JS数组对象

Question

原始数据：
let arr = [
    {
        name:'lsy',
        map:['1','2','3','4']
    },
     {
        name:'lsy',
        map:['1','2','3','4','5']
    },
    {
        name:'lsy',
        map:['1']
    },
    {
        name:'fan',
        map:['1','2','3','4']
    },
     {
        name:'fan',
        map:['1','2']
    },
    {
        name:'fan',
        map:['1']
    }
]

目标数据：
arr1 = [
    {
        name:'lsy',
        map:['1','2','3','4','5']
    },
    {
        name:'fan',
        map:['1','2','3','4']
    },
]

如上，希望最后只保留name唯一、map.length最大的那个对象，该怎么实现

Answer 1

const res = []
for(const item of arr) {
    const index = res.findIndex(i => i.name === item.name)
    if(index > -1) {
        res[index] = {
            name: item.name,
            map: Array.from(new Set([...res[index].map,...item.map]))
        }
    } else {
        res.push(item)
    }
}

最后的res即是你要的结果

Answer 2

Array
    .from(arr)
    .sort((m, n) => m.name === n.name && m.map.length - n.map.length)
    .filter(({ name: m }, index, { [index + 1]: { name: n } = {} }) => m != n);

Answer 3

Object.values([
    {
        name:'lsy',
        map:['1','2','3','4']
    },
     {
        name:'lsy',
        map:['1','2','3','4','5']
    },
    {
        name:'lsy',
        map:['1']
    },
    {
        name:'fan',
        map:['1','2','3','4']
    },
     {
        name:'fan',
        map:['1','2']
    },
    {
        name:'fan',
        map:['1']
    }
].reduce((res,{name,map}) => {
    if(name in res){
        if(res[name].map.length < map.length) res[name].map = map
    } else res[name] = {name, map}
    return res;
}, {}))