从哈希数组中过滤重复的哈希 - Javascript答案

【问题标题】：Filtering duplicate hashes from array of hashes - Javascript从哈希数组中过滤重复的哈希 - Javascript
【发布时间】：2019-05-21 18:48:40
【问题描述】：

我有一个哈希数组，如下所示：

[{id: "4bf58dd8d48988d110941735", name: "italy"},
 {id: "4bf58dd8d48988d1c6941735", name: "skandi"},
 {id: "4bf58dd8d48988d147941735", name: "diner"},
 {id: "4bf58dd8d48988d110941735", name: "italy"},
 {id: "4bf58dd8d48988d1c4941735", name: "resto"},
 {id: "4bf58dd8d48988d14a941735", name: "vietnam"},
 {id: "4bf58dd8d48988d1ce941735", name: "fish"},
 {id: "4bf58dd8d48988d1c4941735", name: "resto"},
 {id: "4bf58dd8d48988d1c4941735", name: "resto"}]

我想丢弃重复的哈希值。 Set 不起作用，因为哈希是唯一的对象。

我觉得卡住了，需要好好思考一下。请指教！

【问题讨论】：

将数组缩减为一个对象，使用id作为key，然后使用Object.values()转换回数组。
你试过什么？
您必须遍历数组以获取每个可能的哈希值并检查、删除。或者更改数据格式，使用带有哈希字符串的对象作为键，名称和其他任何内容作为值的单独对象
只允许纯JS？
@hindmost，是的，这是我的 react 应用程序的一部分

标签： javascript arrays hash

【解决方案1】：

你也可以使用reduce

//I added comma to each object
const data= [{id: "4bf58dd8d48988d110941735", name: "italy"},
    {id: "4bf58dd8d48988d1c6941735", name: "skandi"},
    {id: "4bf58dd8d48988d147941735", name: "diner"},
    {id: "4bf58dd8d48988d110941735", name: "italy"},
    {id: "4bf58dd8d48988d1c4941735", name: "resto"},
    {id: "4bf58dd8d48988d14a941735", name: "vietnam"},
    {id: "4bf58dd8d48988d1ce941735", name: "fish"},
    {id: "4bf58dd8d48988d1c4941735", name: "resto"},
    {id: "4bf58dd8d48988d1c4941735", name: "resto"}]

const result= data.reduce((current,next)=>{   
    if(!current.some(a=> a.name === next.name)){
        current.push(next);
    }
    return current;
},[])
console.log(result);

【讨论】：

【解决方案2】：

试试这个

h.filter(( t={}, a=>!(t[a.id]=a.id in t) ))

输入数组h，时间复杂度O(n)，解释here .

let h = [{id: "4bf58dd8d48988d110941735", name: "italy"},
 {id: "4bf58dd8d48988d1c6941735", name: "skandi"},
 {id: "4bf58dd8d48988d147941735", name: "diner"},
 {id: "4bf58dd8d48988d110941735", name: "italy"},
 {id: "4bf58dd8d48988d1c4941735", name: "resto"},
 {id: "4bf58dd8d48988d14a941735", name: "vietnam"},
 {id: "4bf58dd8d48988d1ce941735", name: "fish"},
 {id: "4bf58dd8d48988d1c4941735", name: "resto"},
 {id: "4bf58dd8d48988d1c4941735", name: "resto"}]
 
 let t; // declare t to avoid use global (however works without it too)
 let r= h.filter(( t={}, a=>!(t[a.id]=a.id in t) ))

 
 console.log(JSON.stringify(r));

【讨论】：

【解决方案3】：

空间换时间

let arr = [
    { id: '4bf58dd8d48988d110941735', name: 'italy' },
    { id: '4bf58dd8d48988d1c6941735', name: 'skandi' },
    { id: '4bf58dd8d48988d147941735', name: 'diner' },
    { id: '4bf58dd8d48988d110941735', name: 'italy' },
    { id: '4bf58dd8d48988d1c4941735', name: 'resto' },
    { id: '4bf58dd8d48988d14a941735', name: 'vietnam' },
    { id: '4bf58dd8d48988d1ce941735', name: 'fish' },
    { id: '4bf58dd8d48988d1c4941735', name: 'resto' },
    { id: '4bf58dd8d48988d1c4941735', name: 'resto' }
]

let map = {};
let rest = arr.filter((item) => {
    if(map[item.id] === void 0) {
        map[item.id] = item.id;
        return true;
    }
});
map = null;

console.log(rest);

【讨论】：

【解决方案4】：

我建议使用关联数组的方法，这使得重复删除更容易。如果可以的话，您应该首先将您的数组构建为关联数组，这样您就不必转换它。这是你的做法：

var array = [{
    id: "4bf58dd8d48988d110941735",
    name: "italy"
  },
  {
    id: "4bf58dd8d48988d1c6941735",
    name: "skandi"
  }, {
    id: "4bf58dd8d48988d147941735",
    name: "diner"
  }, {
    id: "4bf58dd8d48988d110941735",
    name: "italy"
  }, {
    id: "4bf58dd8d48988d1c4941735",
    name: "resto"
  }, {
    id: "4bf58dd8d48988d14a941735",
    name: "vietnam"
  }, {
    id: "4bf58dd8d48988d14a941735",
    name: "fish"
  }, {
    id: "4bf58dd8d48988d1c4941735",
    name: "resto"
  }, {
    id: "4bf58dd8d48988d1c4941735",
    name: "resto"
  }
];

// you can access the array with arrayAssociative[id], where the id is the real id like "4bf58dd8d48988d110941735"
var arrayAssociative = {};
for (item in array) {
  // first get the unique id's
  var addedNode = arrayAssociative[array[item].id] = arrayAssociative[array[item].id] || {};
  if (addedNode.names == null)
    addedNode.names = {};
  // now get the unique names
  var addedName = arrayAssociative[array[item].id].names[array[item].name] = arrayAssociative[array[item].id].names[array[item].name] || {};
}
console.log(arrayAssociative);

我不知道确切的原因，为什么这条线

var 元素 = arrayAssociative[id] =arrayAssociative[id] || {};

为此工作，但让我们接受功能原样:)

【讨论】：

arrayAssociative[array[item].id] || {} 如果不是 null/undefined/0/false，则此行在 || 左侧给出值，如果相反，则在右侧给出值 - stackoverflow.com/q/2100758/860099