Answer 29

C (gcc), 71 49 48 bytes

Many thanks to the commenters for the suggestions. Apparently gcc is happy with just a partial ordering of {equal, greater than} for qsort (just to check, clang seems to be happy with this as well.) Although I'm not happy with hardcoding type sizes, it does reduce 7 bytes from sizeof b: change 4 to 8 for systems with a 64-bit int.

q(int*a){a=1-*a&1;}f(a,b)int*a;{qsort(a,b,4,q);}

Try it online!

Alternative version using two passes through the array (79 bytes):

i,j;f(a,b)int*a;{for(i=2;~--i;)for(j=b;~--j;)!(a[j]%2)-i&&printf("%d\t",a[j]);}

Try it online!

Answer 30

Elixir, 37 35 bytes

Code:

fn x->Enum.sort_by x,&(-rem&1,2)end

Expanded version:

fn x -> Enum.sort_by(x, fn y -> -rem(y, 2) end) end

Try it online!

Answer 31

Java 8, 46 29 27 bytes

l->l.sort((i,j)->1-(i&1)*2)

Uses Java 8's List.sort(), thanks to JollyJoker for making me aware of it and for saving me 19 bytes. Try it online here.

Ungolfed:

l -> // lambda taking a List of Integers and modifying it
        l.sort( // sort the List according to ...
        (i, j) ->                // ... a Comparator<Integer> with a compare() method ...
                 1 - (i & 1) * 2 // ... that finds a number smaller than another if it's odd, and greater otherwise

Answer 32

C++, 174 165 164 149 129 bytes

• Replace the two include lines with a single include, regex, that happens to include both required headers

 

#include <regex>
int main(int c,char**v){std::sort(v+1,v+c,[](auto s,auto t){return atoi(s)%2>atoi(t)%2;});for(;*++v;)puts(*v);}

149 Bytes

• For loop trick for printing. This uses the fact that, on Linux at least, the argv array is null-terminated.
• replace &1 with %2 to remove need for some brackets. Thank you Toby Speight!

 

#include<algorithm>
#include<cstdio>
int main(int c,char**v){std::sort(v+1,v+c,[](auto s,auto t){return atoi(s)%2>atoi(t)%2;});for(;*++v;)puts(*v);}

Older: 164 bytes

#include<algorithm>
#include<cstdio>
int main(int c,char**v){std::sort(v+1,v+c,[](auto s,auto t){return(atoi(s)&1)>(atoi(t)&1);});for(auto i=1;i<c;i++)puts(v[i]);}

Compile as C++14 or higher. Input is taken on the command line, and prints to standard output.

EDIT: removed un-needed return 0; statement

EDIT 2: Removed space after 'return'

Answer 33

Clojure - 35 bytes

(defn o[c](sort(fn[p _](odd? p))c))

Ungolfed:

(defn oddsort [col]
  (sort (fn [p _] (odd? p)) col))

Answer 34

Retina 0.8.2, 21 bytes

O$`.*(([02468])|.)
$2

Try it online! Link includes header that splits on commas for convenience. Explanation: The last digit of the number is captured if it is even, otherwise it is just matched (this avoids degenerate matches). The $ option then uses the matched digit as the sort key. Odd numbers don't match and end up with an empty sort key, causing them to be sorted first (in their original order), then the even numbers are sorted in order of last digit.

Answer 35

T-SQL, 33 bytes

SELECT*FROM t ORDER BY 1-ABS(i)%2

Per our IO standards, input is taken via pre-existing table t with integer column i.

The ABS() function is annoyingly necessary here because the T-SQL modulus operator % returns negative values for negative numbers, which in this case meant I was getting three possible results for i%2: -1, 0, and 1.

Subtracting from 1 is a couple bytes shorter than using DESC.

Note that per the question, I'm not doing any secondary sort, so the order of the odds and evens isn't predictable. If you want a secondary sort, add 2 bytes by adding ,i to the end.

Answer 36

sed 4.2.2, 51 43 bytes

Includes +1 for -r

-8 Thanks to Toby Speight

:
s/([^,]*[02468]),([^,]*[13579])/\2,\1/
t

Try it online!

Answer 37

Python 2, 34 bytes

lambda a:sorted(a,lambda x,y:x%-2)

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This seems to work, but I don't actually understand why it works. If anyone can explain why my half-witted compare function (It returns -1 if the first argument is odd and 0 if the first argument is even, and that's it) works, I'd greatly appreciate it.

Answer 38

><>, 40 bytes

l\ /nao01.
:<}\?%2:$-1/?(1
l0=?;naof2.\~

Try it online!

