如何使用数组中的嵌套对象创建 Object

Question

我有一个数组[1, 2, 3] ，我想将它转移到 object ，嵌套父子对象的系列如下：

{ value: 1, rest: { value: 2, rest: { value: 3, rest: null } } 

如果我有一个数组[1, 2, 3, 4] ，结果将是这样的：

{ value: 1, rest: { value: 2, rest: { value: 3, rest: { value:4, rest:null } } 

我最大的努力是这段代码：

 const arrayToList = (array) => { let list = { value: null, rest: null }; for (let e of array) { array.indexOf(e) === 0 && (list.value = e); array.indexOf(e) >= 1 && (list.rest = { value: e }); } return list; }; console.log(arrayToList([1, 2, 3]));

Answer 1

您可以像这样使用reduceRight ：

let obj = arr.reduceRight((rest, value) => ({ value, rest }), null);

它开始从内到外构建 object； it starts by creating the innermost object and then it uses that object as the rest property for the next outer object and so on until there are no more items in the array.

演示：

 let obj = [1, 2, 3, 4].reduceRight((rest, value) => ({ value, rest }), null); console.log(obj);

Answer 2

您可以通过在递归 function 下面运行来创建这样的 object：

 let arr = [1, 2, 3, 4]; let transform = (arr, obj) => { if(arr.length === 0){ return obj; } else { let last = arr[arr.length - 1]; let newArr = arr.slice(0, arr.length - 1); return transform(newArr, { value: last, rest: obj || null }) } }; console.log(transform(arr));

Answer 3

使用递归 function：

 let array = [1, 2, 3]; function arrayToL(array) { let el = array.splice(0, 1)[0]; let rtn = { value: el } rtn.rest = (array.length > 0)? arrayToL(array): null; return rtn; } console.log(arrayToL(array));

Answer 4

我建议使用spread operator并反转数组并从数组末端开始构建 object 的另一种解决方案：

 let arr = [1, 2, 4, 5] let obj = {} //object to be built arr.slice().reverse().forEach(item => { //i used the slice method //in order to avoid mutating //the original variable obj = {...obj, ...{ value: item, rest: obj } }; }) console.log(obj)