如何在javascript中查找重复值并将重复值存储到新数组中

How to find repeated values and store repeated values into new array in javascript

提问人：Jonas 提问时间：9/14/2021 更新时间：9/14/2021 访问量：130

问：

我正在尝试实现一个逻辑，其中我有一个数组。我想找到所有重复的元素，我想将这些重复的元素存储到一个新数组中。我非常努力，但没有找到解决方案。[3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6]

如果有人编写简单的代码并解释代码的作用，那就太好了。

谢谢

JavaScript 数组 ecmascript-2017

评论

1赞 evolutionxbox 9/14/2021
你能分享你尝试过的代码吗？
1赞 9/14/2021
我发现 3 种可能的解决方案不到 1 分钟，你尝试了什么？
0赞 Jonas 9/14/2021
我是初学者，我不知道发生了什么
0赞 Andy 9/14/2021
如果您不将其添加到问题中，我们将无法帮助调试您的代码。欢迎来到 SO。您可能会发现阅读网站帮助部分在提出好问题和此问题清单时很有用。你为解决问题而编写的代码应该包括一个最小的可重现示例，并包含在你的问题中。
2赞 Harshit Rastogi 9/14/2021
这回答了你的问题吗？获取 JavaScript 数组中的所有唯一值（删除重复项）

答：

0赞 user4431269 9/14/2021 #1

首先，您需要了解一些基础知识。

阵列映射 - MDN 信息

数组映射创建一个新数组，并且不会更改当前数组。

//So if we do:
[3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6].map((element, index, array) => { return 0; } );
//We get
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

Array.indexOf（element） MDN 信息

返回数组中元素的索引，-1 未找到，从 0 开始。

//So if we do:
[3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6].map( (element, index, array) => { return array.indexOf(element); } );
// we get 
// [0, 1, 2, 3, 3, 5, 3, 7, 2, 7, 1, 3, 12, 7, 1, 15, 16, 7, 18, 5, 1, 3, 22, 0, 12]

//so if we do use the second parameter of map in combination with indexOf
[3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6].map( (element, index, array) => { return array.indexOf(element) === index; } );
// we get 
// [true, true, true, true, false, true, false, true, false, false, false, false, true, false, false, true, true, false, true, false, false, false, true, false, false]

总结一下：

[3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6].map( (element, index, array) => { 
    if ( array.indexOf(element) !== index ) { 
       return element; 
    } else { 
        return undefined; 
    }
});
// will give us
// [undefined, undefined, undefined, undefined, 5, undefined, 5, undefined, 63, 2, 4, 5, undefined, 2, 4, undefined, undefined, 2, undefined, 1, 4, 5, undefined, 3, 6]

MDN 上的数组过滤器信息

// so now we have our array of elements that are duplicated with undefined values in it, now we filter.

[undefined, undefined, undefined, undefined, 5, undefined, 5, undefined, 63, 2, 4, 5, undefined, 2, 4, undefined, undefined, 2, undefined, 1, 4, 5, undefined, 3, 6].filter( (element) => { 
    return element !== undefined; 
});
//result
[5, 5, 63, 2, 4, 5, 2, 4, 2, 1, 4, 5, 3, 6]

这是我能给出的最基本的解释，不使用条件运算符，存储在变量中等。如果你在编程方面更高级一点，你可以编写较少的短代码，称为 1 行代码。但我希望这能给你带来灵感，让你继续下去。但是你首先要了解基本知识，这意味着学习、阅读、多学习和多练习。

评论

0赞 sp00m 9/14/2021
您可以直接在原始数组上。.filter((e, i, a) => a.indexOf(e) !== i)
0赞 sp00m 9/14/2021
哇，很简单，我只是指出中间步骤没有任何意义，因为您的最终步骤可以直接在原始数组上调用，逻辑完全相同。这不是关于单行代码，而是关于不要在数组上迭代两次。.map.filter.indexOf
0赞 9/14/2021
是的，优化你的代码，最有趣的部分是追逐那些μs。但我们都必须从某个地方开始。我怀疑这个问题会被删除。我应该删除它。但是新人对编程有一些最基本的问题，不知道该去哪里或如何开始。
0赞 Karma Blackshaw 9/14/2021
哈哈哈哈轻松的家伙
0赞 Ranjan 9/14/2021 #2

如果你只需要数组的重复项，你可以使用这个：

var arry = [3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6];
var rep_arry = [];
   
for (const [key,val] of Object.entries(arry)) {
  for (const [key1,val1] of Object.entries(arry)) {
     if(val == val1 && !rep_arry.includes(val) && key != key1){
         rep_arry.push(val);
     }
  }    
}
console.log(rep_arry);

它返回这个

[3, 4, 63, 5, 1, 2, 6]

上面的代码使用两个 for 循环，它检查每个单独的元素，每个元素使用两个 for 循环。在推送之前，我放置了 3 个条件，1 个用于检查等于，2 个用于检查项目是否可用于新数组，第三个条件用于防止使用键检查自己的元素。它的工作量也很大。请参阅代码片段。

<html>

    <script>
        var arry = [3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6,3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6];
        var rep_arry = [];
       
        for (const [key,val] of Object.entries(arry)) {
           
            for (const [key1,val1] of Object.entries(arry)) {
                if(val == val1 && !rep_arry.includes(val) && key != key1){
                    rep_arry.push(val);
                }
            }
            
        }
        console.log(rep_arry);
    </script>
</html>

评论

0赞 9/14/2021
这在一个小数组上工作，当你有 5000 个元素时，你会得到 troubels。OP 要求解释代码在做什么。
0赞 Ranjan 9/14/2021
你认为它会在数组中遇到更多麻烦吗？,
0赞 Ranjan 9/14/2021
@PhilAndelhofs，回答您的问题，请查看此 codepen.io/ranjandaswani/pen/RwgZEev
0赞 Hayk 9/14/2021 #3

我们可以通过保留唯一元素列表来解决这个问题。 如果我们保留唯一元素的列表，那么收集重复元素将变得容易。

这可能不是效率的最佳解决方案，但它解决了当前的问题。我注释了代码，所以它可能会帮助你了解发生了什么。

要解决这类问题，最好学习算法。

const arr = [3,4,63,5,5,1,5,2,63,2,4,5,6,2,4,56,74,2,671,1,4,5,7,3,6];

let uniqueElmts = [];  // here we gonna store unique elements
let duplicates = [];   // here we gonna store duplicate elements 

for(let i = 0; i < arr.length; i++){
   if(uniqueElmts.includes(arr[i])){  // if 'uniqueElmts' already contains arr[i], then arr[i] is a duplicate 
        duplicates.push(arr[i]);  // push arr[i] element to 'duplicates' array, because it's duplicate
   }
   else {
        uniqueElmts.push(arr[i]);     // if 'uniqueElmts' doesn't contain arr[i] then it's unique element so we push arr[i] to 'uniqueElmts' 
   }
}


console.log(duplicates); // [ 5, 5, 63, 2, 4, 5, 2, 4, 2, 1, 4, 5, 3, 6 ]

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