Consolidate an Array

Question

The task is simple: consolidate an array of ints. Consolidating this array consists of the following:

• All instances of 0 need to be moved to the end of the array.
• There should be no 0s between the non-zero integers.
• All non-zero indices should retain their order.

Challenge

Consolidate an array in the least amount of bytes.

You are consolidating an array of random length with a size up to your language's max with random integers. Input may be any natural way for your language.

Examples

Input

0 5 8 8 3 5 1 6 8 4 0 3 7 5 6 4 4 7 5 6 7 4 4 9 1 0 5 7 9 3 0 2 2 4 3 0 4 8 7 3 1 4 7 5 1 2 1 8 7 8 7 7 2 6 3 1 2 8 5 1 4 2 0 5 0 6 0 3

Output

5 8 8 3 5 1 6 8 4 3 7 5 6 4 4 7 5 6 7 4 4 9 1 5 7 9 3 2 2 4 3 4 8 7 3 1 4 7 5 1 2 1 8 7 8 7 7 2 6 3 1 2 8 5 1 4 2 5 6 3 0 0 0 0 0 0 0 0

Input

-1 -7 -6 5 1 -5 -2 7 -3 -8 0 8 9 1 -8 -1 6 -4 1 -2 1 -7 5 4 -6 7 -3 9 8 3 -1 0 -5 -7 3 8 1 1 3 -3 -2 -2 0 -7 0 -4 8 6 -3 6 0 5 3 2 2 2 -2 -7 -3 9 -1 6 0 6 -7 9 4 -2 8 -8 -4 1 -8 4 3 7 3 5 1 0 3 3 7 -1 -5 1 -3 4 -7 0 3 2 -2 7 -3 0 0 2 -5 8 -3 -2 -7 -5 7 -3 -9 -7 5 8 -3 9 6 7 -2 4 7

Output

-1 -7 -6 5 1 -5 -2 7 -3 -8 8 9 1 -8 -1 6 -4 1 -2 1 -7 5 4 -6 7 -3 9 8 3 -1 -5 -7 3 8 1 1 3 -3 -2 -2 -7 -4 8 6 -3 6 5 3 2 2 2 -2 -7 -3 9 -1 6 6 -7 9 4 -2 8 -8 -4 1 -8 4 3 7 3 5 1 3 3 7 -1 -5 1 -3 4 -7 3 2 -2 7 -3 2 -5 8 -3 -2 -7 -5 7 -3 -9 -7 5 8 -3 9 6 7 -2 4 7 0 0 0 0 0 0 0 0 0 0

Example Code (Java)

public class Consolidate {
    public static void main(String[] args) throws Exception {
        int[] toConsolidate = new int[args.length];
        for (int i=0; i<args.length; i++){
            toConsolidate[i]=Integer.parseInt(args[i]);
        }
        for (int i=0; i<toConsolidate.length; i++) {
            for (int k=0; k<toConsolidate.length-1; k++) {
                if (toConsolidate[k] == 0){
                    toConsolidate[k] = toConsolidate[k+1];
                    toConsolidate[k+1] = 0;
                }
            }
        }
        for (int i:toConsolidate)
            System.out.print(i+" ");
    }
}

Answer 1

Pyth, 3 bytes

!DQ

Explanation:

  Q    Input
!D     Sort by logical NOT

Try it here.

Answer 2

Jelly, 3 bytes

¬Ụị

Sorts the list by the logical NOT of its values. Try it online!

How it works

¬Ụị    Main link. Input: A (list)

¬      Compute the logical NOT of each element of A.
 Ụ     Grade up; sort the resulting list's indices by their corresponding values.
  ị    Retrieve the elements of A at that indices.

Answer 3

Octave, 18 bytes

@(A)[A(~~A) A(~A)]

sort() takes too many bytes. I'll just use logical indexing.

Examples on ideone.

Answer 4

R, 29 23 21 bytes

As noted by MarcoBreitig, we can shorten it to 21 bytes if we don't need to provide it as a function:

x=scan();x[order(!x)]

---

Previous versions:

function(x)x[order(!x)]

The function takes a vector as input and orders by the logical vector that results from negating the input.

Original answer:

function(x)c(x[x!=0],x[x==0])

The function takes a vector as input and the concatenates (c()) the non-zero values and then the zero-values.

Answer 5

Retina, 15

Simple repeated regex substitution:

+`\b0 (.*)
$1 0

Try it online.

Answer 6

Python, 32 bytes

lambda x:sorted(x,key=0..__eq__)

Takes argument as any iterable (list, tuple, etc.). Thanks to @xnor for teaching me a new trick!

