How to get unique values in an array
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;
}
});
}