javascript - 对象的嵌套reduce函数

Question

我正在尝试减少已经减少的数组，并且在第二次减少功能后我的对象出现问题。我在第二个函数运行之前的对象如下所示:

const object = {
     groupA: [
       {name: "Adam", age: 25, Job: "Pharmacist", group: 'groupA'},
       {name: "Eric", age: 25, Job: "Vet", group: 'groupA'},
       {name: "Bob", age: 25, Job: "Pharmacist", group: 'groupA'},
       {name: "Peter", age: 25, Job: "Vet", group: 'groupA'},
     ],
    groupB: [
       {name: "Adam", age: 25, Job: "Pharmacist", group: 'groupB'},
       {name: "Eric", age: 25, Job: "Vet", group: 'groupB'},
       {name: "Bob", age: 25, Job: "Pharmacist", group: 'groupB'},
       {name: "Peter", age: 25, Job: "Vet", group: 'groupB'},
     ],    


 }

如果我想得到这样的结果，我应该使用什么函数。

const object = {
     groupA: {
       pharmacist: [
         {name: "Adam", age: 25, Job: "Pharmacist", group: 'groupA'},
         {name: "Bob", age: 25, Job: "Pharmacist", group: 'groupA'},
      ],
      vet: [
       {name: "Eric", age: 25, Job: "Vet", group: 'groupA'},
       {name: "Peter", age: 25, Job: "Vet", group: 'groupA'},
      ]
     },
     groupB: {
       pharmacist: [
         {name: "Adam", age: 25, Job: "Pharmacist", group: 'groupB'},
         {name: "Bob", age: 25, Job: "Pharmacist", group: 'groupB'},
      ],
       vet: [
        {name: "Eric", age: 25, Job: "Vet", group: 'groupB'},
        {name: "Peter", age: 25, Job: "Vet", group: 'groupB'},
      ]
     }, 


 }

Answer 1

您可以创建一个单独的函数来 reduce() 每个组，然后使用 for..in 将每个组的未缩减数组替换为缩减对象

const object = {"groupA":[{"name":"Adam","age":25,"Job":"Pharmacist","group":"groupA"},{"name":"Eric","age":25,"Job":"Vet","group":"groupA"},{"name":"","age":25,"Job":"Pharmacist","group":"groupA"},{"name":"","age":25,"Job":"Vet","group":"groupA"}],"groupB":[{"name":"Adam","age":25,"Job":"Pharmacist","group":"groupB"},{"name":"Eric","age":25,"Job":"Vet","group":"groupB"},{"name":"","age":25,"Job":"Pharmacist","group":"groupB"},{"name":"","age":25,"Job":"Vet","group":"groupB"}]}

function groupByJob(arr){
  return arr.reduce((ac, a) => {
    if(!ac[a.Job]){
      ac[a.Job] = [];
    }
    ac[a.Job].push(a);
    return ac;
  }, {});
}

for(let k in object){
  object[k] = groupByJob(object[k]);
}
console.log(object)

如果您不想拆分，可以直接将 map() 应用于 object 的条目，然后直接在 map() 上使用每个值。

const object = {"groupA":[{"name":"Adam","age":25,"Job":"Pharmacist","group":"groupA"},{"name":"Eric","age":25,"Job":"Vet","group":"groupA"},{"name":"","age":25,"Job":"Pharmacist","group":"groupA"},{"name":"","age":25,"Job":"Vet","group":"groupA"}],"groupB":[{"name":"Adam","age":25,"Job":"Pharmacist","group":"groupB"},{"name":"Eric","age":25,"Job":"Vet","group":"groupB"},{"name":"","age":25,"Job":"Pharmacist","group":"groupB"},{"name":"","age":25,"Job":"Vet","group":"groupB"}]}

const res = Object.fromEntries(
               Object.entries(object)
                 .map(([k, v]) => 
                    [  
                       k, 
                       v.reduce((ac, a) => 
                           (ac[a.Job] = (ac[a.Job] || []).concat(a), ac), 
                       {})
                    ]
                  )
                )

console.log(res)