Answer 7

ES6, 23 bytes

a=>a.sort((x,y)=>!x-!y)

It used to be the case that sort wasn't stable, in which case you needed 41 bytes:

a=>a.filter(x=>x).concat(a.filter(x=>!x))

Answer 8

Python byte code (2.7.9), 252 bytes, 33 opcodes, 0.0228 seconds

This was build when the contest was still a fastest-code contest

Opens a file in the current directory called 'SourceArray' for use

LOAD_CONST          ''
STORE_FAST          no_zeroes#  no_zeroes = ''

LOAD_NAME           open
LOAD_CONST          'SourceArray'
CALL_FUNCTION       0,1#  open('SourceArray')
LOAD_ATTR           read
CALL_FUNCTION       0,0#  .read()

LOAD_ATTR           split
CALL_FUNCTION       0,0#  .split()

DUP_TOP
DUP_TOP             #Start if
BUILD_LIST          0
COMPARE_OP          ==
POP_JUMP_IF_TRUE    35#  if list == [], GOTO 35
LOAD_ATTR           pop
LOAD_CONST          0
CALL_FUNCTION       0,1#  list.pop(0)
DUP_TOP
LOAD_CONST          '0'
COMPARE_OP          ==
POP_JUMP_IF_TRUE    28#  if list.pop(0) == '0', GOTO 28
PRINT_ITEM          #  print list.pop(0)
JUMP_ABSOLUTE       13

POP_TOP
LOAD_CONST          '0%_'#  '0 '
LOAD_FAST           no_zeroes
INPLACE_ADD
STORE_FAST          no_zeroes#  no_zeroes = no_zeroes + '0 '
JUMP_ABSOLUTE       13

LOAD_FAST           no_zeroes
PRINT_ITEM          #  print no_zeroes

LOAD_CONST          None
RETURN_VALUE

The co_code (The actual codey bit)

'd\x01\x00}\x00\x00\te\x00\x00\x83\x00\x00\tj\x01\x00\x83\x00\x00\t\x04\x04g\x00\x00k\x02\x00sG\x00j\x02\x00d\x02\x00\x83\x01\x00\x04d\x03\x00k\x02\x00s8\x00Gq\x15\x00\t\x01d\x04\x00|\x00\x007}\x00\x00q\x15\x00\t|\x00\x00G\td\x00\x00S'

Or a .pyc file version 03F3

03 F3 0D 0A 40 FD B0 56 63 00 00 00 00 01 00 00 00 03 00 00 00 00 00 00 00 73 59 00 00 00 64 01 00 7D 00 00 09 65 00 00 64 02 00 83 01 00 6A 01 00 83 00 00 09 6A 02 00 83 00 00 09 04 04 67 00 00 6B 02 00 73 50 00 6A 03 00 64 03 00 83 01 00 04 64 04 00 6B 02 00 73 41 00 47 71 1E 00 09 01 64 05 00 7C 00 00 37 7D 00 00 71 1E 00 09 7C 00 00 47 09 64 00 00 53 28 06 00 00 00 4E 74 00 00 00 00 74 0B 00 00 00 53 6F 75 72 63 65 41 72 72 61 79 69 00 00 00 00 74 01 00 00 00 30 73 02 00 00 00 30 20 28 04 00 00 00 74 04 00 00 00 6F 70 65 6E 74 04 00 00 00 72 65 61 64 74 05 00 00 00 73 70 6C 69 74 74 03 00 00 00 70 6F 70 28 01 00 00 00 74 09 00 00 00 6E 6F 5F 7A 65 72 6F 65 73 28 00 00 00 00 28 00 00 00 00 74 09 00 00 00 70 79 6B 65 5F 63 6F 64 65 52 08 00 00 00 01 00 00 00 52 00 00 00 00

You can try to compile my source code yourself using my library on github. I just posted a commit to it that allowed comments so I hope this is still competing as far as fastest-code goes ;)

Roughly equivalent to

no_zeroes = ''
unamed_variable = open('SourceArray').read().split()
while unamed_variable != []:
    unamed_variable_2 = unamed_variable.pop()
    if unamed_variable_2 == '0':
        no_zeroes += '0 '
    else:
        print unamed_variable_2,
print no_zeroes,

Answer 9

Matlab: 21 bytes

@(a)[a(a~=0),a(a==0)]

Prints nonzero elements first, then concatenates with zero elements

@(a)____ create an anonymous function with one input argument a

[___,___] concatenates horizontally vectors inside brackets, separated by commas

a(a~=0) returns vector with all nonzero elements of vector a

a(a==0) returns vector with all zero elements of vector a

Answer 10

APL: 8 bytes

(⍴a)↑a~0

a~0 remove zeros from a (read "a without 0")
(⍴a) original length of a (read "shape of a")
↑ pad a without zeros to a's original length

Try it in http://ngn.github.com/apl/web/index.html

Test data: a←1 0 1 2 3 4 0 1 0 0 0 0 1 2 3 4 5

Answer 11

Zsh, 22 bytes

(input passed as arguments to the script/function ($@ aka $argv array), output on stdout as space separated list, newline terminated)

<<<${@:#0}\ ${(M)@:#0}

• <<< string: here-string here passed as stdin to the $NULLCMD command (cat by default).
• ${@:#0} $@ except elements being 0.
• ${(M)@:#0} reverse of the above

That assumes (like several other answers here) that zeroes in the input are all expressed as 0 (no 00 nor 0x0 nor 36#0).

Answer 12

Haskell, 26 bytes

f x=filter(/=0)x++[0|0<-x]

Take all non-zero numbers followed by all zeros. Filtering constants (here: 0) is quite short when using a list comprehension: [0|0<-x].

