JavaScript中Reduce10個(gè)常用場(chǎng)景技巧

更新時(shí)間：2022年06月16日 14:28:06 作者：掘金安東尼

這篇文章主要為大家介紹了JavaScript中Reduce10個(gè)常用場(chǎng)景和技巧，有需要的朋友可以借鑒參考下，希望能夠有所幫助，祝大家多多進(jìn)步，早日升職加薪

累加/累積

累加我們可能是最熟悉 Reduce 的一種用法，除此之外，還可以用做累積。

// adder
const sum = (...nums) => {
  return nums.reduce((sum, num) => sum + num);
};
console.log(sum(1, 2, 3, 4, 10)); // 20
// accumulator
const 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，無(wú)可厚非，Reduce 也能實(shí)現(xiàn)同樣的效果。

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

格式化搜索參數(shù)

獲取 url 上的參數(shù)是我們經(jīng)常面臨的需求，用 forEach 遍歷可以，用 Reduce 累加更可以，這樣可以減少聲明 query 對(duì)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;
    }, {});
};

反序列化搜索參數(shù)

有了獲取 url 參數(shù)，就有把參數(shù)重新掛在到 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

拉平嵌套數(shù)組

我們都會(huì)用 .flat(Infinity) 無(wú)限拉平所有多維數(shù)組成一維數(shù)組，只用 reduce 和 flat 也是可以做到這一點(diǎn)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]

實(shí)現(xiàn) flat

如果想實(shí)現(xiàn) flat，用 reduce 沒(méi)錯(cuò)了，又是一個(gè)手寫原生 api 內(nèi)部實(shí)現(xiàn)，妥妥的剛。

// Expand one layer by default
Array.prototype.flat2 = function (n = 1) {
  const len = this.length
  let count = 0
  let current = this
  if (!len || n === 0) {
    return current
  }
  // Confirm whether there are array items in current
  const hasArray = () => current.some((it) => Array.isArray(it))
  // Expand one layer after each cycle
  while (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 layer
console.log(array.flat()) // [ 1, 2, [ 3, [ 4, [ 5 ] ] ] ] 
console.log(array.flat2()) // [ 1, 2, [ 3, [ 4, [ 5 ] ] ] ] 
// Expand all
console.log(array.flat(Infinity))
console.log(array.flat2(Infinity))

數(shù)組去重

數(shù)組去重，用 reduce 竟然也可以，寫法如下：

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 ], [])

數(shù)組計(jì)數(shù)

將數(shù)組的項(xiàng)進(jìn)行計(jì)數(shù)，返回一個(gè) map，分別是每個(gè)項(xiàng)重復(fù)的次數(shù)，reduce 一行代碼搞定，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, …}

獲取對(duì)象多個(gè)屬性

獲取對(duì)象的多個(gè)屬性，然后賦給新的對(duì)象，比較笨的做法如下：

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

用 Reduce 這樣解決，就顯得明智了許多：

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)

反轉(zhuǎn)字符串

除了 reverse 做數(shù)組的翻轉(zhuǎn)，Reduce 也可以，再加上 split，就可以反轉(zhuǎn)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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