最好的转换方式是什么:
['a','b','c']
to:
{
0: 'a',
1: 'b',
2: 'c'
}
最好的转换方式是什么:
['a','b','c']
to:
{
0: 'a',
1: 'b',
2: 'c'
}
当前回答
ES5 -解决方案:
使用数组原型函数“push”和“apply”,你可以用数组元素填充对象。
Var arr = ['a','b','c']; var obj = new Object(); Array.prototype.push。应用(obj, arr); console.log (obj);// {'0': 'a', '1': 'b', '2': 'c', length: 3} console.log (obj [2]);/ / c
其他回答
我曾多次遇到过这个问题,并决定编写一个尽可能通用的函数。看一看,可以随意修改吗
function typeOf(obj) {
if ( typeof(obj) == 'object' ) {
if (obj.length)
return 'array';
else
return 'object';
} else
return typeof(obj);
}
function objToArray(obj, ignoreKeys) {
var arr = [];
if (typeOf(obj) == 'object') {
for (var key in obj) {
if (typeOf(obj[key]) == 'object') {
if (ignoreKeys)
arr.push(objToArray(obj[key],ignoreKeys));
else
arr.push([key,objToArray(obj[key])]);
}
else
arr.push(obj[key]);
}
}else if (typeOf(obj) == 'array') {
for (var key=0;key<obj.length;key++) {
if (typeOf(obj[key]) == 'object')
arr.push(objToArray(obj[key]));
else
arr.push(obj[key]);
}
}
return arr;
}
如果你正在使用jquery:
$.extend({}, ['a', 'b', 'c']);
更面向对象的方法:
Array.prototype.toObject = function() {
var Obj={};
for(var i in this) {
if(typeof this[i] != "function") {
//Logic here
Obj[i]=this[i];
}
}
return Obj;
}
.reduce((o,v,i)=>(o[i]=v,o), {})
(文档)
或者更冗长
var trAr2Obj = function (arr) {return arr.reduce((o,v,i)=>(o[i]=v,o), {});}
or
var transposeAr2Obj = arr=>arr.reduce((o,v,i)=>(o[i]=v,o), {})
最短的一个香草JS
JSON.stringify([["a", "X"], ["b", "Y"]].reduce((o,v,i)=>{return o[i]=v,o}, {}))
=> "{"0":["a","X"],"1":["b","Y"]}"
更复杂的例子
[["a", "X"], ["b", "Y"]].reduce((o,v,i)=>{return o[v[0]]=v.slice(1)[0],o}, {})
=> Object {a: "X", b: "Y"}
甚至更短(通过使用函数(e) {console.log(e);} === (e)=>(console.log(e),e))
nodejs
> [[1, 2, 3], [3,4,5]].reduce((o,v,i)=>(o[v[0]]=v.slice(1),o), {})
{ '1': [ 2, 3 ], '3': [ 4, 5 ] }
[/ docs]
令I = 0; let myArray = ["first", "second", "third", "fourth"]; const arrayToObject = (arr) => 对象。分配(arr{},……。Map (item => ({[i++]: item}))); console.log (arrayToObject (myArray));
或使用
myArray = ["first", "second", "third", "fourth"] console.log (myArray{…})