Pick random element

Pick random element
You are encouraged to solve this task according to the task description, using any language you may know.

Demonstrate how to pick a random element from a list.

Ada

The following program generates three 20-letter words. Each vowel and each consonant is picked randomly from a list of vowels resp. a list of consonants.

<lang Ada>with Ada.Text_IO, Ada.Numerics.Float_Random;

procedure Pick_Random_Element is

  package Rnd renames Ada.Numerics.Float_Random;
  Gen: Rnd.Generator; -- used globally

  type Char_Arr is array (Natural range <>) of Character;

  function Pick_Random(A: Char_Arr) return Character is
     -- Chooses one of the characters of A (uniformly distributed)
  begin
     return A(A'First + Natural(Rnd.Random(Gen) * Float(A'Last)));
  end Pick_Random;

  Vowels    : Char_Arr := ('a', 'e', 'i', 'o', 'u');
  Consonants: Char_Arr := ('t', 'n', 's', 'h', 'r', 'd', 'l');
  Specials  : Char_Arr := (',', '.', '?', '!');

begin

  Rnd.Reset(Gen);
  for J in 1 .. 3 loop
     for I in 1 .. 10 loop
        Ada.Text_IO.Put(Pick_Random(Consonants));
        Ada.Text_IO.Put(Pick_Random(Vowels));
     end loop;
     Ada.Text_IO.Put(Pick_Random(Specials) & " ");
  end loop;
  Ada.Text_IO.New_Line;

end Pick_Random_Element;</lang>

Sample Output:

horanohesuhodinahiru. desehonirosedisinelo, losihehederidonolahe?

Aime

<lang aime>list l;

l_append(l, 'a'); l_append(l, 'b'); l_append(l, 'c'); l_append(l, 'd'); l_append(l, 'e'); l_append(l, 'f');

o_byte(l_query(l, drand(5))); o_byte('\n');</lang>

AutoHotkey

<lang AutoHotkey>list := "abc,def,gh,ijklmnop,hello,world" StringSplit list, list, `, Random, randint, 1, %list0% MsgBox % List%randint%</lang>

C

<lang C>#include <stdio.h>

1. include <stdlib.h>
2. include <time.h>

int main(){

 char array[] = { 'a', 'b', 'c' };

 srand( time( 0 ) );

 printf( "%c\n", array[ rand() % 3 ] );

 return 0;

}</lang>

D

<lang D>import std.stdio, std.random;

void main() {

   auto items = ["foo", "bar", "baz"];
   auto r = items[uniform(0, items.length)];
   writeln(r);

}</lang>

Euphoria

<lang euphoria>constant s = {'a', 'b', 'c'} puts(1,s[rand($)])</lang>

Go

<lang go>package main

import (

   "fmt"
   "rand"
   "time"

)

var list = []string{"bleen", "fuligin", "garrow", "grue", "hooloovoo"}

func main() {

   rand.Seed(time.Nanoseconds())
   fmt.Println(list[rand.Intn(len(list))])

}</lang>

Haskell

Creating a custom function:

<lang haskell>import Random (randomRIO)

pick :: [a] -> IO a pick xs = randomRIO (0, length xs - 1) >>= return . (xs !!)

x <- pick [1 2 3]</lang>

Using the random-extras library:

<lang haskell>import Data.Random import Data.Random.Source.DevRandom import Data.Random.Extras

x <- runRVar (choice [1 2 3]) DevRandom</lang>

Icon and Unicon

The unary operator '?' selects a random element from its argument which may be a string, list, table, or set.

<lang Icon>procedure main()

  L := [1,2,3]  # a list
  x := ?L       # random element

end</lang>

J

<lang j> ({~ ?@#) 'abcdef' b</lang>

Java

<lang java>import java.util.Random; ... int[] array = {1,2,3}; return array[new Random().nextInt(array.length)]; // if done multiple times, the Random object should be re-used</lang>

For a List object rather than an array, substitute list.get(...) for array[...]. If preserving the order of the List isn't important, you could call Collections.shuffle(list); and then list.get(0);. You would need to shuffle each time unless you removed the item from the list.

JavaScript

<lang javascript>var array = [1,2,3]; return array[Math.floor(Math.random() * array.length)];</lang>

Liberty BASIC

The natural way to hold an array of text is in a space- or comma-delimited string, although an array could be used. <lang lb> list$ ="John Paul George Ringo Peter Paul Mary Obama Putin" wantedTerm =int( 10 *rnd( 1)) print "Selecting term "; wantedTerm; " in the list, which was "; word$( list$, wantedTerm, " ") </lang>

Selecting term 5 in the list, which was Peter

OCaml

With a list: <lang ocaml>let list_rand lst =

 let len = List.length lst in
 List.nth lst (Random.int len)</lang>

# list_rand [1;2;3;4;5] ;;
- : int = 3

With an array: <lang ocaml>let array_rand ary =

 let len = Array.length ary in
 ary.(Random.int len)</lang>

# array_rand [|1;2;3;4;5|] ;;
- : int = 3

PARI/GP

<lang parigp>pick(v)=v[random(#v)+1]</lang>

PHP

<lang php>$arr = array('foo', 'bar', 'baz'); $x = $arr[array_rand($arr)];</lang>

Perl

<lang perl>my @array = ('a', 'b', 'c'); print $array[ rand @array ];</lang>

Perl 6

Perl 6 has two functions to return random elements depending on whether you are doing selection with or without replacement.

Selection with replacement: (roll of a die) <lang perl6>(1..6).roll; # return 1 random value in the range 1 through 6 (1..6).roll(3); # return a list of 3 random values in the range 1 through 6 (1..6).roll(*); # return a lazy infinite list of random values in the range 1 through 6</lang>

Selection without replacement: (pick a card from a deck) <lang perl6>( 2..9, <J Q K A> X~ <♠ ♣ ♥ ♦> ).pick; # Pick a card ( 2..9, <J Q K A> X~ <♠ ♣ ♥ ♦> ).pick(5); # Draw 5 ( 2..9, <J Q K A> X~ <♠ ♣ ♥ ♦> ).pick(*); # Get a shuffled deck</lang>

PicoLisp

<lang PicoLisp>(get Lst (rand 1 (length Lst)))</lang>

Python

<lang python>>>> import random >>> random.choice(['foo', 'bar', 'baz']) 'baz'</lang>

Ruby

Works with: Ruby version 1.9

<lang ruby>irb(main):001:0> %w(north east south west).sample => "west" irb(main):002:0> (1..100).to_a.sample(2) => [17, 79]</lang>

Works with: Ruby version 1.8, but not 1.9

<lang ruby>irb(main):001:0> %w(north east south west).choice => "south"</lang>

Seed7

<lang seed7>$ include "seed7_05.s7i";

const proc: main is func

 begin
   writeln(rand([] ("foo", "bar", "baz")));
 end func;</lang>

Smalltalk

<lang smalltalk>x := #(1 2 3) atRandom.</lang>

Tcl

Random selection from a list is implemented by composing lindex (for selection of an item from a list) and the pattern for generating an integral random number from the range $[0,n)$. It's simpler to use when wrapped up as a helper procedure: <lang tcl>proc randelem {list} {

   lindex $list [expr {int(rand()*[llength $list])}]

} set x [randelem {1 2 3 4 5}]</lang>