为了完整起见，这里有一个O(1) ES2015方法。

var arr = [1, 2, 3, 4, 5]; // array, already an object
Object.setPrototypeOf(arr, Object.prototype); // now no longer an array, still an object

这是我刚刚写的递归函数。这很简单，而且效果很好。

// Convert array to object
var convArrToObj = function(array){
    var thisEleObj = new Object();
    if(typeof array == "object"){
        for(var i in array){
            var thisEle = convArrToObj(array[i]);
            thisEleObj[i] = thisEle;
        }
    }else {
        thisEleObj = array;
    }
    return thisEleObj;
}

下面是一个例子(jsFiddle):

var array = new Array();
array.a = 123;
array.b = 234;
array.c = 345;
var array2 = new Array();
array2.a = 321;
array2.b = 432;
array2.c = 543;
var array3 = new Array();
array3.a = 132;
array3.b = 243;
array3.c = 354;
var array4 = new Array();
array4.a = 312;
array4.b = 423;
array4.c = 534;
var array5 = new Array();
array5.a = 112;
array5.b = 223;
array5.c = 334;

array.d = array2;
array4.d = array5;
array3.d = array4;
array.e = array3;


console.log(array);

// Convert array to object
var convArrToObj = function(array){
    var thisEleObj = new Object();
    if(typeof array == "object"){
        for(var i in array){
            var thisEle = convArrToObj(array[i]);
            thisEleObj[i] = thisEle;
        }
    }else {
        thisEleObj = array;
    }
    return thisEleObj;
}
console.log(convArrToObj(array));

结果:

如果你正在使用jquery:

$.extend({}, ['a', 'b', 'c']);

最短的答案:(使用解构)

const obj = { ...input }

例子:

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

更多的浏览器支持和更灵活的方法是使用一个正常的循环，比如:

const arr = ['a', 'b', 'c'],
obj = {};

for (let i=0; i<arr.length; i++) {
   obj[i] = arr[i];
}

但现代的方法也可以使用展开运算符，比如:

{...arr}

或对象赋值:

Object.assign({}, ['a', 'b', 'c']);

两者都会返回:

{0: "a", 1: "b", 2: "c"}

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}), {})