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' ] } ]

@metakungfu answer略有不同，主要区别在于它从结果对象中省略了原始键，因为在某些情况下对象本身不再需要它，因为它现在在父对象中可用。

const groupBy = (_k, a) => a.reduce((r, {[_k]:k, ...p}) => ({
    ...r, ...{[k]: (
        r[k] ? [...r[k], {...p}] : [{...p}]
    )}
}), {});

考虑到您的原始输入对象:

console.log(groupBy('make', cars));

会导致:

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

我喜欢写它没有依赖/复杂性，只是纯粹的简单js。

const mp = {} const cars = [ { model: 'Imaginary space craft SpaceX model', year: '2025' }, { 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' } ] cars.forEach(c => { if (!c.make) return // exit (maybe add them to a "no_make" category) if (!mp[c.make]) mp[c.make] = [{ model: c.model, year: c.year }] else mp[c.make].push({ model: c.model, year: c.year }) }) console.log(mp)

使用lodash/fp，你可以使用_.flow()创建一个函数，它首先按键分组，然后映射每个组，并从每个项中省略一个键:

const { flow, groupBy, mapValues, map, omit } = _; const groupAndOmitBy = key => flow( groupBy(key), mapValues(map(omit(key))) ); 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 groupAndOmitMake = groupAndOmitBy('make'); const result = groupAndOmitMake(cars); console.log(result); .as-console-wrapper { max-height: 100% !important; top: 0; } <script src='https://cdn.jsdelivr.net/g/lodash@4(lodash.min.js+lodash.fp.min.js)'></script>

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>

提莫的答案是我会怎么做。简单的_。groupBy，并允许在分组结构中的对象中有一些重复。

然而，OP还要求删除重复的make键。如果你想从头到尾:

var grouped = _.mapValues(_.groupBy(cars, 'make'),
                          clist => clist.map(car => _.omit(car, 'make')));

console.log(grouped);

收益率:

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

如果你想使用Underscore.js来实现这个功能，请注意它的_. js版本。mapValues被称为_.mapObject。