How to get unique values in an array [duplicate]

Question

Here's a much cleaner solution for ES6 that I see isn't included here. It uses the Set and the spread operator: ...

var a = [1, 1, 2];

[... new Set(a)]

Which returns [1, 2]

Or for those looking for a one-liner (simple and functional) compatible with current browsers:

let a = ["1", "1", "2", "3", "3", "1"];
let unique = a.filter((item, i, ar) => ar.indexOf(item) === i);
console.log(unique);

Update 2021 I would recommend checking out Charles Clayton's answer, as of recent changes to JS there are even more concise ways to do this.

Update 18-04-2017

It appears as though 'Array.prototype.includes' now has widespread support in the latest versions of the mainline browsers (compatibility)

Update 29-07-2015:

There are plans in the works for browsers to support a standardized 'Array.prototype.includes' method, which although does not directly answer this question; is often related.

Usage:

["1", "1", "2", "3", "3", "1"].includes("2");     // true

Pollyfill (browser support, source from mozilla):

// https://tc39.github.io/ecma262/#sec-array.prototype.includes
if (!Array.prototype.includes) {
  Object.defineProperty(Array.prototype, 'includes', {
    value: function(searchElement, fromIndex) {

      // 1. Let O be ? ToObject(this value).
      if (this == null) {
        throw new TypeError('"this" is null or not defined');
      }

      var o = Object(this);

      // 2. Let len be ? ToLength(? Get(O, "length")).
      var len = o.length >>> 0;

      // 3. If len is 0, return false.
      if (len === 0) {
        return false;
      }

      // 4. Let n be ? ToInteger(fromIndex).
      //    (If fromIndex is undefined, this step produces the value 0.)
      var n = fromIndex | 0;

      // 5. If n ≥ 0, then
      //  a. Let k be n.
      // 6. Else n < 0,
      //  a. Let k be len + n.
      //  b. If k < 0, let k be 0.
      var k = Math.max(n >= 0 ? n : len - Math.abs(n), 0);

      // 7. Repeat, while k < len
      while (k < len) {
        // a. Let elementK be the result of ? Get(O, ! ToString(k)).
        // b. If SameValueZero(searchElement, elementK) is true, return true.
        // c. Increase k by 1.
        // NOTE: === provides the correct "SameValueZero" comparison needed here.
        if (o[k] === searchElement) {
          return true;
        }
        k++;
      }

      // 8. Return false
      return false;
    }
  });
}

Since I went on about it in the comments for @Rocket's answer, I may as well provide an example that uses no libraries. This requires two new prototype functions, contains and unique

Array.prototype.contains = function(v) {
  for (var i = 0; i < this.length; i++) {
    if (this[i] === v) return true;
  }
  return false;
};

Array.prototype.unique = function() {
  var arr = [];
  for (var i = 0; i < this.length; i++) {
    if (!arr.contains(this[i])) {
      arr.push(this[i]);
    }
  }
  return arr;
}

var duplicates = [1, 3, 4, 2, 1, 2, 3, 8];
var uniques = duplicates.unique(); // result = [1,3,4,2,8]

console.log(uniques);

For more reliability, you can replace contains with MDN's indexOf shim and check if each element's indexOf is equal to -1: documentation

One Liner, Pure JavaScript

With ES6 syntax

list = list.filter((x, i, a) => a.indexOf(x) === i)

x --> item in array
i --> index of item
a --> array reference, (in this case "list")

With ES5 syntax

list = list.filter(function (x, i, a) { 
    return a.indexOf(x) === i; 
});

Browser Compatibility: IE9+

Using EcmaScript 2016 you can simply do it like this.

 var arr = ["a", "a", "b"];
 var uniqueArray = Array.from(new Set(arr)); // Unique Array ['a', 'b'];

Sets are always unique, and using Array.from() you can convert a Set to an array. For reference have a look at the documentations.

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/from https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Set