下面是您自己的groupBy函数，它是来自https://github.com/you-dont-need/You-Dont-Need-Lodash-Underscore的代码的泛化

函数groupBy(xs, f) { 返回x。减少((r, v, i, a、k = f (v)) = > ((r [k] | | (r [k] = [])) .push (v), r), {}); ｝ Const cars = [{make: 'audi'，型号:'r8'，年份:'2012'}，{make: 'audi'，型号:'rs5'，年份:'2013'}，{make: 'ford'，型号:'mustang'，年份:'2012'}，{make: 'ford'，型号:'fusion'，年份:'2015'}，{make: 'kia'，型号:'optima'，年份:'2012'}]; const result = groupBy(cars， (c) => c.make); console.log(结果);

const groupBy = (array, callback) => {
  const groups = {};
  
  array.forEach((element) => {
    const groupName = callback(element);
    if (groupName in groups) {
      groups[groupName].push(element);
    } else {
      groups[groupName] = [element];
    }
  });
  
  return groups;
};

或者花哨的裤子:

(() => {
  Array.prototype.groupBy = function (callback) {
    const groups = {};
    this.forEach((element, ...args) => {
      const groupName = callback(element, ...args);
      if (groupName in groups) {
        groups[groupName].push(element);
      } else {
        groups[groupName] = [element];
      }
    });

    return groups;
  };
})();

const res = [{ name: 1 }, { name: 1 }, { name: 0 }].groupBy(({ name }) => name);

// const res = {
//   0: [{name: 0}],
//   1: [{name: 1}, {name: 1}]
// }

这是一个MDN阵列的填充。groupBy函数。

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'));

下面是您自己的groupBy函数，它是来自https://github.com/you-dont-need/You-Dont-Need-Lodash-Underscore的代码的泛化

函数groupBy(xs, f) { 返回x。减少((r, v, i, a、k = f (v)) = > ((r [k] | | (r [k] = [])) .push (v), r), {}); ｝ Const cars = [{make: 'audi'，型号:'r8'，年份:'2012'}，{make: 'audi'，型号:'rs5'，年份:'2013'}，{make: 'ford'，型号:'mustang'，年份:'2012'}，{make: 'ford'，型号:'fusion'，年份:'2015'}，{make: 'kia'，型号:'optima'，年份:'2012'}]; const result = groupBy(cars， (c) => c.make); console.log(结果);

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' ] } ]

我制定了一个基准测试不使用外部库的每个解决方案的性能。

JSBen.ch

由@Nina Scholz发布的reduce()选项似乎是最佳选项。