高效地将一个数组中的字段合并到另外两个数组中

Question

假设你有三个对象数组：

```html ```

let a1 = [
  { id: 1, name: 'foo' },
  { id: 2, name: 'bar' },
  { id: 3, name: 'baz' }
]

let a2 = [
  { name: 'foo' },
  { name: 'bar' }
]

let a3 = [
  { name: 'bar' },
  { name: 'baz' }
]

目标是使用a1作为源，并在a2和a3元素中添加一个id字段，该字段与a1中的相应name字段对应。有什么高效的方法可以实现这一目标吗？（注：此处的“高效”意味着“比循环嵌套更优雅的方法”） 结果应如下所示：

a2: [
  { id: 1, name: 'foo' },
  { id: 2, name: 'bar' }
]

a3: [
  { id: 2, name: 'bar' },
  { id: 3, name: 'baz' }
]

- ken.dunnington

Answer 1

你可以使用 Map 来引用给定名称的 id。然后进行分配。

var a1 = [{ id: 1, name: 'foo' }, { id: 2, name: 'bar' }, { id: 3, name: 'baz' }], 
    a2 = [{ name: 'foo' }, { name: 'bar' }],
    a3 = [{ name: 'bar' }, { name: 'baz' }],
    map = new Map(a1.map(o => [o.name, o.id]));
    
[a2, a3].forEach(a => a.forEach(o => o.id = map.get(o.name)));

console.log(a2);
console.log(a3);

.as-console-wrapper { max-height: 100% !important; top: 0; }

- Nina Scholz

---

这些都是很好的答案，但我喜欢使用Map的简洁性。整个程序还包括更复杂的部分（Node + MongoDB + 文件系统读写），所以每一点帮助都很重要！ :) - ken.dunnington

Answer 2

作为另一种答案，它可能是这样的。

它不包括循环，可能是答案中最短的代码。

const a1 = [{ id: 1, name: 'foo' }, { id: 2, name: 'bar' }, { id: 3, name: 'baz' }];
const a2 = [{ name: 'foo' }, { name: 'bar' }];
const a3 = [{ name: 'bar' }, { name: 'baz' }];

let f = x => a1.filter(a => x.some(y => y.name === a.name));

console.log(f(a2));
console.log(f(a3));

.as-console-wrapper { max-height: 100% !important; top: 0; }

- Gaslan

Answer 3

a2.forEach((a2Elem) => a2Elem.id = a1.filter((a1Elem) => a1Elem.name === a2Elem.name)[0].id)

- Ivan Minakov

Answer 4

我会先获取给定名称的索引，然后将要合并的数组映射到这些索引上：

function combine(mergeInto, base) {
    let indexes = base.map(e => e.name);
    return mergeInto.map(e => ({
        name: e.name,
        id: base[indexes.indexOf(e.name)].id
    }));
}

let a1 = [
    { id: 1, name: 'foo' },
    { id: 2, name: 'bar' },
    { id: 3, name: 'baz' }
]

let a2 = [
    { name: 'foo' },
    { name: 'bar' }
]

let a3 = [
    { name: 'bar' },
    { name: 'baz' }
]

function combine(mergeInto, base) {
    let indexes = base.map(e => e.name);
    return mergeInto.map(e => ({
        name: e.name,
        id: base[indexes.indexOf(e.name)].id
    }));
}

console.log(combine(a3, a1));

- baao

Answer 5

一个循环的建议 - 创建一个哈希表，然后将字段合并到数组中 - 示例如下：

let a1=[{id:1,name:'foo'},{id:2,name:'bar'},{id:3,name:'baz'}], a2=[{name:'foo'},{name:'bar'}], a3=[{name:'bar'},{name:'baz'}];

// create a hash table
let hash = a1.reduce(function(p,c){
  p[c.name] = c;
  return p;
},Object.create(null))

// merge the results
function merge(arr) {
  Object.keys(arr).map(function(e){
    arr[e]['id'] = hash[arr[e].name]['id'];
  });
  return arr;
}

console.log(merge(a2), merge(a3));

.as-console-wrapper{top:0;max-height:100%!important;}

- kukkuz