Question

我有一系列的哈希，就像这样：

[{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"}]

我想扔掉重复的哈希。设置无效，因为哈希是唯一的对象。

我感到被困住了，需要踢一下去思考。请指教！

Answer 1

您也可以使用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);

Answer 2

尝试一下

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

以h为单位的输入数组，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));

Answer 3

时间充裕

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);

Answer 4

我建议使用关联数组的方法，这使重复删除变得更容易。如果可以的话，首先应该将数组构建为关联数组，这样就不必转换它。这是您的操作方式：

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，为什么行

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

可以做到这一点，但让我们直接接受其功能性：）

从哈希数组中过滤重复的哈希-Javascript

4 个答案: