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>

我喜欢@metakunfu的答案，但它并没有提供预期的输出。 下面是在最终的JSON有效负载中去除“make”的更新。

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'
    },
];

result = cars.reduce((h, car) => Object.assign(h, { [car.make]:( h[car.make] || [] ).concat({model: car.model, year: car.year}) }), {})

console.log(JSON.stringify(result));

输出:

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

这是另一个解决方案。按照要求。

我想创建一个新的汽车对象数组，由make分组:

function groupBy() {
  const key = 'make';
  return cars.reduce((acc, x) => ({
    ...acc,
    [x[key]]: (!acc[x[key]]) ? [{
      model: x.model,
      year: x.year
    }] : [...acc[x[key]], {
      model: x.model,
      year: x.year
    }]
  }), {})
}

输出:

console.log('Grouped by make key:',groupBy())

您可以尝试在调用per iteration的函数中修改对象_。groupBy func。 注意，源数组改变了它的元素!

var res = _.groupBy(cars,(car)=>{
    const makeValue=car.make;
    delete car.make;
    return makeValue;
})
console.log(res);
console.log(cars);

Prototype version using ES6 as well. Basically this uses the reduce function to pass in an accumulator and current item, which then uses this to build your "grouped" arrays based on the passed in key. the inner part of the reduce may look complicated but essentially it is testing to see if the key of the passed in object exists and if it doesn't then create an empty array and append the current item to that newly created array otherwise using the spread operator pass in all the objects of the current key array and append current item. Hope this helps someone!.

Array.prototype.groupBy = function(k) {
  return this.reduce((acc, item) => ((acc[item[k]] = [...(acc[item[k]] || []), item]), acc),{});
};

const projs = [
  {
    project: "A",
    timeTake: 2,
    desc: "this is a description"
  },
  {
    project: "B",
    timeTake: 4,
    desc: "this is a description"
  },
  {
    project: "A",
    timeTake: 12,
    desc: "this is a description"
  },
  {
    project: "B",
    timeTake: 45,
    desc: "this is a description"
  }
];

console.log(projs.groupBy("project"));

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);