js两个数组比较去重

Question

请问怎么list2怎么根据list1里的id过滤相同的这条数据，有没有简便的方法

list1:[
    {
        id:0
        name:'xiaomin'
    },
    {
        id:1
        name:'xiaohong'
    },
]

list2:[
    {
        id:0
        name:'xiaomin'
    },
    {
        id:1
        name:'xiaohong'
    },
            {
        id:3
        name:'xiaomin'
    },
    {
        id:4
        name:'xiaohong'
    },
]

Answer 1

如果追求性能，可以采用空间换时间的做法，在JS中最常见的处理方式就是构建Object，因为Object查找key的时间复杂度是O(1)，而数组是O(n)

const list1 = [
    {id: 0, name: 'xiaomin'},
    {id: 1, name: 'xiaohong'},
];
const list2 = [
    {id: 0, name: 'xiaomin'},
    {id: 1, name: 'xiaohong'},
    {id: 3, name: 'xiaomin'},
    {id: 4, name: 'xiaohong'},
];
// 首先构造Object
const idSet = list1.reduce((acc, v) => {
    acc[v.id] = true;
    return acc;
}, {});
// 遍历list2，去掉在idSet中存在的id
const result = list2.filter(v => !idSet[v.id]);

Answer 2

var result = list2.filter(function(item1) {
     return list1.every(function(item2) {
       return item2.id !== item1.id
     })
 })
 console.log(result);

可以这样看看

Answer 3

楼主，可以参考下

const newData = list2.filter((l2) => list1.findIndex((l1) => l2.id === l1.id) !== -1);