jQuery function to get all unique elements from an array?

You can use array.filter to return the first item of each distinct value-

var a = [ 1, 5, 1, 6, 4, 5, 2, 5, 4, 3, 1, 2, 6, 6, 3, 3, 2, 4 ];

var unique = a.filter(function(itm, i, a) {
    return i == a.indexOf(itm);
});

console.log(unique);

If supporting IE8 and below is primary, don't use the unsupported filter method.

Otherwise,

if (!Array.prototype.filter) {
    Array.prototype.filter = function(fun, scope) {
        var T = this, A = [], i = 0, itm, L = T.length;
        if (typeof fun == 'function') {
            while(i < L) {
                if (i in T) {
                    itm = T[i];
                    if (fun.call(scope, itm, i, T)) A[A.length] = itm;
                }
                ++i;
            }
        }
        return A;
    }
}

Based on @kennebec's answer, but fixed for IE8 and below by using jQuery wrappers around the array to provide missing Array functions filter and indexOf:

$.makeArray() wrapper might not be absolutely needed, but you'll get odd results if you omit this wrapper and JSON.stringify the result otherwise.

var a = [1,5,1,6,4,5,2,5,4,3,1,2,6,6,3,3,2,4];

// note: jQuery's filter params are opposite of javascript's native implementation :(
var unique = $.makeArray($(a).filter(function(i,itm){ 
    // note: 'index', not 'indexOf'
    return i == $(a).index(itm);
}));

// unique: [1, 5, 6, 4, 2, 3]

Just use this code as the basis of a simple JQuery plugin.

$.extend({
    distinct : function(anArray) {
       var result = [];
       $.each(anArray, function(i,v){
           if ($.inArray(v, result) == -1) result.push(v);
       });
       return result;
    }
});

Use as so:

$.distinct([0,1,2,2,3]);