Question

假设您有三个对象数组：

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作为来源，并在id和a2的元素中添加a3字段，其中包含相应的name字段a1。什么是实现这一目标的有效方法？ （注意：'高效'在这里意味着'比循环内循环更优雅'。

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' }
]

Answer 1

您可以使用Map来引用给定名称的ID。然后分配。

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);

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Answer 2

对于另一种答案，它可能是这样的。

它不包含循环，可能是答案中最短的代码。

＆＃13;

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));

＆＃13;

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＆＃13;

Answer 3

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

Answer 4

我首先获取给定名称的索引，然后将要合并的数组映射到：

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));

Answer 5

单循环提议 - 在数组中创建hash table然后合并字段 - 演示如下：

＆＃13;

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));

＆＃13;

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＆＃13;

有效地将字段从一个数组合并到另外两个数组中

5 个答案: