0

我怎样才能array.reduce改变数组内容的方式。我不想对内容做任何数学运算。

原始数组：

var myArray = [
    {id:1, name:'name01', value:11},
    {id:2, name:'name02', value:22},
    {id:3, name:'name03', value:33},
    {id:4, name:'name04', value:44},
    {id:5, name:'name05', value:55}
]

我希望将其更改为：

[
    {1:{id:1, name:'name01', value:11}},
    {2:{id:2, name:'name02', value:22}},
    {3:{id:3, name:'name03', value:33}},
    {4:{id:4, name:'name04', value:44}},
    {5:{id:5, name:'name05', value:55}}
]

所以 the是从as aid中弹出的，而它是整个.objectkeyvalueobject

这可以通过array.reduce不使用任何for loopor来实现groupBy吗？

4

4 回答 4

6

您最好使用Array#map，因为您需要为数组中的每个项目创建一个新对象。

var array = [{ id: 1, name: 'name01', value: 11 }, { id: 2, name: 'name02', value: 22 }, { id: 3, name: 'name03', value: 33 }, { id: 4, name: 'name04', value: 44 }, { id: 5, name: 'name05', value: 55 }],
    result = array.map(a => ({ [a.id]: a }));

console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }

想要的减少样式，我不推荐。

var array = [{ id: 1, name: 'name01', value: 11 }, { id: 2, name: 'name02', value: 22 }, { id: 3, name: 'name03', value: 33 }, { id: 4, name: 'name04', value: 44 }, { id: 5, name: 'name05', value: 55 }],
    result = array.reduce((r, a) => (r.push({ [a.id]: a }), r), []);

console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }

使用 concat，我也不推荐。

var array = [{ id: 1, name: 'name01', value: 11 }, { id: 2, name: 'name02', value: 22 }, { id: 3, name: 'name03', value: 33 }, { id: 4, name: 'name04', value: 44 }, { id: 5, name: 'name05', value: 55 }],
    result = array.reduce((r, a) => r.concat({ [a.id]: a }), []);

console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }

于 2017-09-19T17:38:24.213 回答
2

您可以使用.map()ES6 计算的属性名称来代替，如下所示：

var myArray = [
    {id:1, name:'name01', value:11},
    {id:2, name:'name02', value:22},
    {id:3, name:'name03', value:33},
    {id:4, name:'name04', value:44},
    {id:5, name:'name05', value:55}
];

var result = myArray.map((elem) => ({[elem.id]: elem}));

console.log(result);

或者，如果您坚持使用.reduce()：

var myArray = [
    {id:1, name:'name01', value:11},
    {id:2, name:'name02', value:22},
    {id:3, name:'name03', value:33},
    {id:4, name:'name04', value:44},
    {id:5, name:'name05', value:55}
];

var result = myArray.reduce((res, curr) => res.concat({[curr.id]: curr}), []);

console.log(result);

或者，ES5 语法（为了清楚起见，因为您正在尝试学习）

var result = myArray.reduce(function (res, curr) {
    var newObj = {};
    newObj[curr.id] = curr;
    return res.concat(newObj);
}, []);
于 2017-09-19T17:39:16.683 回答
1 

var myArray = [
    {id:1, name:'name01', value:11},
    {id:2, name:'name02', value:22},
    {id:3, name:'name03', value:33},
    {id:4, name:'name04', value:44},
    {id:5, name:'name05', value:55}
];

let result = myArray.map(x => ({[x.id]: x }));
console.log(result)

于 2017-09-19T17:40:09.613 回答
1

你没有减少任何东西，所以尝试 array.reduce 也没有意义。

您可以array.map非常轻松地使用和执行此操作。

var myArray = [
{id:1, name:'name01', value:11},
{id:2, name:'name02', value:22},
{id:3, name:'name03', value:33},
{id:4, name:'name04', value:44},
{id:5, name:'name05', value:55}
]

var newArray = myArray.map((function(val) { return { [val.id]: val }}));
console.log(newArray);

于 2017-09-19T17:49:19.527 回答