How do I check for vowels in JavaScript?

Question

I\'m supposed to write a function that takes a character (i.e. a string of length 1) and returns true if it is a vowel, false otherwise. I came up with two functions, but do

Answer 1

Personally, I would define it this way:

function isVowel( chr ){ return 'aeiou'.indexOf( chr[0].toLowerCase() ) !== -1 }

You could also use ['a','e','i','o','u'] and skip the length test, but then you are creating an array each time you call the function. (There are ways of mimicking this via closures, but those are a bit obscure to read)

Answer 2

I kind of like this method which I think covers all the bases:

const matches = str.match(/aeiou/gi];
return matches ? matches.length : 0;

Answer 3

I created a simplified version using Array.prototype.includes(). My technique is similar to @Kunle Babatunde.

const isVowel = (char) => ["a", "e", "i", "o", "u"].includes(char);

console.log(isVowel("o"), isVowel("s"));

Answer 4

This is a rough RegExp function I would have come up with (it's untested)

function isVowel(char) {
    return /^[aeiou]$/.test(char.toLowerCase());
}

Which means, if (char.length == 1 && 'aeiou' is contained in char.toLowerCase()) then return true.

Answer 5

benchmark

I think you can safely say a for loop is faster.

I do admit that a regexp looks cleaner in terms of code. If it's a real bottleneck then use a for loop, otherwise stick with the regular expression for reasons of "elegance"

If you want to go for simplicity then just use

function isVowel(c) {
    return ['a', 'e', 'i', 'o', 'u'].indexOf(c.toLowerCase()) !== -1
}

Answer 6

cycles, arrays, regexp... for what? It can be much quicker :)

function isVowel(char)
{
    return char === 'a' || char === 'e' || char === 'i' || char === 'o' || char === 'u' || false;
}