对象数组中的Concat数组

Question

我建立了这个数组：

const array = [
  {
    title: 'something',
    list: ['a', 'b', 'c', 'd']
  },
  {
    title: 'dog',
    list: ['aa']
  },
  {
    title: 'cat',
    list: ['aaa', 'b', 'cccc']
  },
  {
    title: 'apple',
    list: [],
  }
]

我想要一个包含其他数组中所有值的数组，所以：

const res = ['a', 'b', 'c', 'd', 'aa', 'aaa', 'b', 'cccc']

我可以这样做吗？我可以使用concat，但是怎么用？

Answer 1

您可以减少用于平整属性的数组。

const
    array = [{ title: 'something', list: ['a', 'b', 'c', 'd'] }, { title: 'dog', list: ['aa'] }, { title: 'cat', list: ['aaa', 'b', 'cccc'] }, { title: 'apple', list: [] }],
    list = array.reduce((r, { list }) => [...r, ...list], []);

console.log(list);

或者采取即将到来的flatMap。

const
    array = [{ title: 'something', list: ['a', 'b', 'c', 'd'] }, { title: 'dog', list: ['aa'] }, { title: 'cat', list: ['aaa', 'b', 'cccc'] }, { title: 'apple', list: [] }],
    list = array.flatMap(({ list }) => list);

console.log(list);

Answer 2

一起使用concat和reduce

var a=[
  {
    title: 'something',
    list: ['a', 'b', 'c', 'd']
  },
  {
    title: 'dog',
    list: ['aa']
  },
  {
    title: 'cat',
    list: ['aaa', 'b', 'cccc']
  },
  {
    title: 'apple',
    list: [],
  }
];
console.log(a.reduce((acc,e)=>acc.concat(e.list),[]))

Answer 3

您可以在concat的空数组上使用map并扩展语法...。

const array = [{"title":"something","list":["a","b","c","d"]},{"title":"dog","list":["aa"]},{"title":"cat","list":["aaa","b","cccc"]},{"title":"apple","list":[]}]

const res = [].concat(...array.map(({list}) => list))
console.log(res)

Answer 4

忘记MyClass, MyClass1, YourClass2, RepositoryClass,和flatmap。让事情简单明了。

reduce

Answer 5

使用reduce和...（散布运算符）非常容易。

const array = [{
    title: 'something',
    list: ['a', 'b', 'c', 'd']
  },
  {
    title: 'dog',
    list: ['aa']
  },
  {
    title: 'cat',
    list: ['aaa', 'b', 'cccc']
  },
  {
    title: 'apple',
    list: [],
  }
]

const result = array.reduce((accum, item) => {
  return [...accum, ...item.list]
}, [])

console.log(result);