如何使一个数组的属性成为另一个数组的属性（代码优化）

Question

这就是我所拥有的

const users = [
    { id: 1, name: 'Mike', postIds: [11, 22] },
    { id: 2, name: 'Dan', postIds: [33] },
    { id: 3, name: 'Lance', postIds: [44] },
];

const posts = [
    { id: 11, title: 'How good is he' },
    { id: 22, title: 'How fast is he' },
    { id: 33, title: 'How to make it faster' },
    { id: 44, title: 'How can he do it' },
];

这就是我需要在输出中获得的内容

const expectedResult = [
  {
    id: 1,
    name: 'Mike',
    posts: [
      { id: 11, title: 'How good is he' },
      { id: 22, title: 'How fast is he' },
    ]
  },
  {
    id: 2,
    name: 'Dan',
    posts: [{ id: 33, title: 'How to make it faster' }]
  },
  {
    id: 3,
    name: 'Lance',
    posts: [{ id: 44, title: 'How can he do it' }]
  },
]

这就是我尝试过的。它有效，但它很愚蠢，我认为它可以在一次操作中完成。请检查我可以做些什么使它更干净

const users = [
    { id: 1, name: 'Mike', postIds: [11, 22] },
    { id: 2, name: 'Dan', postIds: [33] },
    { id: 3, name: 'Lance', postIds: [44] },
];

const posts = [
    { id: 11, title: 'How good is he' },
    { id: 22, title: 'How fast is he' },
    { id: 33, title: 'How to make it faster' },
    { id: 44, title: 'How can he do it' },
];

let updUsers = users.map(obj => ({ ...obj,
  posts: [...posts]
}))
const output = updUsers.map(
  user => ({
    ...user,
    posts: user.posts.filter(
      post => user.postIds.includes(post.id)
    )
  })
);
const expectedOut = output.map(({
  id,
  name,
  posts
}) => ({
  id,
  name,
  posts
}))
console.log(expOut)

Answer 1

将帖子数组转换为帖子 ID 的映射 -> 帖子对象以加快查找速度：

const postMap = new Map(posts.map((p) => [p.id, p]));

const expectedResult = users.map((u) => {
  const newU = { ...u, posts: u.postIds.map((id) => postMap.get(id)) };
  delete newU.postIds;  // Remove the undesired `postIds` property from the copy
  return newU;
});

console.log(expectedResult);

Answer 2

您可以在映射users 时过滤posts，而不是将其作为第二次传递。

const users = [
    { id: 1, name: 'Mike', postIds: [11, 22] },
    { id: 2, name: 'Dan', postIds: [33] },
    { id: 3, name: 'Lance', postIds: [44] },
];

const posts = [
    { id: 11, title: 'How good is he' },
    { id: 22, title: 'How fast is he' },
    { id: 33, title: 'How to make it faster' },
    { id: 44, title: 'How can he do it' },
];

let expOut = users.map(({id, name, postIds}) => ({ id, name,
  posts: posts.filter(({id}) => postIds.includes(id))
}))

console.log(expOut)

Answer 3

• 这是另一种幼稚的方法，但时间复杂度更高。
• P.S.：时间复杂度为 n^3

const users = [
    { id: 1, name: 'Mike', postIds: [11, 22] },
    { id: 2, name: 'Dan', postIds: [33] },
    { id: 3, name: 'Lance', postIds: [44] },
];

const posts = [
    { id: 11, title: 'How good is he' },
    { id: 22, title: 'How fast is he' },
    { id: 33, title: 'How to make it faster' },
    { id: 44, title: 'How can he do it' },
];

let updUsers = users.map(function(obj){
    let postsArr = [];
    for(i=0; i<obj.postIds.length; i++){
        const postArr = posts.find((item) => item.id == obj.postIds[i]);
        postsArr.push(postArr);
    };


    return{'id':obj.id,'name':obj.name,'posts':postsArr};});

console.log(updUsers);

Answer 4

您可以先将您的帖子数组转换为键值对并使用Object.fromEntries 获取对象，现在只需映射它。这是一个实现：

const users = [ { id: 1, name: 'Mike', postIds: [11, 22] }, { id: 2, name: 'Dan', postIds: [33] }, { id: 3, name: 'Lance', postIds: [44] },];

const posts = [ { id: 11, title: 'How good is he' }, { id: 22, title: 'How fast is he' }, { id: 33, title: 'How to make it faster' }, { id: 44, title: 'How can he do it' },];

//convert posts to object and then map it:

const postObjects = Object.fromEntries(posts.map(p=>[p.id, p]));

const result = users.map(({id, name, postIds})=>({id, name, posts:postIds.map(p=>postObjects[p])}));

console.log(result);

Answer 5

我们可以使用Array.prototype.reduce 和Array.prototype..filter 来简化这一过程，并使用Set 为用户找到合适的帖子以便更快地查找：

const users = [
    { id: 1, name: 'Mike', postIds: [11, 22] },
    { id: 2, name: 'Dan', postIds: [33] },
    { id: 3, name: 'Lance', postIds: [44] },
];

const posts = [
    { id: 11, title: 'How good is he' },
    { id: 22, title: 'How fast is he' },
    { id: 33, title: 'How to make it faster' },
    { id: 44, title: 'How can he do it' },
];

const mapUsersByPost = (users, posts) => {
  return users.reduce((acc, {id, name, postIds}) => {
   const filteredPosts = posts.filter(({id}) => new Set(postIds).has(id));
   acc.push({ id, name, posts: filteredPosts});
   return acc;
  }, []);
}

console.log(mapUsersByPost(users, posts));