Answer 13

Mathematica, 14 bytes

Sort[#,#!=0&]&

Answer 14

Common Lisp, 46 bytes

(lambda(a)(stable-sort a(lambda(_ b)(= 0 b))))

Sort the array so that for each couple (a,b), we have a < b if b is zero. When neither a < b or b < a, the sort is stable: the original order between elements is retained.

I also tried with adjust-array and remove, but this was too long:

(lambda(a)(adjust-array(remove 0 a)(length a):initial-element 0))

Answer 15

Javascript, 52 54 51 bytes

s=>s.replace(/\b0 /g,x=>++i&&'',i=0)+' 0'.repeat(i)

Answer 16

Java 7, 78 bytes

void g(int[]a){int c=0;for(int o:a)a[o==0?c:c++]=o;for(;c<a.length;a[c++]=0);}

I'm not sure why the other Java entries are using strings. If you want to filter an integer array, it seems best to use an integer array. This modifies the input in place by keeping two indices, then just filling the remaining slots with zeros.

Answer 17

Vyxal, 3 2 bytes

µ¬

Try it Online!

Sort by logical negation. The header puts the input into a list format.

Answer 18

PHP, 73 71 70 52 49 48 46 bytes - BIG thanks to Ismael Miguel

// Assuming
$a = array(4,8,6,1,0,8,0,0,0,0,0,-4,'-5',-1,564,0);

// Produces a notice-level error
foreach($a as$v)$v?print"$v ":$b.="0 ";echo$b;

Answer 19

Bash + GNU utilities, 23

grep -v ^0 a
grep ^0 a

Assumes input is newline-separated entries in a file called a. Score includes +1 for this filename.

Answer 20

Perl 5, 26 bytes

23 plus three for -an (-E is free)

say for sort{!$a-!$b}@F

Thanks to Dennis for reminding me of -a, saving two bytes.

Answer 21

JavaScript ES6, 16 bytes

x=>x.sort(t=>!t)

Works on firefox

Answer 22

AWK, 34 bytes

{for(;a++-NF;c?$++b=c:0)$a-=c=$a}1

Try it online!

This code runs through all the commandline arguments, shifts non-zero values to the left and clears arguments as it processes them.

The top of loop "should we iteration" condition expression is:

a++-NF

which increments a until it hits the number of commandline arguments NF.

The body of the loop,

$a-=c=$a

saves the value of the current commandline argument in variable c, then sets the current argument to 0.

At the end of each iteration, the ternery:

c?$++b=c:0

checks to see if the current argument c is non-zero and if so, increments the b (the number of non-zero argument found so far) and stores the current argument value in that slot.

After the code block processes the commandline arguments, the only left to do is trigger AWK to print all the commandline arguments with a truthy test without a code block.

1

Answer 23

Uiua, 5 bytes

⊏⍏=0.

Try it!

⊏⍏=0.
    .  # duplicate
  =0   # where is it equal to zero?
 ⍏     # rise (grade up)
⊏      # select

Answer 24

CJam, 6 bytes

{{!}$}

An anonymous function. Sort using “whether or not an element is zero” as a key.

Answer 25

Clojure/ClojureScript, 18 bytes

#(sort-by zero? %)

Answer 26

MATL, 7 bytes

t~FT#S)

Try it online!

t      % input array. Duplicate
~      % logical negate: nonzero values become false, zeros become true
FT#S   % sort (false, then true) and output a vector with the indices of the sorting
)      % apply that vector of indices to original array

Answer 27

Seriously, 12 bytes

4,n`Y`M@░)░+

Try it online!

Explanation:

4,n`Y`M@░)░+
4,n           push 4 copies of input
   `Y`M       map logical negate
       @░)    filter (take zeroes) and push to bottom of stack
          ░   filter (take non-zeroes)
           +  append zeroes

Answer 28

Swift, 13 bytes

a.sort{$1==0}

Answer 29

Perl6, 11 bytes

{.sort(!*)}

Produces a Block - which can be called on an array:

{.sort(!*)}.([1,2,0,3]).say

Although it would be more natural (and shorter) to write:

[1,2,0,3].sort(!*).say

How it works: if the perl6 sort routine is called with a block which accepts only one argument, the list elements are sorted according to by($a) cmp by($b). In this case, the block is !*, i.e. a negation of the whatever operator.

I notice that:

• The example in the question is a class which provides a method, not including boilerplate required to read in
• The description of the task does not require printing, and, except for the fact that the example prints, implies that an array might be returned

Answer 30

TeX (Plain format), 160 bytes

Make the 0 character active (that is, make the interpreter process it as a command), then define that command to skip the character and increment a counter. At the end of the string, print as many zeros as were counted.

Save this as zero.tex and give the input through the command line with this command:

pdftex "\def\I{0 1 0 3 2 0 0 8 0 5 0 1 9 4}\input zero"

\def\I{}\newcount\Z\def\L{\loop\advance\Z by-1\ifnum\Z>00 \repeat}
\begingroup\catcode`\013 \def0{\advance\Z by1}
\scantokens\expandafter{\I\empty}\endgroup\L\bye

(Newlines added for clarity)