Randomize the scalars of an array

Question

You must fill an array with every number from 0-n inclusive. No numbers should repeat. However they must be in a random order.

Rules

All standard code-golf rules and standard loopholes are banned

The array must be generated pseudo-randomly. Every possible permutation should have a equal probability.

Input

n in any way allowed in the I/O post on meta.

Output

The array of numbers scrambled from 0-n inclusive.

Answer 1

Perl 6, 14 bytes

{pick *,0..$_}

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Expanded:

{           # bare block lambda with implicit parameter ｢$_｣

  pick      # choose randomly without repeats
    *,      # Whatever (all)
    0 .. $_ # Range from 0, to the input (inclusive)
}

Answer 2

05AB1E, 3 bytes

Ý.r

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Ý   # Make a list from 0 to input
 .r # Shuffle it randomly

Answer 3

Pyth, 3 bytes

.Sh

Demonstration

.S is shuffle. It implicitly casts an input integer n to the range [0, 1, ..., n-1]. h is +1, and the input is taken implicitly.

Answer 4

R, 16 bytes

sample(0:scan())

reads from stdin. sample randomly samples from the input vector, returning a (pseudo)random sequence.

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Answer 5

Jelly, 3 bytes

0rẊ

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Explanaion:

0rẊ 
0r  Inclusive range 0 to input.
  Ẋ Shuffle.
    Implicit print.

Alternate solution, 3 bytes

‘ḶẊ

Explanation:

‘ḶẊ
‘   Input +1
 Ḷ  Range 0 to argument.
  Ẋ Shuffle.

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Answer 6

Python 2, 51 bytes

lambda n:sample(range(n+1),n+1)
from random import*

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There is random.shuffle() but it modifies the argument in place instead of returning it...

Answer 7

PHP, 42 Bytes

$r=range(0,$argn);shuffle($r);print_r($r);

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Answer 8

Bash, 18 11 bytes

shuf -i0-$1

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Answer 9

Japt, 4 bytes

ò öx

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---

    :Implicit input of integer U
ò   :Generate array of 0 to U.
öx  :Generate random permutation of array.
    :Implicit output of result.

Answer 10

MATL, 4 bytes

QZ@q

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Explanation

Q     % Implicitly input n. Add 1
Z@    % Random permutation of [1 2 ... n+1]
q     % Subtract 1, element-wise. Implicitly display

Answer 11

CJam, 7 6 bytes

1 byte removed thanks to Erik the Outgolfer.

{),mr}

This is an anonymous block (function) that takes an integer from the stack and replaces it with the result. Try it online!

Explanation

{     e# Begin block
)     e# Increment: n+1
,     e# Range: [0 1 ... n]
mr    e# Shuffle
}     e# End block

Answer 12

Mathematica, 24 bytes

RandomSample@Range[0,#]&

Answer 13

Brachylog, 2 bytes

⟦ṣ

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Explanation

⟦     Range from 0 to Input
 ṣ    Shuffle

Answer 14

C#, 76 bytes

using System.Linq;i=>new int[i+1].Select(x=>i--).OrderBy(x=>Guid.NewGuid());

This returns an IOrderedEnumerable, I hope that's okay, or else I need a few more bytes for a .ToArray()

Answer 15

Java 8, 114 111 97 bytes

import java.util.*;n->{List l=new Stack();for(;n>=0;l.add(n--));Collections.shuffle(l);return l;}

-3 bytes and bug-fixed thanks to @OlivierGrégoire.
-4 bytes thanks to @Jakob.
-10 bytes by removing .toArray().

Explanation:

Try it here.

import java.util.*;        // Required import for List, Stack and Collections
n->{                       // Method with integer parameter and Object-array return-type
  List l=new Stack();      //  Initialize a List
  for(;n>=0;l.add(n--));   //  Loop to fill the list with 0 through `n`
  Collections.shuffle(l);  //  Randomly shuffle the List
  return l;                //  Convert the List to an Object-array and return it
}                          // End of method

Answer 16

Ohm, 2 bytes

#╟

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Answer 17

Pyth, 4 Bytes

.S}0

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Answer 18

C, 75 bytes

a[99],z,y;f(n){if(n){a[n]=--n;f(n);z=a[n];a[n]=a[y=rand()%(n+1)];a[y]=z;}}

Recursive function that initializes from the array's end on the way in, and swaps with a random element before it on the way out.

Answer 19

Charcoal, 33 bytes

Ａ…·⁰ＮβＦβ«ＡβδＡ‽δθＰＩθ↓Ａ⟦⟧βＦδ¿⁻θκ⊞βκ

Try it online! Link is to verbose version of code.

Apparently it takes 17 bytes to remove an element from a list in Charcoal.

