从数组中选择一个值作为键并按该值分组[重复]

Question

我有一个对象数组。我想按一个值对对象进行分组，然后将该值用作该对象中其余值的键。

例如我有这样的数组：

{
  0: {prize: "Foo", first_name: "John", last_name: "Smith"},
  1: {prize: "Foo", first_name: "Mary", last_name: "Smith"},
  2: {prize: "Bar", first_name: "Jane", last_name: "Doe"},
  3: {prize: "Bar", first_name: "Jack", last_name: "Jones"},
  4: {prize: "Foo", first_name: "Judy", last_name: "Alvarez"},
}

我想要的结果是这样的：

{
  Foo: [
   {first_name: "John", last_name: "Smith"},
   {first_name: "Mary", last_name: "Smith"},
   {first_name: "Judy", last_name: "Alvarez"}
  ],
  Bar: [
   {first_name: "Jane", last_name: "Doe"}, 
   {first_name: "Jack", last_name: "Jones"}
  ]
}

我正在使用 TypeScript，我得到的最接近的是使用我发现的这段代码 sn-p：

console.log(
  _.chain(res.data)
  .groupBy("prize")
  .map((value: any, key: any) => ({prize: key, winners: value}))
  .value()
);

这给了我这样的结果：

[
  0: {
      prize: "Foo", 
      winners: [
       {prize: "Foo", first_name: "John", last_name: "Smith"}, 
       {prize: "Foo", first_name: "Mary", last_name: "Smith"},
       {prize: "Foo", first_name: "Judy", last_name: "Alvarez"}
      ]
     },
  1: {
      prize: "Bar", 
      winners: [
       {prize: "Bar", first_name: "Jane", last_name: "Doe"}, 
       {prize: "Bar", first_name: "Jack", last_name: "Jones"}
      ]
     },
]

我将如何修改我的代码以有效地实现我需要的格式？我什至不完全确定我是否走在正确的轨道上，也许代码需要完全改变？

这似乎是一个很常见的问题，之前可能已经回答过，但是因为我什至很难描述我想要的东西，所以我找不到任何东西。如果这是重复的，我很抱歉。

Answer 1

您可以在单个循环中按prize 解构对象和分组。

const
    data = [{ prize: "Foo", first_name: "John", last_name: "Smith" }, { prize: "Foo", first_name: "Mary", last_name: "Smith" }, { prize: "Bar", first_name: "Jane", last_name: "Doe" }, { prize: "Bar", first_name: "Jack", last_name: "Jones" }, { prize: "Foo", first_name: "Judy", last_name: "Alvarez" }],
    result =  data.reduce(
        (r, { prize, ...rest }) => ((r[prize] ??= []).push(rest), r),
        {}
    );

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 2

您可以使用for...in 遍历原始对象，并且由于您使用prize 属性作为新对象的新键，您可以将对象（将放入键的数组中）存储为名为t 的变量并将密钥存储为变量k。

 var k = obj[i].prize
 var t = obj[i]

请参阅下面的 sn-p 以获取工作示例。

var obj = {
  0: {prize: "Foo", first_name: "John", last_name: "Smith"},
  1: {prize: "Foo", first_name: "Mary", last_name: "Smith"},
  2: {prize: "Bar", first_name: "Jane", last_name: "Doe"},
  3: {prize: "Bar", first_name: "Jack", last_name: "Jones"},
  4: {prize: "Foo", first_name: "Judy", last_name: "Alvarez"},
}

const constructObj = () =>{
  var newObj = {}
  
  for (var i in obj){
    var k = obj[i].prize
    var t = obj[i]
    //remove below line to include all keys
    delete t.prize
    if (newObj[k]){
      var vals = newObj[k]
      vals.push(t)
    } else {
      newObj[k] = [t]
    }
  }
  return newObj
};

var final = constructObj()

console.log(final)