Es6 based solution...

var arr = [2, 3, 4, 2, 3, 4, 2];
const result = [...new Set(arr)];
console.log(result);

Finding unique Array values in simple method

function arrUnique(a){
  var t = [];
  for(var x = 0; x < a.length; x++){
    if(t.indexOf(a[x]) == -1)t.push(a[x]);
  }
  return t;
}
arrUnique([1,4,2,7,1,5,9,2,4,7,2]) // [1, 4, 2, 7, 5, 9]

Array.prototype.getUnique = function() {
    var o = {}, a = []
    for (var i = 0; i < this.length; i++) o[this[i]] = 1
    for (var e in o) a.push(e)
    return a
}

[...new Set(duplicates)]

This is the simplest one and referenced from MDN Web Docs.

const numbers = [2,3,4,4,2,3,3,4,4,5,5,6,6,7,5,32,3,4,5]
console.log([...new Set(numbers)]) // [2, 3, 4, 5, 6, 7, 32]

Now using sets you can remove duplicates and convert them back to the array.

var names = ["Mike","Matt","Nancy", "Matt","Adam","Jenny","Nancy","Carl"];

console.log([...new Set(names)])

Another solution is to use sort & filter

var names = ["Mike","Matt","Nancy", "Matt","Adam","Jenny","Nancy","Carl"];
var namesSorted = names.sort();
const result = namesSorted.filter((e, i) => namesSorted[i] != namesSorted[i+1]);
console.log(result);

This prototype getUnique is not totally correct, because if i have a Array like: ["1",1,2,3,4,1,"foo"] it will return ["1","2","3","4"] and "1" is string and 1 is a integer; they are different.

Here is a correct solution:

Array.prototype.unique = function(a){
    return function(){ return this.filter(a) }
}(function(a,b,c){ return c.indexOf(a,b+1) < 0 });

using:

var foo;
foo = ["1",1,2,3,4,1,"foo"];
foo.unique();

The above will produce ["1",2,3,4,1,"foo"].

我们可以使用ES6集来做到这一点：

var duplicatedArray = [1, 2, 3, 4, 5, 1, 1, 1, 2, 3, 4];
var uniqueArray = Array.from(new Set(duplicatedArray));

console.log(uniqueArray);

//输出将是

uniqueArray = [1,2,3,4,5];

["Defects", "Total", "Days", "City", "Defects"].reduce(function(prev, cur) {
  return (prev.indexOf(cur) < 0) ? prev.concat([cur]) : prev;
 }, []);

[0,1,2,0,3,2,1,5].reduce(function(prev, cur) {
  return (prev.indexOf(cur) < 0) ? prev.concat([cur]) : prev;
 }, []);

最简单，最快的方法（在Chrome中）：

Array.prototype.unique = function() {
    var a = [];
    for (var i=0, l=this.length; i<l; i++)
        if (a.indexOf(this[i]) === -1)
            a.push(this[i]);
    return a;
}

只需遍历数组中的每个项目，测试该项目是否已经在列表中，如果不是，则将其推入要返回的数组。

根据jsPerf的说法，此功能是我在任何地方都能找到的最快的功能- 可以随意添加自己的功能。

非原型版本：

function uniques(arr) {
    var a = [];
    for (var i=0, l=arr.length; i<l; i++)
        if (a.indexOf(arr[i]) === -1 && arr[i] !== '')
            a.push(arr[i]);
    return a;
}

排序

当还需要对数组进行排序时，以下是最快的：

Array.prototype.sortUnique = function() {
    this.sort();
    var last_i;
    for (var i=0;i<this.length;i++)
        if ((last_i = this.lastIndexOf(this[i])) !== i)
            this.splice(i+1, last_i-i);
    return this;
}

或非原型：

function sortUnique(arr) {
    arr.sort();
    var last_i;
    for (var i=0;i<arr.length;i++)
        if ((last_i = arr.lastIndexOf(arr[i])) !== i)
            arr.splice(i+1, last_i-i);
    return arr;
}

This is also faster than the above method in most non-chrome browsers.

最简单的解决方案：

var arr = [1, 3, 4, 1, 2, 1, 3, 3, 4, 1];
console.log([...new Set(arr)]);

要么：

var arr = [1, 3, 4, 1, 2, 1, 3, 3, 4, 1];
console.log(Array.from(new Set(arr)));

一线纯JavaScript

使用ES6语法

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")

使用ES5语法

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

浏览器兼容性：IE9 +

您还可以使用underscore.js。

console.log(_.uniq([1, 2, 1, 3, 1, 4]));

<script src="http://underscorejs.org/underscore-min.js"></script>

它将返回：

[1, 2, 3, 4]

ES6 / ES2015的更新答案：使用Set，单行解决方案是：

var items = [4,5,4,6,3,4,5,2,23,1,4,4,4]
var uniqueItems = Array.from(new Set(items))

哪个返回

[4, 5, 6, 3, 2, 23, 1]

正如le_m建议的那样，也可以使用spread操作符来缩短它，例如

var uniqueItems = [...new Set(items)]

从那以后，我发现了一个使用jQuery的不错的方法

arr = $.grep(arr, function(v, k){
    return $.inArray(v ,arr) === k;
});

注意：此代码是从Paul Irish的鸭打孔帖子中提取的 -我忘了给您以谢意了：P