Edit: These days it only takes three bytes, assuming you want to remove all occurrences of the item from the list. This plus other Charcoal changes cut the answer down to 21 bytes: Try it online!

Answer 20

APL (Dyalog), 5 bytes

?⍨1+⊢

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Assumes ⎕IO←0, which is default on many machines.

Explanation

⊢ the right argument

1+ add 1 to it

?⍨ generate numbers 0 .. 1+⊢-1 and randomly deal them in an array so that no two numbers repeat

Answer 21

q/kdb+, 11 bytes

Solution:

{(0-x)?1+x}

Example:

q){(0-x)?1+x}10
5 9 7 1 2 4 8 0 3 10
q){(0-x)?1+x}10
6 10 2 8 4 5 9 0 7 3
q){(0-x)?1+x}10
9 6 4 1 10 8 2 7 0 5

Explanation:

Use the ? operator with a negative input to give the full list of 0->n without duplicates:

{(0-x)?1+x} / solution
{         } / lambda expression
         x  / implicit input
       1+   / add one
      ?     / rand
 (0-x)      / negate x, 'dont put item back in the bag'

Answer 22

TI-83 BASIC, 5 bytes (boring)

randIntNoRep(0,Ans

Yep, a builtin. randIntNoRep( is a two-byte token, and Ans is one byte.

More fun, 34 bytes:

Ans→N
seq(X,X,0,N→L₁
rand(N+1→L₂
SortA(L₂,L₁
L₁

Straight from tibasicdev. Probably golfable, but I haven't found anything yet.

What this does: Sorts a random array, moving elements of the second arg (L₁ here) in the same way as their corresponding elements.

Answer 23

JavaScript (ES6), 51 bytes

n=>[...Array(n+1).keys()].sort(_=>.5-Math.random())

Answer 24

Aceto, 15 14 16 bytes

@lXp
Y!`n
zi&
0r

Push zero on the stack, read an integer, construct a range and shuffle it:

Y
zi
0r

Set a catch mark, test length for 0, and (in that case) exit:

@lX
 !`

Else print the value, a newline, and jump back to the length test:

   p
   n
  &

(I had to change the code because I realized I misread the question and had constructed a range from 1-n, not 0-n.)

Answer 25

Go, 92 bytes

Mostly losing to the need to seed the PRNG.

import(."fmt";."math/rand";."time")
func f(n int){Seed(Now().UnixNano());Println(Perm(n+1))}

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Answer 26

Ruby, 20 bytes

->n{[*0..n].shuffle}

Answer 27

8th, 42 36 34 bytes

Code

>r [] ' a:push 0 r> loop a:shuffle

SED (Stack Effect Diagram) is n -- a

Usage and example

ok> 5 >r [] ' a:push 0 r> loop a:shuffle .
[2,5,0,3,1,4]

Answer 28

Javascript (ES6), 68 bytes

n=>[...Array(n+1)].map((n,i)=>[Math.random(),i]).sort().map(n=>n[1])

Creates an array of form

[[Math.random(), 0],
 [Math.random(), 1],
 [Math.random(), 2],...]

Then sorts it and returns the last elements in the new order

Answer 29

Tcl, 90 bytes

proc R n {lsort -c {try expr\ rand()>.5 on 9} [if [incr n -1]+2 {concat [R $n] [incr n]}]}

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Tcl, 96 bytes

proc R n {proc f a\ b {expr rand()>.5}
set i -1
while \$i<$n {lappend L [incr i]}
lsort -c f $L}

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Answer 30

J, 11 Bytes

(?@!A.i.)>:

Explanation:

         >:   | Increment
(?@!A.i.)     | Fork, (f g h) n is evaluated as (f n) g (h n)
      i.      | Integers in range [0,n) inclusive
 ?@!          | Random integer in the range [0, n!)
    A.        | Permute right argument according to left

Examples:

    0 A. i.>:5
0 1 2 3 4 5
    1 A. i.>:5
0 1 2 3 5 4
    (?@!A.i.)>: 5
2 3 5 1 0 4
    (?@!A.i.)>: 5
0 3 5 1 2 4

Answer 31

JavaScript ES6, 60 bytes, uniform distribute

f=n=>n?(z=f(n-1),z[n]=z[y=Math.random()*-~n|0],z[y]=n,z):[0]

s={};
for(i=0; i<1000; i++) k=f(2), s[k]=-~s[k];
for(i in s) console.log(i, s[i]);

Answer 32

AWK, 82 bytes

{srand();for(n=$1+1;j++<n;$j=j-1);for(;n;n--){t=$(i=int(n*rand()+1));$i=$n;$n=t}}1

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AWK doesn't have a built-in shuffle, so this is the best I could come up with.