JavaScript中Reduce10个常用场景技巧 JavaScript中Reduce10个常用场景技

累加/累积
求最大/最小值
格式化搜索参数
反序列化搜索参数
拉平嵌套数组
实现 flat
数组去重
数组计数
获取对象多个属性
反转字符串

【JavaScript中Reduce10个常用场景技巧】不知道大家平常用 Reduce 多不多，反正本瓜用的不多。但实际上，Reduce 能做的，比我们能想到的要多得多，本篇带来 10 个Reduce 常用场景和技巧，一定有你不知道~
冲ヾ(?°?°?)??

累加/累积累加我们可能是最熟悉 Reduce 的一种用法，除此之外，还可以用做累积。

// adderconst sum = (...nums) => {return nums.reduce((sum, num) => sum + num); }; console.log(sum(1, 2, 3, 4, 10)); // 20// accumulatorconst accumulator = (...nums) => {return nums.reduce((acc, num) => acc * num); }; console.log(accumulator(1, 2, 3)); // 6

求最大/最小值如果你用原生 api 求最大/最小值，无可厚非，Reduce 也能实现同样的效果。

const array = [-1, 10, 6, 5]; const max = Math.max(...array); // 10const min = Math.min(...array); // -1

const array = [-1, 10, 6, 5]; const max = array.reduce((max, num) => (max > num ? max : num)); const min = array.reduce((min, num) => (min < num ? min : num));

格式化搜索参数获取 url 上的参数是我们经常面临的需求，用 forEach 遍历可以，用 Reduce 累加更可以，这样可以减少声明 query 对象。

// url https://qianlongo.github.io/vue-demos/dist/index.html?name=fatfish&age=100#/home// format the search parameters{"name": "fatfish","age": "100"}

const parseQuery = () => {const search = window.location.search; let query = {}; search.slice(1).split("&").forEach((it) => {const [key, value] = it.split("="); query[key] = decodeURIComponent(value); }); return query; };

const parseQuery = () => {const search = window.location.search; return search.slice(1).split("&").reduce((query, it) => {const [key, value] = it.split("="); query[key] = decodeURIComponent(value); return query; }, {}); };

反序列化搜索参数有了获取 url 参数，就有把参数重新挂在到 url 上面，好用，收藏。

const searchObj = {name: "fatfish",age: 100,// ...}; const link = `https://medium.com/?name=${searchObj.name}&age=${searchObj.age}`; // https://medium.com/?name=fatfish&age=100

const stringifySearch = (search = {}) => {return Object.entries(search).reduce((t, v) => `${t}${v[0]}=${encodeURIComponent(v[1])}&`,Object.keys(search).length ? "?" : "").replace(/&$/, ""); }; const search = stringifySearch({name: "fatfish",age: 100,}); const link = `https://medium.com/${search}`; console.log(link); // https://medium.com/?name=fatfish&age=100

拉平嵌套数组我们都会用 .flat(Infinity) 无限拉平所有多维数组成一维数组，只用 reduce 和 flat 也是可以做到这一点的。

const array = [1, [2, [3, [4, [5]]]]]; // expected output [ 1, 2, 3, 4, 5 ]const flatArray = array.flat(Infinity); // [1, 2, 3, 4, 5]

const flat = (array) => {return array.reduce((acc, it) => acc.concat(Array.isArray(it) ? flat(it) : it),[]); }; const array = [1, [2, [3, [4, [5]]]]]; const flatArray = flat(array); // [1, 2, 3, 4, 5]

实现 flat 如果想实现 flat，用 reduce 没错了，又是一个手写原生 api 内部实现，妥妥的刚。

// Expand one layer by defaultArray.prototype.flat2 = function (n = 1) {const len = this.lengthlet count = 0let current = thisif (!len || n === 0) {return current}// Confirm whether there are array items in currentconst hasArray = () => current.some((it) => Array.isArray(it))// Expand one layer after each cyclewhile (count++ < n && hasArray()) {current = current.reduce((result, it) => {result = result.concat(it)return result}, [])}return current}const array = [ 1, [ 2, [ 3, [ 4, [ 5 ] ] ] ] ]// Expand one layerconsole.log(array.flat()) // [ 1, 2, [ 3, [ 4, [ 5 ] ] ] ] console.log(array.flat2()) // [ 1, 2, [ 3, [ 4, [ 5 ] ] ] ] // Expand allconsole.log(array.flat(Infinity))console.log(array.flat2(Infinity))

数组去重数组去重，用 reduce 竟然也可以，写法如下：

const array = [ 1, 2, 1, 2, -1, 10, 11 ]const uniqueArray1 = [ ...new Set(array) ]const uniqueArray2 = array.reduce((acc, it) => acc.includes(it) ? acc: [ ...acc, it ], [])

数组计数将数组的项进行计数，返回一个 map，分别是每个项重复的次数，reduce 一行代码搞定，收藏！

const count = (array) => {return array.reduce((acc, it) => (acc.set(it, (acc.get(it) || 0) + 1), acc), new Map())}const array = [ 1, 2, 1, 2, -1, 0, '0', 10, '10' ]console.log(count(array)) // Map(7) {1 => 2, 2 => 2, -1 => 1, 0 => 1, '0' => 1, …}

获取对象多个属性获取对象的多个属性，然后赋给新的对象，比较笨的做法如下：

// There is an object with many propertiesconst obj = {a: 1,b: 2,c: 3,d: 4,e: 5// ...}// We just want to get some properties above it to create a new objectconst newObj = {a: obj.a,b: obj.b,c: obj.c,d: obj.d// ...}// Do you think this is too inefficient?

用 Reduce 这样解决，就显得明智了许多：

const getObjectKeys = (obj = {}, keys = []) => {return Object.keys(obj).reduce((acc, key) => (keys.includes(key) && (acc[key] = obj[key]), acc), {}); }const obj = {a: 1,b: 2,c: 3,d: 4,e: 5// ...}const newObj = getObjectKeys(obj, [ 'a', 'b', 'c', 'd' ])console.log(newObj)

反转字符串除了 reverse 做数组的翻转，Reduce 也可以，再加上 split，就可以反转字符串啦。

const reverseString = (string) => {return string.split("").reduceRight((acc, s) => acc + s)}const string = 'fatfish'console.log(reverseString(string)) // hsiftaf

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