Question

我有以下数组。

原始数组：

var array =[
  { "id": 1, "name": "abc", "gender": "m","age": "15" },
  { "id": 2, "name": "a", "gender": "m", "age": "25"  },
  { "id": 3,"name": "efg", "gender": "f","age": "5" },
  { "id": 4,"name": "hjk","gender": "m","age": "35" },
  {  "id": 5, "name": "ikly","gender": "m","age": "41" },
  { "id": 6, "name": "ert", "gender": "f", "age": "30" },
  { "id": 7, "name": "qwe", "gender": "f", "age": "31" },
  { "id":8, "name": "bdd",  "gender": "m", "age": " 8" }
];

数组删除：

var arrayCopy =[
  { "id": 1, "name": "abc", "gender": "m","age": "15" },
  { "id": 3,"name": "efg", "gender": "f","age": "5" },
  { "id": 7, "name": "qwe", "gender": "f", "age": "31" },
  { "id":8, "name": "bdd",  "gender": "m", "age": " 8" }
];

我必须删除remove数组中包含的原始数组中的元素，但是我遇到了一些问题。

我考虑过从数组副本中获取所有ID，然后进行排除，但我不确定这是最好的方法。

一些建议？

Answer 1

创建一个Set个ID，将其排除在arrayCopy之外。然后，您可以过滤数组，并保留ID中未出现在Set中的所有项目：

const array = [{"id":1,"name":"abc","gender":"m","age":"15"},{"id":2,"name":"a","gender":"m","age":"25"},{"id":3,"name":"efg","gender":"f","age":"5"},{"id":4,"name":"hjk","gender":"m","age":"35"},{"id":5,"name":"ikly","gender":"m","age":"41"},{"id":6,"name":"ert","gender":"f","age":"30"},{"id":7,"name":"qwe","gender":"f","age":"31"},{"id":8,"name":"bdd","gender":"m","age":" 8"}];
const arrayCopy = [{"id":1,"name":"abc","gender":"m","age":"15"},{"id":3,"name":"efg","gender":"f","age":"5"},{"id":7,"name":"qwe","gender":"f","age":"31"},{"id":8,"name":"bdd","gender":"m","age":" 8"}];

const idsToExclude = new Set(arrayCopy.map(({ id }) => id));

const result = array.filter(({ id }) => !idsToExclude.has(id));

console.log(result);

Answer 2

我想从数组副本中获取所有ID，然后进行排除...

是的，那可能是最好的方法。将它们存储在Set（ES2015 +）或对象中作为属性（例如ids[id] = true），然后在原始数组上使用filter，仅保留ID不在条目中的条目。集合/对象。

或者，如果数组很短，则可以在外部数组上使用filter，然后在“删除”数组上使用some，以确定它是否具有外部数组条目的ID。这意味着要反复遍历“删除”数组（对于外部数组中的每个条目一次），但是对于较小的数组来说这不是问题。

Js按数组对象过滤数组对象

2 个答案: