# 118 List to set
Implementation
Perl

Be concise.

Be useful.

All contributions dictatorially edited by webmasters to match personal tastes.

Please do not paste any copyright violating resource.

Please try to avoid dependencies to third-party libraries and frameworks.

Other implementations
import java.util.HashSet;
import java.util.List;
Set<T> y = new HashSet<>(x);
y := make(map[T]struct{}, len(x))
for _, v := range x {
	y[v] = struct{}{}
}
y = set(x)
bool[typeof(x[0])] y;

foreach (e ; x)
    y[e] = true;
import std.container;
auto y = redBlackTree(x);
import qualified Data.Set as Set
y = Set.fromList x
var y = new Set(x);
require 'set'
y = x.to_set
y = x |> Enum.uniq |> List.to_tuple
y = MapSet.new(x)
use std::collections::HashSet;
let y = x.iter().cloned().collect::<HashSet<_>>();
local hash = {}
local y = {}
for _,v in ipairs(x) do
   if (not hash[v]) then
       y[#y+1] = v
       hash[v] = true
   end
end
#include<unordered_set>
std::unordered_set<T> y (x.begin (), x.end ());
#include <set>
std::set<T> y (x.begin (), x.end ());
$y = array_unique($x);
for i := Low(X) to High(X) do Include(Y,X[i]);
val y = x.toSet
void main() {
  List<int> list = [1,2,3,4,5];
  list.toSet(); 
  print(list);
}
using System.Collections.Generic;
static void Main(string[] args)
{
    string[] x = new string[] { "a", "b", "c", "b" };
    HashSet<string> y = new HashSet<string>(x);
}
(setf y (remove-duplicates x))
(def y (set x))
set(x)
Y = sets:from_list(X).