过滤包含数组的对象数组

Question

这是我拥有的数组的较小版本，但结构相同

使用下面的 const arr，我想创建 2 个具有唯一值的新数组，这些数组按升序排序

const arr = [{
    tags: ['f', 'b', 'd'],
    weight: 7,
    something: 'sdfsdf'
  },
  {
    tags: ['a', 'b', 'c', 'd', 'e'],
    weight: 6,
    something: 'frddd'
  },
  {
    tags: ['f', 'c', 'e', 'a'],
    weight: 7,
    something: 'ththh'
  },
  {
    tags: ['a', 'c', 'g', 'e'],
    weight: 5,
    something: 'ghjghj'
  }
];

const finalTags = [];
const finalWeight = [];

// TODO:  find a better way to do this
arr.forEach(v => {
  if (finalWeight.indexOf(v.weight) === -1) finalWeight.push(v.weight);
  v.tags.forEach(val => {
    if (finalTags.indexOf(val) === -1) finalTags.push(val);
  });
});

// Ascending order
finalTags.sort();
finalWeight.sort();

我上面的作品，但似乎有点凌乱，如果有更好/更整洁的方式来做这件事，我会徘徊

Answer 1

您可以将Array.prototype.reduce() 与Set 结合使用，以获取具有排序数组{tags: [], weights: []} 的对象：

const arr = [{tags: ['f', 'b', 'd'],weight: 7,something: 'sdfsdf'},{tags: ['a', 'b', 'c', 'd', 'e'],weight: 6,something: 'frddd'},{tags: ['f', 'c', 'e', 'a'],weight: 7,something: 'ththh'},{tags: ['a', 'c', 'g', 'e'],weight: 5,something: 'ghjghj'}];
const obj = arr.reduce((a, {tags, weight}) => {
  a.tags = [...new Set(a.tags.concat(tags))];
  a.weights = [...new Set(a.weights.concat(weight))];
  return a;
}, {tags: [], weights: []});

// final result i want
console.log('finalTags:', obj.tags.sort()); // ['a', 'b', 'c', 'd', 'e', 'f', 'g'];
console.log('finalWeight:', obj.weights.sort()); // [5, 6, 7];

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Answer 2

一种解决方案是使用Array.reduce() 创建两组，一组使用tags，另一组使用weights。在此之后，您可以将sets 转换为arrays 并在它们上使用Array.sort()：

const arr = [
  {
    tags: ['f', 'b', 'd'],
    weight: 7,
    something: 'sdfsdf'
  },
  {
    tags: ['a', 'b', 'c', 'd', 'e'],
    weight: 6,
    something: 'frddd'
  },
  {
    tags: ['f', 'c', 'e', 'a'],
    weight: 7,
    something: 'ththh'
  },
  {
    tags: ['a', 'c', 'g', 'e'],
    weight: 5,
    something: 'ghjghj'
  }
];

let res = arr.reduce((acc, {tags, weight}) =>
{
    acc.tags = new Set([...acc.tags, ...tags]);
    acc.weights.add(weight);
    return acc;
}, {tags: new Set(), weights: new Set()});

let sortedWeigths = [...res.weights].sort();
let sortedTags = [...res.tags].sort((a, b) => a.localeCompare(b));
console.log("weights: ", sortedWeigths, "tags: ", sortedTags);

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Answer 3

您可以使用以下代码。这基本上将arr 拆分为finalTags 和finalWeights。

.flat() 将数组展平（[1, [2, [3]]] 将变为 [1, 2, 3]）

finalTags.filter((item, index) => finalTags.indexOf(item) >= index).sort(); 删除重复项。

const arr = [{
    tags: ['f', 'b', 'd'],
    weight: 7,
    something: 'sdfsdf'
  },
  {
    tags: ['a', 'b', 'c', 'd', 'e'],
    weight: 6,
    something: 'frddd'
  },
  {
    tags: ['f', 'c', 'e', 'a'],
    weight: 7,
    something: 'ththh'
  },
  {
    tags: ['a', 'c', 'g', 'e'],
    weight: 5,
    something: 'ghjghj'
  }
];

let finalTags = arr.map(e => e.tags);
finalTags = finalTags.flat();
finalTags = finalTags.filter((item, index) => finalTags.indexOf(item) >= index).sort();

let finalWeight = arr.map(e => e.weight);
finalWeight = finalWeight.filter((item, index) => finalWeight.indexOf(item) >= index).sort();

console.log(finalTags);
console.log(finalWeight);

来源：

删除重复项：https://gomakethings.com/removing-duplicates-from-an-array-with-vanilla-javascript/