概述
Array.map() 方法返回一个新数组，数组中的元素为原始数组元素调用函数处理后的值，同时不会改变原来的数组。

用法

Array.map(callback);

示例

//简单数组
const arr = [1, 3, 4, 5, 6, 7, 8, 10];

const cube = (num) => {
return num * num;
}

const res = arr.map(cube);//[ 1, 9, 16, 25, 36, 49, 64, 100 ]

// or

const res = arr.map((num)=>{
return num * num;
})
//[ 1, 9, 16, 25, 36, 49, 64, 100 ]

//对象数组和构造器
const OjbArray = [{
name: 'a',
age: 18,
isLikeEat: true,
isLikeSleep: true
}, {
name: 'b',
age: 19,
isLikeEat: true,
isLikeSleep: true
}, {
name: 'c',
age: 22,
isLikeEat: true,
isLikeSleep: true
}];

const Person = (target) => {
return {
name: target.name || 'default',
age: target.age || 18,
_eat: function() {
console.log(target.isLikeEat ? 'i like eat' : 'i dont like eat');
},
_sleep: function() {
console.log(target.isLikeSleep ? 'i like sleep' : 'i dont like sleep')
}
}
}

let newPersons = OjbArray.map(Person);
console.log((newPerso// [ { name: 'a// age: 18,

// _eat: [Function: _eat],
// _sleep: [Function: _sleep] },
// { name: 'b',
// age: 19,
// _eat: [Function: _eat],
// _sleep: [Function: _sleep] },
// { name: 'c',
// age: 22,
// _eat: [Function: _eat],
// _sleep: [Function: _sleep] } ]

注意 ⚠️

当和parseInt()函数配合使用 将字符转成数字的时候，可直接

['1', '2'].map(str => parseInt(str));
//or
['1', '2'].map(Number)

本质上是用元素作为函数参数去调用函数，所以无需加上参数

//这种写法是错误的
const arr = [1, 3, 4, 5, 6, 7, 8, 10];

const cube = (num) => {
return num * num;
}

const res = arr.map(cube(num))

原文链接：https://blog.csdn.net/ruffaim/article/details/82260280