快速和肮脏的#2:

var i = 0
  , s = {}
  , a = ['A', 'B', 'C'];

while( i < a.length ) { s[i] = a[i++] };

如果你使用ES6，你可以使用Object。赋值运算符和展开运算符

{ ...['a', 'b', 'c'] }

如果你有嵌套数组

var arr=[[1,2,3,4]]
Object.assign(...arr.map(d => ({[d[0]]: d[1]})))

这不是直接相关的，但我来这里寻找一个合并嵌套对象如一行

const nodes = {
    node1: {
        interfaces: {if1: {}, if2: {}}
    },
    node2: {
        interfaces: {if3: {}, if4: {}}
    },
    node3: {
        interfaces: {if5: {}, if6: {}}
    },
}

解决方案是结合使用reduce和对象扩展:

const allInterfaces = nodes => Object.keys(nodes).reduce((res, key) => ({...res, ...nodes[key].interfaces}), {})

快速和肮脏的#2:

var i = 0
  , s = {}
  , a = ['A', 'B', 'C'];

while( i < a.length ) { s[i] = a[i++] };

import books from "./books.json";

export const getAllBooks = () => {
  return {
    data: books,
    // a=accoumulator, b=book (data itelf), i=index
    bookMap: books.reduce((a, book, i) => {
      // since we passed {} as initial data, initially a={}
      // {bookID1:book1, bookID2:i}
      a[book.id] = book;
      // you can add new property index
      a[book.id].index=i
      return a;
      // we are passing initial data structure
    }, {}),
  };
};

最短的答案:(使用解构)

const obj = { ...input }

例子:

const inputArray = ["a", "b", "c"]
const outputObj = { ...inputArray }