# 118 List to set

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.

Implementation edit is for fixing errors and enhancing with metadata.

Instead of changing the code of the snippet, consider creating another Haskell implementation.

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);
var y = new Set(x);
require 'set'
y = x.to_set
my %y = map {$_=>0} @x;
y = x |> Enum.uniq |> List.to_tuple
y =
use std::collections::HashSet;
let y: HashSet<_> = x.into_iter().collect();
local hash = {}
local y = {}
for _,v in ipairs(x) do
   if (not hash[v]) then
       y[#y+1] = v
       hash[v] = true
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];
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))
Y = sets:from_list(X).