摘要：看完本文，希望可以寫出更加漂亮簡潔函數式的代碼可以用來匯總數據的初始值把一個對象數組變成一個以數組中各個對象的為屬性名，對象本身為屬性值的對象。

Handle javascript data structures with map/reduce

看完本文，希望可以寫出更加漂亮、簡潔、函數式的代碼?

reduce 可以用來匯總數據

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

把一個對象數組變成一個以數組中各個對象的 id 為屬性名，對象本身為屬性值的對象。haoduoshipin

let products = [
  {
    id: "123",
    name: "蘋果"
  },
  {
    id: "345",
    name: "橘子"
  }
];

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

console.log("result", productsById);

map 可以理解為是數組的轉換器，依次對數組中的每個元素做變換進而得到一個新的數組。

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數組并不會受到影響

篩選出數組中的個別元素

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才發現有這么好用的東西，譯者傻缺還像下面這么寫過呢

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

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

篩選出數組中某些符合條件的元素組成新的數組

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

請大家自行思考下filter和find的區別

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]

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}

對象是引用傳參的，所以函數內部應該盡可能的保證傳入的參數不受到污染。

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

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"

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);

更加優雅的寫法呱呱呱提供：

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)

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

這里我認為原作者有點為了函數式編程而函數式了，下面是我的解決方案：

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

更現代的寫法YiHzo提供: ?????

const user = {name: "Shivek Khurana", age: 23, password: "SantaCl@use"};
// 利用對象的解構，取出非password的所有字段
const {password, ...newUser} = user

以上代碼片段的共同原則：不改變原數據。希望大家的代碼都可以盡可能的簡潔，可維護?。

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