const reGroup = (list, key) => {
    const newGroup = {};
    list.forEach(item => {
        const newItem = Object.assign({}, item);
        delete newItem[key];
        newGroup[item[key]] = newGroup[item[key]] || [];
        newGroup[item[key]].push(newItem);
    });
    return newGroup;
};

const animals = [
  {
    type: 'dog',
    breed: 'puddle'
  },
  {
    type: 'dog',
    breed: 'labradoodle'
  },
  {
    type: 'cat',
    breed: 'siamese'
  },
  {
    type: 'dog',
    breed: 'french bulldog'
  },
  {
    type: 'cat',
    breed: 'mud'
  }
];
console.log(reGroup(animals, 'type'));

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'
  },
];

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

完全没有理由下载第三方库来解决这个简单的问题，就像上面的解决方案所建议的那样。

在es6中按特定键对对象列表进行分组的单行版本:

const groupByKey = (list, key) => list.reduce((hash, obj) => ({...hash, [obj[key]]:( hash[obj[key]] || [] ).concat(obj)}), {})

较长的版本过滤掉没有键的对象:

function groupByKey(array, key) { return array .reduce((hash, obj) => { if(obj[key] === undefined) return hash; return Object.assign(hash, { [obj[key]]:( hash[obj[key]] || [] ).concat(obj)}) }, {}) } 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'}]; console.log(groupByKey(cars, 'make'))

注意:原来的问题似乎是问如何按制造商对汽车进行分组，但省略了每组中的制造商。因此，如果没有第三方库，简单的回答是这样的:

const groupByKey = (list, key, {omitKey=false}) => list.reduce((hash, {[key]:value, ...rest}) => ({...hash, [value]:( hash[value] || [] ).concat(omitKey ? {...rest} : {[key]:value, ...rest})} ), {}) 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'}]; console.log(groupByKey(cars, 'make', {omitKey:true}))

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);

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。

你也可以像这样使用数组#forEach()方法:

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' }]; let newcars = {} cars.forEach(car => { newcars[car.make] ? // check if that array exists or not in newcars object newcars[car.make].push({model: car.model, year: car.year}) // just push : (newcars[car.make] = [], newcars[car.make].push({model: car.model, year: car.year})) // create a new array and push }) console.log(newcars);

根据@Jonas_Wilms的回答，如果你不想输入所有的字段:

    var result = {};

    for ( let { first_field, ...fields } of your_data ) 
    { 
       result[first_field] = result[first_field] || [];
       result[first_field].push({ ...fields }); 
    }

我没有做任何基准测试，但我相信使用for循环会比这个答案中建议的任何方法都更有效。