我们如何使用 reduce 和 rest 运算符将以下数组转换为对象

Posted 2023-03-16

【中文标题】我们如何使用 reduce 和 rest 运算符将以下数组转换为对象

【英文标题】：How can we convert below arrays into objects using reduce and rest operator

【发布时间】：2021-12-24 02:55:02

【问题描述】：

谁能帮我将以下数组转换为对象：

let data = [
   name: "A", job: "soft", address: "ggn" ,
   name: "B", job: "software", address: "GGN" ,
   name: "C", job: "software", address: "GGN" ,
];

Output:  
“A”:job: “software”, address: “GGN”,
“B”:job: “software”, address: “GGN”,
“C”:job: “software”, address: “GGN”
;

这里我不想要对象的名称及其值。

我尝试使用以下解决方案：

const output = data.reduce((object, x) => 
    return  ...object, [x.name]: x ;
, );

但我得到了不同的答案。 有人可以通过提供最佳解决方案来提供帮助。我们可以使用 reduce 和 rest 运算符来解决它。

【参考方案1】：

你可以这样使用它：

let data = [
     name: "A", job: "soft", address: "ggn" ,
     name: "B", job: "software", address: "GGN" ,
     name: "C", job: "software", address: "GGN" ,
];

const output = data.reduce((acc, curr) => 
    const  name, ...rest  = curr;
    acc[name] = rest
    return acc;
, );

或者：

const output = data.reduce((acc, curr) => 
    const  name, ...rest  = curr;
    return  ...acc, [name]: rest ;
, );

【参考方案2】：

最简单的方法是首先对对象进行解构，这样您就可以将名称分开。然后你可以把它们和累加器一起返回。

const data = [
   name: "A",  job: "software",  address: "GGN"  ,
   name: "B",  job: "software",  address: "GGN"  ,
   name: "C",  job: "software",  address: "GGN"  ,
];

const output = data.reduce((acc, item) => 
  const  name, ...rest  = item;
  return 
    ...acc,
    [name]: rest,
  
, )

console.log(output)

【参考方案3】：

这里不需要reduce。

Object.fromEntries() 救援 - 您可以使用 .map 和解构将对象映射到对数组中，fromEntries 可以变成普通对象：

> let data = [
   name: "A", job: "soft", address: "ggn" ,
   name: "B", job: "software", address: "GGN" ,
   name: "C", job: "software", address: "GGN" ,
];

> Object.fromEntries(data.map((name, ...rest) => [name, rest]));

  A: job: 'soft', address: 'ggn'
  B: job: 'software', address: 'GGN',
  C: job: 'software', address: 'GGN'

【参考方案4】：

您可以将.map() 和.reduce() 与Object.assign() 结合使用，如下所示：

let output = data.map((name,...rest) => ([name]:rest))
.reduce((acc,cur) => Object.assign(acc,cur),);

演示

let data = [
name: "A", job: "soft", address: "ggn",
name: "B", job: "software", address: "GGN",
name: "C", job: "software", address: "GGN",
];

let output = data.map((name,...rest) => ([name]:rest))
    .reduce((acc,cur) => Object.assign(acc,cur),);

console.log( output );

或者，您可以使用.map() 和.reduce()，如下所示：

let output = data.map((name,...rest) => ([name]:rest))
.reduce((acc,cur) => (...acc,...cur),);

演示

let data = [
name: "A", job: "soft", address: "ggn",
name: "B", job: "software", address: "GGN",
name: "C", job: "software", address: "GGN",
];

let output = data.map((name,...rest) => ([name]:rest))
    .reduce((acc,cur) => (...acc,...cur),);

console.log( output );

或者，如果您的对象没有太多属性：

let output = data.reduce((acc,cur) => (...acc,...[cur.name]: job:cur.job,address:cur.address),);

演示

let data = [
name: "A", job: "soft", address: "ggn",
name: "B", job: "software", address: "GGN",
name: "C", job: "software", address: "GGN",
];

let output = data.reduce((acc,cur) => (...acc,...[cur.name]: job:cur.job,address:cur.address),);

console.log( output );

【参考方案5】：

let data = [
    name: 'A', job: 'soft', address: 'ggn',
    name: 'B', job: 'software', address: 'GGN',
    name: 'C', job: 'software', address: 'GGN',
];

const output = ;

const obj = data.reduce((acc, el) => 
    const  name, job, address  = el;
    return  ...acc, [name]:  job, address ;
, output);

console.log(obj);

【参考方案6】：

首先你需要改变双引号，它们不是标准的双引号 你可以用 map 方法用更短的代码编写它

let data = [
name: "A", job: "soft", address: "ggn",
name: "B", job: "software", address: "GGN",
name: "C", job: "software", address: "GGN",
];


const toto = data.map ( element =>  

const name = element.name 
delete element.name 
return [name] : element)

console.log(toto)

查看地图方法以了解更多信息：https://learnjsx.com/category/2/posts/es6-mapFunction

【参考方案7】：

你可以这样做

let data = [
    name: "A",
    job: "soft",
    address: "ggn"
  ,
  
    name: "B",
    job: "software",
    address: "GGN"
  ,
  
    name: "C",
    job: "software",
    address: "GGN"
  ,
];


const output = data.reduce((object, x) => 
  const 
    name,
    job,
    address
   = x;
  object[name] = 
    job,
    address
  

  return object;
, );
console.log(output)