复杂度为O（n）的Vanilla JS解决方案（对于这个问题最快）。如果需要，修改hashFunction以区分对象（例如1和“1”）。第一种解决方案避免了隐藏循环（在Array提供的函数中常见）。

var dedupe = function(a) 
{
    var hash={},ret=[];
    var hashFunction = function(v) { return ""+v; };
    var collect = function(h)
    {
        if(hash.hasOwnProperty(hashFunction(h)) == false) // O(1)
        {
            hash[hashFunction(h)]=1;
            ret.push(h); // should be O(1) for Arrays
            return;
        }
    };

    for(var i=0; i<a.length; i++) // this is a loop: O(n)
        collect(a[i]);
    //OR: a.forEach(collect); // this is a loop: O(n)

    return ret;
}

var dedupe = function(a) 
{
    var hash={};
    var isdupe = function(h)
    {
        if(hash.hasOwnProperty(h) == false) // O(1)
        {
            hash[h]=1;
            return true;
        }

        return false;
    };

    return a.filter(isdupe); // this is a loop: O(n)
}

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

for(var i = 0; i < lines.length; i++)
{
    if(uniqueNames.indexOf(lines[i]) == -1)
        uniqueNames.push(lines[i]);
}
if(uniqueNames.indexOf(uniqueNames[uniqueNames.length-1])!= -1)
    uniqueNames.pop();
for(var i = 0; i < uniqueNames.length; i++)
{
    document.write(uniqueNames[i]);
      document.write("<br/>");
}

另一种不用编写太多代码就可以做到这一点的方法是使用ES5 Object.keys-method：

var arrayWithDuplicates = ['a','b','c','d','a','c'],
    deduper = {};
arrayWithDuplicates.forEach(function (item) {
    deduper[item] = null;
});
var dedupedArray = Object.keys(deduper); // ["a", "b", "c", "d"]

在函数中提取

function removeDuplicates (arr) {
    var deduper = {}
    arr.forEach(function (item) {
        deduper[item] = null;
    });
    return Object.keys(deduper);
}

以下脚本返回仅包含唯一值的新数组。它适用于字符串和数字。不需要额外的库，只需要普通的JS。

浏览器支持：

Feature Chrome  Firefox (Gecko)     Internet Explorer   Opera   Safari
Basic support   (Yes)   1.5 (1.8)   9                   (Yes)   (Yes)

https://jsfiddle.net/fzmcgcxv/3/

var duplicates = ["Mike","Matt","Nancy","Adam","Jenny","Nancy","Carl","Mike","Mike","Nancy","Carl"]; 
var unique = duplicates.filter(function(elem, pos) {
    return duplicates.indexOf(elem) == pos;
  }); 
alert(unique);

如果你在使用

D3.js

你可以的

d3.set(["foo", "bar", "foo", "baz"]).values() ==> ["foo", "bar", "baz"]

https://github.com/mbostock/d3/wiki/Arrays#set_values

我知道我有点晚了，但这里有另一个使用jinqJ的选项

参见Fiddle

var result = jinqJs().from(["Mike","Matt","Nancy","Adam","Jenny","Nancy","Carl"]).distinct().select();