JS 对象数组去重的方法

先模拟一条数据

let arrays = [
    {"id": 1, "name": "李四1"},
    {"id": 2, "name": "李四"},
    {"id": 2, "name": "李四"},
    {"id": 1, "name": "李四1"},
    {"id": 5, "name": "李四5"}
];

方法一：根据某一个属性是否重复

let obj = {};
arrays = arrays.reduce((item, next) => {
    obj[next.id] ? '' : obj[next.id] = true && item.push(next);
    return item;
}, []);

console.log(arrays);

方法二：借助于 for 循环实现某个属性的去重

let result = [];
let obj = {};
for(let i = 0; i < arrays.length; i++){
    if(!obj[arrays[i].id]){
        result.push(arrays[i]);
        obj[arrays[i].id] = true;
    }
}

console.log(result);

方法三：借助于 for 实现所有属性值相同的情况下去重

let result = [];
let obj = [];
let str;
for(let i = 0; i < arrays.length; i++){
    str = JSON.stringify(arrays[i]);
    if(obj.indexOf(str) == -1){
        result.push(arrays[i]);
        obj.push(str);
    }
}

console.log(result)

方法四：对象的所有属性相同才能认为相同

let obj = [];
let str;
arrays = arrays.reduce((item, next) => {
    str = JSON.stringify(next);
    obj.indexOf(str) == -1 ? item.push(next) && obj.push(str) : '';
    return item;
}, []);

console.log(arrays)

二维数组去重

对象的所有属性相同才能认为相同
let arrays = [
    ["id1", "name李四1"],
    ["id2", "name李四1"],
    ["id2", "name李四"],
    ["id1", "name李四1"],
    ["id5", "name李四5"]
];

let obj = [];
let str;
arrays = arrays.reduce((item, next) => {
    str = next.toString();
    obj.indexOf(str) == -1 ? item.push(next) && obj.push(str) : '' ;
    return item;
}, []);

console.log(arrays.toString())