如何在JavaScirpt中展开数组中的对象

Question

我正在尝试在JavaScript中扩展数组。object↓

const tests = [
    {
        id: 1,
        name: 'taro',
        designs: [
            {
                designId: 1,
                designName: "design1"
            },
            {
                designId: 2,
                designName: "design2"
            }
        ]
    },
    {
        id: 2,
        name: 'John',
        designs: [
            {
                designId: 3,
                designName: "design3"
            },
            {
                designId: 4,
                designName: "design4"
            }
        ]
    },
{
        id: 3,
        name: 'Lisa',
        designs: []
    },
];

[
  { id: 1, name: 'taro', designId: 1, designName: 'design1' },
  { id: 1, name: 'taro', designId: 2, designName: 'design2' },
  { id: 2, name: 'John', designId: 3, designName: 'design3' },
  { id: 2, name: 'John', designId: 4, designName: 'design4' },
  { id: 3, name: 'Lisa', designId: null, designName: null },
]

使用double for很容易做到这一点，但我想将其用于高阶函数。

我写的代码

for (let i = 0; i < tests.length; i++) {
    for (let j = 0; j < tests[i].designs.length; j++) {
        const id = tests[i].id
        const name = tests[i].name
        result.push({
            id,
            name,
            designId: tests[i].designs[j].designId,
            designName: tests[i].designs[j].designName
        })
    }
}

此外，如果您能另外解释double for函数和高阶函数之间的性能差异，将不胜感激。

Answer 1

您可以在测试数组上使用.flatMap()，每个designs数组上都有一个内部.map()。设计数组上的内部映射将从当前迭代的设计对象中获取属性，并将其与父对象中的属性合并。然后可以使用外部.flatMap()将所有返回的映射连接到一个数组中：

​

const tests = [ { id: 1, name: 'taro', designs: [ { designId: 1, designName: "design1" }, { designId: 2, designName: "design2" } ] }, { id: 2, name: 'John', designs: [ { designId: 3, designName: "design3" }, { designId: 4, designName: "design4" } ] }, ];

const res = tests.flatMap(({designs, ...rest}) => designs.map(design => ({
  ...rest,
  ...design
})));
console.log(res);

​

编辑:如果在设计数组为空的情况下需要为设计对象显示null值，则可以将键显式添加到新对象中，该对象可以在设计数组为空时返回：

​

const tests = [ { id: 1, name: 'taro', designs: [] }, { id: 2, name: 'John', designs: [] }, ];

const res = tests.flatMap(({designs, ...rest}) => 
  designs.length 
    ? designs.map(design => ({
       ...rest,
       ...design
      }))
    : {...rest, designId: null, designName: null}
);
console.log(res);

​

Answer 2

您可以使用带有Array.map的Array.reduce函数来生成数组：

const results = tests.reduce((acc, { designs, ...rest }) => [
  ...acc,
  ...designs.map(e => ({ ...rest, ...e }))
], []);

​

const tests = [
    {
        id: 1,
        name: 'taro',
        designs: [
            {
                designId: 1,
                designName: "design1"
            },
            {
                designId: 2,
                designName: "design2"
            }
        ]
    },
    {
        id: 2,
        name: 'John',
        designs: [
            {
                designId: 3,
                designName: "design3"
            },
            {
                designId: 4,
                designName: "design4"
            }
        ]
    },
];

const results = tests.reduce((acc, { designs, ...rest }) => [
  ...acc,
  ...designs.map(e => ({ ...rest, ...e }))
], []);

console.log(results);

​

Answer 3

您可以在Array.prototype.map()中使用高阶函数Array.prototype.reduce()

const newArr = tests.reduce((prev, {designs, ...current}) => [
   ...prev, ...designs.map(design => ({...design,...current}));
]      
, []);

您的方法和这种高阶方法的性能是相同的，因为Array.prototype.reduce遍历整个数组，只是简化了我们的initialValue方法。