如何按某些属性值对对象数组进行分组

Question

我有两个数组。 一个包含字符串的数组，其中包含名称

let companies = ['Google', 'Coca Cola', 'Jonson & Jonson'];

另一个数组包含带有人物的物体

let employees = [
  {name: 'Alina' company: 'Google', id : 1},
  {name: 'Vika' company: 'Coca Cola', id : 2},
  {name: 'Alex' company: 'Jonson & Jonson', id : 3},
  {name: 'Vlad' company: 'Google', id : 4},
  {name: 'Fibi' company: 'Coca Cola', id : 5},
  {name: 'Joey' company: 'Google', id : 6},
]

我的任务是按名字对这些人进行分组

const groups = [
 {'Google': [
   {name: 'Alina' company: 'Google', id : 1},
   {name: 'Vlad' company: 'Google', id : 4},
 ]},
 'Jonson & Jonso': [
   {name: 'Alex' company: 'Jonson & Jonson', id : 3},
 ]},
 ...
]

也许有人知道如何以最简单的方式做到这一点，并且无需额外的 JS 迭代？ 我可以使用嵌套循环，但它太复杂了。 也许可以用

lodash

来做？ 另请注意，公司名称的字符串键可能包含空格。 非常感谢您的任何建议。

Answer 1

回到未来的答案：

许多浏览器尚未支持，但很快就会推出（TC39 的第 3 阶段），并且已经在 polyfill core-js 中可用）是数组对象上的新

group

方法。

这将允许您直接这样做：

employees.group(employee => employee.company);

甚至：

employees.group(({company}) => company);

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/group

Answer 2

let employees = [
  {name: 'Alina', company: 'Google', id : 1},
  {name: 'Vika', company: 'Coca Cola', id : 2},
  {name: 'Alex', company: 'Jonson & Jonson', id : 3},
  {name: 'Vlad', company: 'Google', id : 4},
  {name: 'Fibi', company: 'Coca Cola', id : 5},
  {name: 'Joey', company: 'Google', id : 6},
]
function groupBy(arr, property) {
    return arr.reduce(function (memo, x) {
        if (!memo[x[property]]) { memo[x[property]] = []; }
        memo[x[property]].push(x);
        return memo;
    }, {});
};
console.log(groupBy(employees,'company'));

Answer 3

最简单的方法是：

let employees = [
  {name: 'Alina' company: 'Google', id : 1},
  {name: 'Vika' company: 'Coca Cola', id : 2},
  {name: 'Alex' company: 'Jonson & Jonson', id : 3},
  {name: 'Vlad' company: 'Google', id : 4},
  {name: 'Fibi' company: 'Coca Cola', id : 5},
  {name: 'Joey' company: 'Google', id : 6},
]

const grouped = groupBy(employees, employee => employee.company);

Answer 4

伙计，如果您在提问之前甚至不检查您的结构是否没有错误，这似乎是对潜在响应者的不尊重。 NVM：/

所以，有固定的变量：

let companies = ['Google', 'Coca Cola', 'Jonson & Jonson'];
let employees = [
  {name: 'Alina', company: 'Google', id : 1},
  {name: 'Vika', company: 'Coca Cola', id : 2},
  {name: 'Alex', company: 'Jonson & Jonson', id : 3},
  {name: 'Vlad', company: 'Google', id : 4},
  {name: 'Fibi', company: 'Coca Cola', id : 5},
  {name: 'Joey', company: 'Google', id : 6}]

请注意，

companies

是多余的，因为所有需要的信息都在

employees

中。

您正在寻找的结构可能是

Map

。你只需做：

let map = new Map()
employees.forEach((currentValue) => {
    map.has(currentValue.company) 
    ? map.get(currentValue.company).push({name: currentValue.name, id: currentValue.id})
    : map.set(currentValue.company, [{name: currentValue.name, id: currentValue.id}])
});

要获得此结果（

company

对象中不再需要字段

employee

）：

{
  Coca Cola: [{
  id: 2,
  name: "Vika"
}, {
  id: 5,
  name: "Fibi"
}],
  Google: [{
  id: 1,
  name: "Alina"
}, {
  id: 4,
  name: "Vlad"
}, {
  id: 6,
  name: "Joey"
}],
  Jonson & Jonson: [{
  id: 3,
  name: "Alex"
}]
}

Answer 5

由于您走的是简单路线，所以会有点长。

let companies = ['Google', 'Coca Cola,' 'Jonson & Jonson',];

let employees = [
  {name: 'Alina' company: 'Google', id : 1},
  {name: 'Vika' company: 'Coca Cola', id : 2},
  {name: 'Alex' company: 'Jonson & Jonson', id : 3},
  {name: 'Vlad' company: 'Google', id : 4},
  {name: 'Fibi' company: 'Coca Cola', id : 5},
  {name: 'Joey' company: 'Google', id : 6},
]

//Let's create an intermediate object 
let interm = {};
/*This would create an object like 
 {
 Key:[],
 Key2:[],
 ...
 }

*/
companies.forEach((each)=>{
    interm[`${each}`] = [];
})

/*filling that interm a
Object with values like

{
'Google':[
{name: 'Alina' company: 'Google', id : 1}, 
{name: 'Vlad' company: 'Google', id : 4},
{name: 'Joey' company: 'Google', id : 6}
],
 Coca Cola:[
{name: 'Vika' company: 'Coca Cola', id : 2},
  {name: 'Fibi' company: 'Coca Cola', id : 5},
],
"Jonson & Jonson":[
 {name: 'Alex' company: 'Jonson & Jonson', id : 3},
]    
}

*/

employee.forEach((each)=>{
    if(companies.indexOf(each.company) != -1)
    interm[`${each.company}`].push(each)

})

//Now our intermediate data is ready 
//We need to convert to our   desirable format
let finalArray = []
Object.keys(interm).forEach((each)=>{
    finalArray.push({each:interm[`${each}`]})
})

console.log(finalArray)

Answer 6

let companies = ['Google', 'Coca Cola', 'Jonson & Jonson'];

let employees = [
  { name: 'Alina', company: 'Google', id: 1 },
  { name: 'Vika', company: 'Coca Cola', id: 2 },
  { name: 'Alex', company: 'Jonson & Jonson', id: 3 },
  { name: 'Vlad', company: 'Google', id: 4 },
  { name: 'Fibi', company: 'Coca Cola', id: 5 },
  { name: 'Joey', company: 'Google', id: 6 },
];

const groups = {};

employees.forEach((employee) => {
  const { company } = employee;
  if (!groups[company]) {
    groups[company] = [];
  }
  groups[company].push(employee);
});

console.log(groups);

Answer 7

可以使用较新的

Object.groupBy

方法来实现此目的。

const employees = [
  {name: 'Alina', company: 'Google', id : 1},
  {name: 'Vika', company: 'Coca Cola', id : 2},
  {name: 'Alex', company: 'Jonson & Jonson', id : 3},
  {name: 'Vlad', company: 'Google', id : 4},
  {name: 'Fibi', company: 'Coca Cola', id : 5},
  {name: 'Joey', company: 'Google', id : 6},
];
const res = Object.groupBy(employees, e => e.company);
console.log(res);