在javascript中包装一个函数答案

【问题标题】：wrap a function in javascript在javascript中包装一个函数
【发布时间】：2021-12-26 13:08:53
【问题描述】：

let str = "i am writing an algorithm.";

//function to count alphabets
const alphabet_count = (str) => str.length;

//function to count words
const word_count = (str) => str.split(" ").length;

//function to count vowel
const vowel_count = (str) => (str.match(/[aeiou]/gi)).length;

//here i am trying to wrap all three functions in one
const sentence_read() = {alphabet_count(), word_count(), vowel_count()};

我正在尝试将所有三个功能合二为一。

【问题讨论】：

标签： javascript string algorithm function

【解决方案1】：

const sentence_read = (str) => [alphabet_count(str), word_count(str), vowel_count(str)]

将返回一个包含 3 个结果的数组。用法：

let str = "a word";
console.log(sentence_read(str)) // output : [6, 2, 2]

【讨论】：

这行得通，谢谢...

【解决方案2】：

使用模板字符串

let str = "i am writing an algorithm.";

// function to count alphabets
const alphabet_count = (str) => str.length;

// function to count words
const word_count = (str) => str.split(" ").length;

//function to count vowel
const vowel_count = (str) => (str.match(/[aeiou]/gi)).length;

const sentence_read = (str) => `a_c : ${alphabet_count(str)}, w_c : ${word_count(str)}, v_c : ${vowel_count(str)}`

console.log(sentence_read(str)) // a_c : 26, w_c : 5, v_c : 8

如果你想对一个对象中的函数进行分组，你可以使用：

const str = "i am writing an algorithm.";

const counter = {
    alphabet: (s) => s.length,
    word: (s) => s.split(" ").length,
    vowel: (s) => (s.match(/[aeiou]/gi)).length
}

const count = (unit, str) => {
    if(!counter[unit]) throw Error('Unit does not exist')
    return counter[unit](str)
}

console.log(count('alphabet', str)) // 26
console.log(count('word', str))     // 5
console.log(count('vowel', str))    // 8

【讨论】：