添加Array.prototype.group和Array.prototype.groupToMap的提案现在处于阶段3!

当它达到阶段4并在大多数主流浏览器上实现时，你将能够这样做:

const cars = [
  { make: 'audi', model: 'r8', year: '2012' },
  { make: 'audi', model: 'rs5', year: '2013' },
  { make: 'ford', model: 'mustang', year: '2012' },
  { make: 'ford', model: 'fusion', year: '2015' },
  { make: 'kia', model: 'optima', year: '2012' }
];

const grouped = cars.group(item => item.make);
console.log(grouped);

这将输出:

{
  audi: [
    { make: 'audi', model: 'r8', year: '2012' },
    { make: 'audi', model: 'rs5', year: '2013' }
  ],
  ford: [
    { make: 'ford', model: 'mustang', year: '2012' },
    { make: 'ford', model: 'fusion', year: '2015' }
  ],
  kia: [
    { make: 'kia', model: 'optima', year: '2012' }
  ]
}

在那之前，你可以使用这个core-js polyfill:

const cars = [ { make: 'audi', model: 'r8', year: '2012' }, { make: 'audi', model: 'rs5', year: '2013' }, { make: 'ford', model: 'mustang', year: '2012' }, { make: 'ford', model: 'fusion', year: '2015' }, { make: 'kia', model: 'optima', year: '2012' } ]; const grouped = cars.group(item => item.make); //console.log(grouped); // Optional: remove the "make" property from resulting object const entriesUpdated = Object .entries(grouped) .map(([key, value]) => [ key, value.map(({make, ...rest}) => rest) ]); const noMake = Object.fromEntries(entriesUpdated); console.log(noMake); <script src="https://unpkg.com/core-js-bundle@3.26.1/minified.js"></script>

letfinaldata=[]

let data =[{id:1,name:"meet"},{id:2,name:"raj"},{id:1,name:"hari"},{id:3,name:"hari"},{id:2,name:"ram"}]

data = data.map((item)=> 
{
    return {...item,
        name: [item.name]
    }
}) // Converting the name key from string to array


let temp = [];

for(let i =0 ;i<data.length;i++)
{
    const index = temp.indexOf(data[i].id) // Checking if the object id is already present
    if(index>=0)
    {
        letfinaldata[index].name = [...letfinaldata[index].name,...data[i].name] // If present then append the name to the name of that object
    }
    else{
        temp.push(data[i].id); // Push the checked object id
        letfinaldata.push({...data[i]}) // Push the object
    }
}

console.log(letfinaldata)

输出

[ { id: 1, name: [ 'meet', 'hari' ] },
  { id: 2, name: [ 'raj', 'ram' ] },
  { id: 3, name: [ 'hari' ] } ]

您正在寻找_.groupBy()。

如果需要，从对象中删除分组的属性应该很简单:

Const cars = [{ “做”:“奥迪”, “模型”:“r8”, “年”:“2012” },{ “做”:“奥迪”, “模型”:“生活费”, “年”:“2013” },{ “做”:“福特”, “模型”:“野马”, “年”:“2012” },{ “做”:“福特”, “模型”:“融合”, “年”:“2015” },{ “做”:“克钦独立军”, “模型”:“最佳状态”, “年”:“2012” }); Const分组= _。groupBy(cars, car => car.make); console.log(分组); < script src = " https://cdn.jsdelivr.net/lodash/4.17.2/lodash.min.js " > < /脚本>

简单的for循环也可以实现:

 const result = {};

 for(const {make, model, year} of cars) {
   if(!result[make]) result[make] = [];
   result[make].push({ model, year });
 }

另一个解决方案:

Var汽车= [ {“使”:“奥迪”,“模型”:“r8”,“年”:“2012”},{“使”:“奥迪”,“模型”:“生活费”,“年”:“2013”}, {“使”:“福特”,“模型”:“野马”,“年”:“2012”},{“使”:“福特”,“模型”:“融合”,“年”:“2015”}, {'make': 'kia'，'model': 'optima'，'year': '2012'}， ]; const reducedCars =汽车。Reduce ((acc， {make, model, year}) => ( ｛ acc, [make]: acc[make] ?[…Acc [make]， {model, year}]: [{model, year}]， ｝ ), {}); console.log (reducedCars);

对于key可以为null的情况，我们希望将它们分组为其他

var cars = [{'make':'audi','model':'r8','year':'2012'},{'make':'audi','model':'rs5','year':'2013'},{'make':'ford','model':'mustang','year':'2012'},{'make':'ford','model':'fusion','year':'2015'},{'make':'kia','model':'optima','year':'2012'},
            {'make':'kia','model':'optima','year':'2033'},
            {'make':null,'model':'zen','year':'2012'},
            {'make':null,'model':'blue','year':'2017'},

           ];


 result = cars.reduce(function (r, a) {
        key = a.make || 'others';
        r[key] = r[key] || [];
        r[key].push(a);
        return r;
    }, Object.create(null));