下面的函数(前面提到的eliminateduplates函数的变体)似乎可以做到这一点，它为输入["test"， "test2"， "test2"， 1,1,2,3,4,5,6,7,7,10,22,43,1,5,5,8]返回test2,1,7,7,8]。

请注意，这个问题在JavaScript中比在大多数其他语言中更奇怪，因为JavaScript数组几乎可以容纳任何东西。注意，使用排序的解决方案可能需要提供适当的排序函数——我还没有尝试过这种方法。

这个特殊的实现(至少)适用于字符串和数字。

function findDuplicates(arr) {
    var i,
        len=arr.length,
        out=[],
        obj={};

    for (i=0;i<len;i++) {
        if (obj[arr[i]] != null) {
            if (!obj[arr[i]]) {
                out.push(arr[i]);
                obj[arr[i]] = 1;
            }
        } else {
            obj[arr[i]] = 0;            
        }
    }
    return out;
}

var a = ["a","a","b","c","c"];

a.filter(function(value,index,self){ return (self.indexOf(value) !== index )})

我试图改善@swilliams的答案，这将返回一个没有重复的数组。

// arrays for testing
var arr = [9, 9, 111, 2, 3, 4, 4, 5, 7];

// ascending order
var sorted_arr = arr.sort(function(a,b){return a-b;}); 

var arr_length = arr.length;
var results = [];
if(arr_length){
    if(arr_length == 1){
        results = arr;
    }else{
        for (var i = 0; i < arr.length - 1; i++) {
            if (sorted_arr[i + 1] != sorted_arr[i]) {
                results.push(sorted_arr[i]);
            }
            // for last element
            if (i == arr.length - 2){
                results.push(sorted_arr[i+1]);
            }
        }
    }
}

alert(results);

使用underscore.js

function hasDuplicate(arr){
    return (arr.length != _.uniq(arr).length);
}

这是我能想到的最简单的解决办法:

const arr =[1、2、2、2 0,0,0,500,1,“,”“,”“) Const filtered = arr。filter((el, index) => arr.indexOf(el) !== index) // => filtered = [2,2,0,0， -1， 'a'， 'a'] Const duplicate =[…]新的(过滤) console.log(副本) // => [2,0， -1， 'a']

就是这样。

注意:

It works with any numbers including 0, strings and negative numbers e.g. -1 - Related question: Get all unique values in a JavaScript array (remove duplicates) The original array arr is preserved (filter returns the new array instead of modifying the original) The filtered array contains all duplicates; it can also contain more than 1 same value (e.g. our filtered array here is [ 2, 2, 0, 0, -1, 'a', 'a' ]) If you want to get only values that are duplicated (you don't want to have multiple duplicates with the same value) you can use [...new Set(filtered)] (ES6 has an object Set which can store only unique values)

希望这能有所帮助。

ES6语法的简单代码(返回重复的排序数组):

let duplicates = a => {d=[]; a.sort((a,b) => a-b).reduce((a,b)=>{a==b&&!d.includes(a)&&d.push(a); return b}); return d};

使用方法:

duplicates([1,2,3,10,10,2,3,3,10]);