Filter through the list, printing odd numbers, keeping even numbers
l                  Get the length of the stack, use it as an iterator
:          /?(1    Check that there is at least one value left to process
        $-1        Decrement the iterator
    \?%2           Test if the next number is divisible by two
    /nao           Print it if so,
   }               Else, move it to the bottom of the stack
            ~      Delete the iterator after exiting the loop

Print the even numbers left on the stack
l0=?               If there are no values left on the stack,
    ;              Terminate
     nao           Print the next value

Answer 39

Clojure, 18 bytes

#(sort-by even? %)

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Answer 40

Haskell, 44 bytes

Not as byte efficient as Laikoni's answer, but a different approach.

import Data.List
f=uncurry(++).partition odd

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Using a sort instead, 47 bytes.

import Data.List
f=sortBy((.even).compare.even)

Try it online!

Answer 41

><>, 25 bytes

l/!?:-1$v?%2:{~
$/<+2oan<

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Input is via the -v flag. Ends with an error.

Explanation:

l\            ~  Push length of stack as the counter
 \<

         ?%2:{   Pull a number from the bottom of the stack and check if it is odd

 /   -1$v  If even, subtract one from the counter
 /<+2oan<  If odd, print the number and a newline and add two to the counter

 /!?:      If the counter is 0, go down
$/+2oan<   Print the even number anyway, and set the counter to 2

For a non-erroring version, you can replace the last ~ on the first line with ;?=1

Answer 42

Perl 5 with -M5.010 + -MList::MoreUtils+(sort_by), 21 bytes

sub{sort_by{$_%-2}@_}

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Answer 43

Wolfram Language (Mathematica), 15 12 bytes

SortBy@EvenQ

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(saved three bytes thanks to JungHwan Min)

---

Previous version:

#~SortBy~EvenQ&

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Answer 44

Awk: 50 bytes

{and($1,1)?(o=o $1 RS):(e=e $1 RS)}END{print o e}

Input on stdin, Output on stdout

I use RS, the record separator, as a stand-in for "\n", saving two characters each time.

Answer 45

Ruby, 19 20 bytes

Works on ruby 2.5.1p57 (2018-03-29 revision 63029) [x86_64-linux] from Arch Linux repos.

->a{a.sort{|i|~i%2}}

Answer 46

Pyke, 4 bytes

.#t1.&

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.#t1.& - sorted(input, key=lambda i: ↓)
  t    -   i-1
   1.& -  ↑ & 1

4 byte hexdump:

A3 74 31 A6

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Answer 47

Rust, 38 bytes

|l:&mut Vec<_>|l.sort_by_key(|e|1^e&1)

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Answer 48

Swift 4, 40 bytes

print(n.filter{$0%2>0}+n.filter{$0%2<1})

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n is the input array

print will be like: [1, 3, 2, 2, 2]

Answer 49

Mathematica 25 bytes

Flatten@GatherBy[#, OddQ] &

Answer 50

Chip -z, 44 bytes

V z.
>~#<
|at|Bb
|A-}]~S
>vvv+Cc
DEFGH
defgh

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Chip works in bytes, so this expects a sequence of byte-width values on stdin. I use 0x00 as a special delimiter, so that can't be one of the values. (This could be modified to use a different delimiter, at the expense of code size).

The TIO link above uses ASCII "123456789" as the input, corresponding to values 0x31, 0x32, 0x33, ..., 0x39.

This solution reads though the input twice (it knows it hit the end when it sees the 0x00 mentioned earlier, which is tacked onto the end by the flag -z). On the first pass, it prints the odd values, on the second the evens. It actually starts a third pass, but the termination logic kicks in before it does anything else.

Let's look at a rough overview of the code (Chip is 2D, so regions are described, each line on its own would do nothing useful):

           The V is a bookmarker of sorts. The mark is set at the start of a
V z.       pass, and then recalled when the input is exhausted. This does the
>~#<       looping (as opposed to using a stack or queue to hold the data).
|at|Bb     The # is a half-adder, it increments on each pass. When it
>A-}]~S    carries, the t is triggered, terminating the program. The S
>vvv+Cc    suppresses the values we don't want to print on this pass.
DEFGH      The remaining A-H letter pairs are a cat program, smooshed around
defgh      so that the input elements (capital letters) can do double duty,
           for things like detecting oddness, and the end of input.

Answer 51

Lua, 142 134 bytes

loadstring'p={...}n={}for i=1,#p*2 do n[#n+1]=i<=#p and p[i]%2==1 and p[i]or i>#p and p[i-#p]%2==0 and p[i-#p]or n[#n+1] end return n'

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Answer 52

k, 8 bytes

{x@>2!x}

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{      } /function
    2!x  /  modulo 2 (mapped through list)
   >     /  get indices to sort descending
 x@      /  return the original list at those indices

Answer 53

Julia 1.0, 27 bytes

f(l)=l[sortperm(l.%2 .==0)]

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This works too and is a lot more elegant:

f(l)=sort(l,lt=(a,b)->b%2==0)

But it's a few bytes longer :(

Answer 54

Kotlin, 33 bytes

{l:List<Int>->l.sortedBy{1-it%2}}

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