我有一个非常简单的JavaScript数组,可能包含也可能不包含重复项。

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

我需要删除重复项并将唯一值放入新数组。

我可以指出我尝试过的所有代码,但我认为它们没有用,因为它们不起作用。我也接受jQuery解决方案。

类似的问题:

获取数组中的所有非唯一值(即:重复/多次出现)


当前回答

使用array.filter和.indexOf函数的单行版本:

arr = arr.filter(function (value, index, array) { 
  return array.indexOf(value) === index;
});

其他回答

VanillaJS:使用像Set这样的Object删除重复项

您可以始终尝试将其放入对象中,然后遍历其关键点:

function remove_duplicates(arr) {
    var obj = {};
    var ret_arr = [];
    for (var i = 0; i < arr.length; i++) {
        obj[arr[i]] = true;
    }
    for (var key in obj) {
        ret_arr.push(key);
    }
    return ret_arr;
}

Vanilla JS:通过跟踪已经看到的值来删除重复项(订单安全)

或者,对于订单安全版本,使用一个对象来存储所有以前看到的值,并在添加到数组之前检查值。

function remove_duplicates_safe(arr) {
    var seen = {};
    var ret_arr = [];
    for (var i = 0; i < arr.length; i++) {
        if (!(arr[i] in seen)) {
            ret_arr.push(arr[i]);
            seen[arr[i]] = true;
        }
    }
    return ret_arr;

}

ECMAScript 6:使用新的Set数据结构(顺序安全)

ECMAScript 6添加了新的Set Data Structure,它允许您存储任何类型的值。Set.values按插入顺序返回元素。

function remove_duplicates_es6(arr) {
    let s = new Set(arr);
    let it = s.values();
    return Array.from(it);
}

示例用法:

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

b = remove_duplicates(a);
// b:
// ["Adam", "Carl", "Jenny", "Matt", "Mike", "Nancy"]

c = remove_duplicates_safe(a);
// c:
// ["Mike", "Matt", "Nancy", "Adam", "Jenny", "Carl"]

d = remove_duplicates_es6(a);
// d:
// ["Mike", "Matt", "Nancy", "Adam", "Jenny", "Carl"]
function removeDuplicates(inputArray) {
            var outputArray=new Array();

            if(inputArray.length>0){
                jQuery.each(inputArray, function(index, value) {
                    if(jQuery.inArray(value, outputArray) == -1){
                        outputArray.push(value);
                    }
                });
            }           
            return outputArray;
        }
var uniqueCompnies = function(companyArray) {
    var arrayUniqueCompnies = [],
        found, x, y;

    for (x = 0; x < companyArray.length; x++) {
        found = undefined;
        for (y = 0; y < arrayUniqueCompnies.length; y++) {
            if (companyArray[x] === arrayUniqueCompnies[y]) {
                found = true;
                break;
            }
        }

        if ( ! found) {
            arrayUniqueCompnies.push(companyArray[x]);
        }
    }

    return arrayUniqueCompnies;
}

var arr = [
    "Adobe Systems Incorporated",
    "IBX",
    "IBX",
    "BlackRock, Inc.",
    "BlackRock, Inc.",
];

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();
var duplicates = function(arr){
     var sorted = arr.sort();
   var dup = [];
   for(var i=0; i<sorted.length; i++){
        var rest  = sorted.slice(i+1); //slice the rest of array
       if(rest.indexOf(sorted[i]) > -1){//do indexOf
            if(dup.indexOf(sorted[i]) == -1)    
         dup.push(sorted[i]);//store it in another arr
      }
   }
   console.log(dup);
}

duplicates(["Mike","Matt","Nancy","Adam","Jenny","Nancy","Carl"]);