How to change if/else statement to loop in javascript

Question

I would like to change if/else statement to loops in a function that receives a string and return to a new string which is changed the vowel letters to the next letter after it.

function changeVowels(str) {
        var alphabet = 'abefijopuvABEFIJOPUV';
        var strNew = ""
        for(var i = 0; i < str.length; i++){
           for(var j = 0; j < alphabet.length; j++){
             if(str[i] === alphabet[j] && j%2 === 0){
               strNew += alphabet[j+1]
            }
          }
          if(str[i]==='a' || str[i]==='i' || str[i]==='u'||str[i]==='e' || str[i]==='o'){}
          else if(str[i]==='A' || str[i]==='I' || str[i]==='U'|| str[i]==='E' || str[i]==='O'){}
          else{
           strNew+=str[i]
          }
        }
        return strNew
      }

console.log(changeVowels('Car'))

how to change if/else statement below to loops ?

if(str[i]==='a' || str[i]==='i' || str[i]==='u'||str[i]==='e' || str[i]==='o'){}
else if(str[i]==='A' || str[i]==='I' || str[i]==='U'|| str[i]==='E' || str[i]==='O'){}
else{strNew+=str[i]}

or i don't mind you change it besides loop, or make it shorter

not sure how you can turn ANY if/else to a loop - the two things are two different programming constructs — Jaromanda X
– Jaromanda X, Commented Sep 19, 2019 at 6:23
return str.split('').map(l => 'aeiou'.includes(l.toLowerCase()) ? String.fromCharCode(l.charCodeAt(0) + 1): l).join(''); — Jaromanda X
– Jaromanda X, Commented Sep 19, 2019 at 6:27

Nina Scholz · Accepted Answer · 2019-09-19 06:28:08Z

2

You could check if the character is a vowel and then take the next letter from a given string or the actual letter.

function changeVowels(str) {
  var alphabet = 'abefijopuvABEFIJOPUV';
  var strNew = ""
  for (var i = 0; i < str.length; i++) {
    if ('aeiou'.includes(str[i].toLowerCase())) {
      strNew += alphabet[alphabet.indexOf(str[i]) + 1];
    } else {
      strNew += str[i];
    }
  }
  return strNew;
}

console.log(changeVowels('Car'))

answered Sep 19, 2019 at 6:28

Nina Scholz

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Comments

Jaromanda X · Accepted Answer · 2019-09-19 06:49:14Z

Here's another fun way to do it

function changeVowels(str) {
    var x = {a:'b', e:'f', i:'j', o: 'p', u: 'v', A: 'B', E: 'F', I: 'J', O: 'P', U: 'V'};
    return str.split('').map(l => x[l] || l).join('');
}
console.log(changeVowels('Car'))

Or if you really want that alphabet var

function changeVowels(str) {
    var alphabet = 'abefijopuvABEFIJOPUV';
    var x = Object.fromEntries(alphabet.match(/../g).map(kv => kv.split('')));
    return str.split('').map(l => x[l] || l).join('');
}
console.log(changeVowels('Car'))

Explanation of var x = Object.fromEntries(alphabet.match(/../g).map(kv => kv.split('')));

alphabet.match(/../g) produces an array ['ab', 'ef', 'ij', ... etc]

.map(kv => kv.split('')) maps the above to [['a', 'b'], ['e', 'f'], ... etc

Object.fromEntries([['a', 'b'], ['e', 'f'] .... etc]) produces

{
    a: 'b',
    e: 'f',
    i: 'j',
    o: 'p',
    u: 'v',
    A: 'B',
    E: 'F',
    I: 'J',
    O: 'P',
    U: 'V'
};

same as the first snippet

VLAZ · Accepted Answer · 2019-09-19 06:30:41Z

0

You don't need loops to simplify this code

Both str[i]==='a' || str[i]==='i' || str[i]==='u'||str[i]==='e' || str[i]==='o' and str[i]==='A' || str[i]==='I' || str[i]==='U'|| str[i]==='E' || str[i]==='O' are about the same letters but with different capitalisation, so you can reduce your code in half by only considering one case (upper or lower) and converting anything to that. If you extract str[i] as a variable, you can further reduce the repetition:

var char = str[i].toUpperCase();
/* ... */
if (char==='A' || char==='I' || char==='U'|| char==='E' || char==='O') {}

you can further reduce the multiple ORs if you use an array and Array#includes alternatively String#includes will do the same but I'll use an array since it's a bit more universal for values, even if a string would be shorter.

if (['A', 'I', 'U', 'E', 'O'].includes(char)) {
  /* code */
} else {
  strNew+=str[i];
}

and if you want to do nothing in the first case, you can further just eliminate the else by inverting the condition:

if (!['A', 'I', 'U', 'E', 'O'].includes(char)) {
//  ^ invert the result
  strNew+=str[i];
}

edited Sep 19, 2019 at 6:30

answered Sep 19, 2019 at 6:29

VLAZ

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2 Comments

Jaromanda X Over a year ago

strings also have 'includes' method

VLAZ Over a year ago

@JaromandaX yeah, I just added a note for why I used an array. basically, more universal, you can also use it for numbers, for example.

thanthuong · Accepted Answer · 2019-09-19 07:00:24Z

0

I will make it simpler for you

function changeVowels(str) {
        var alphabet = 'abefijopuvABEFIJOPUV';
        for(var i = 0; i < alphabet.length; i++){
           str = str.replace(new RegExp(alphabet[i], 'g'), alphabet[i+1]);
           i++
        }
        return str
}

console.log(changeVowels('Facebook'))

answered Sep 19, 2019 at 7:00

thanthuong

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Comments

RamKumar · Accepted Answer · 2019-09-19 07:02:30Z

-1

separate your string and check if that exist in your alphabet variable. if it present means u get the next character or else return the same character and so on

var alphabet = 'abefijopuvABEFIJOPUV';

function changeVowels(str) {
const newArray = str.split('').map(element => {
  let index = alphabet.indexOf(element);
  if(index > -1) {
return alphabet.charAt(index + 1);
  } else {
return element;
  }
})
return newArray.join('');
}

console.log('final ---  ', changeVowels('Car'))

edited Sep 19, 2019 at 7:02

answered Sep 19, 2019 at 6:31

RamKumar

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