Question

以下代码有效，但是如何使它更简洁？在两个数组中，我们有一个公共属性（_id）。有了这个属性，我从数组2获得了一个属性。更加具体：数组“电影”包含具有属性“ _id”和“标题”的电影项目。数组“收藏夹电影”仅包含“ _id”。我想返回用户喜欢的电影列表作为标题。

var favoritemovies = [1111,3333];

const movies = [        
    {_id:1111, title: 'movie1'},    
    {_id:2222, title: 'movie2'},       
    {_id:3333, title: 'movie3'},    
];

var movieTitles = [];

for (var i = 0; i < favoritemovies.length; i++) {      
    var favMovies = movies.filter(movie => movie._id == favoritemovies[i])       

movieTitles.push(favMovies); 

var movieTitlesFlat = movieTitles.flat(); 
} 

for (var i = 0; i < favoritemovies.length; i++) {    
  console.log(movieTitlesFlat[i].title); 
}

Answer 1

归纳为_id索引的对象，然后在.map上调用favoritemovies，总体复杂度为O(N)：

var favoritemovies = [1111,3333];

const movies = [        
    {_id:1111, title: 'movie1'},    
    {_id:2222, title: 'movie2'},       
    {_id:3333, title: 'movie3'},    
];
const moviesById = movies.reduce((a, { _id, title }) => {
  a[_id] = title;
  return a;
}, {});

const titles = favoritemovies.map(id => moviesById[id]);
console.log(titles);

（这也可以通过嵌套循环操作来完成，但是如果不先将movies转换为某种形式的映射，则复杂度将为O(N^2)）

Answer 2

您可以先在filter()上使用movies来获取_id出现在favoritemovies内的所有项目，然后将map()过滤到的数组获得标题。

var favoritemovies = [1111,3333];

const movies = [        
    {_id:1111, title: 'movie1'},    
    {_id:2222, title: 'movie2'},       
    {_id:3333, title: 'movie3'},    
];

const res = movies.filter(x => favoritemovies.includes(x._id)).map(x => x.title);
console.log(res)

上述算法的时间复杂度为O(n ^ 2)。可以通过创建O(n)和Set来使其favoritemovies

var favoritemovies = [1111,3333];
let favmovset = new Set(favoritemovies)
const movies = [        
    {_id:1111, title: 'movie1'},    
    {_id:2222, title: 'movie2'},       
    {_id:3333, title: 'movie3'},    
];

const res = movies.filter(x => favmovset.has(x._id)).map(x => x.title);
console.log(res)

Answer 3

var favoritemovies = [1111,3333];

const movies = [        
    {_id:1111, title: 'movie1'},    
    {_id:2222, title: 'movie2'},       
    {_id:3333, title: 'movie3'},    
];

var movieTitles = movies.filter( (e)=> favoritemovies.includes(e._id));

movieTitles.forEach( (e)=> console.log(e.title));

如何简化此代码？通过第二个数组的共享属性（_id）获取一个数组的属性（标题）

3 个答案: