reduce

reduce 可以用来汇总数据

const customer = [
  {id: 1, count: 2},
  {id: 2, count: 89},
  {id: 3, count: 1}
];
const totalCount = customer.reduce((total, item) =>
  total + item.count,
  0 // total 的初始值
);
// now totalCount = 92

把一个对象数组变成一个以数组中各个对象的 id 为属性名，对象本身为属性值的对象。haoduoshipin

let products = [
  {
    id: '123',
    name: '苹果'
  },
  {
    id: '345',
    name: '橘子'
  }
];

const productsById = products.reduce(
  (obj, product) => {
    obj[product.id] = product
    return obj
  },
  {}
);

console.log('result', productsById);

map

map 可以理解为是数组的转换器，依次对数组中的每个元素做变换进而得到一个新的数组。

const integers = [1, 2, 3, 4, 6, 7];
const twoXIntegers = integers.map(i => i*2);
// twoXIntegers are now [2, 4, 6, 8, 12, 14]
// integers数组并不会受到影响

find🔥

筛选出数组中的个别元素

const posts = [
  {id: 1, title: 'Title 1'},
  {id: 2, title: 'Title 2'},
];
// find the title of post whose id is 1
const title = posts.find(p => p.id === 1).title;

唉~ 使用了半年的 es6才发现有这么好用的东西，译者傻缺还像下面这么写过呢

const posts = [
  {id: 1, title: 'Title 1'},
  {id: 2, title: 'Title 2'},
];

const title = posts.filter(item => item.id === 1)[0].title;

filter

筛选出数组中某些符合条件的元素组成新的数组

const integers = [1, 2, 3, 4, 6, 7];
const evenIntegers = integers.filter(i => i/2 === 0);
// evenIntegers are [2, 4, 6]

请大家自行思考下filter和find的区别

数组concat

const arr1 = [1, 2, 3, 4, 5];
const arr2 = [6, 7, 8, 9, 0];
const arrTarget = [...arr1, ...arr2];
// [1, 2, 3, 4, 5, 6, 7, 8, 9, 0]

对象操作

function operation(query, option = {}) {
    const param = {...query, ...option};
    // ....
    return param;
}
let opt = {startTime: 123455555, endTime: 113345555};
let q = {name: '一步', age: 'xxx'};
operation(q, opt);
// {name: "一步", age: "xxx", startTime: 123455555, endTime: 113345555}

对象是引用传参的，所以函数内部应该尽可能的保证传入的参数不受到污染。

为对象动态地添加字段

const dynamicKey = 'wearsSpectacles';
const user = {name: 'Shivek Khurana'};
const updatedUser = {...user, [dynamicKey]: true};
// updatedUser is {name: 'Shivek Khurana', wearsSpectacles: true}

将对象转换为query字符串👏

const params = {color: 'red', minPrice: 8000, maxPrice: 10000};
const query = '?' + Object.keys(params)
  .map(k =>
    encodeURIComponent(k) + '=' + encodeURIComponent(params[k])
  )
  .join('&')
;
// encodeURIComponent encodes special characters like spaces, hashes
// query is now "color=red&minPrice=8000&maxPrice=10000"

得到对象数组的元素 index

const posts = [
  {id: 13, title: 'Title 221'},
  {id: 5, title: 'Title 102'},
  {id: 131, title: 'Title 18'},
  {id: 55, title: 'Title 234'}
];
// to find index of element with id 131
const requiredIndex = posts.map(p => p.id).indexOf(131);

更加优雅的写法呱呱呱提供：

const posts = [
  {id: 13, title: 'Title 221'},
  {id: 5, title: 'Title 102'},
  {id: 131, title: 'Title 18'},
  {id: 55, title: 'Title 234'}
];
const index = posts.findIndex(p => p.id === 131)

删除对象的某个字段

const user = {name: 'Shivek Khurana', age: 23, password: 'SantaCl@use'};
const userWithoutPassword = Object.keys(user)
  .filter(key => key !== 'password')
  .map(key => {[key]: user[key]})
  .reducer((accumulator, current) =>
    ({...accumulator, ...current}),
    {}
  )
;
// userWithoutPassword becomes {name: 'Shivek Khurana', age: 23}

这里我认为原作者有点为了函数式编程而函数式了，下面是我的解决方案：

const user = {name: 'Shivek Khurana', age: 23, password: 'SantaCl@use'};
const newUser = {...user};
delete newUser.password;
// {name: "Shivek Khurana", age: 23}

以上代码片段的共同原则：不改变原数据。希望大家的代码都可以尽可能的简洁，可维护🙏。