Empresas
Empleos
  • Sobre nosotros
  • Soluciones
    • Publicación de vacantes
      Publica tu vacante y recibe candidatos calificados en 48h.
    • Evaluación de candidatos
      500+ pruebas técnicas y psicológicas, más anti-fraude.
    • Headhunting
      Búsqueda ejecutiva a la medida de principio a fin.
    • Nómina + EOR
      Dispersión de nómina y EOR en más de 15 países de LATAM.
  • Precios
  • Empleos

0

117
Vistas
Count the occurrence of every alphabet from a string in Javascript

I have seen similar questions like this asked before, such as counting characters in a given string. However when it comes to comparing given string to the letters of the alphabet and returning an object with occurences such as:

const letters = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]

const sampleString = "a bee";

const results = {
 a: 1,
 b: 1,
 c: 0,
 d: 0,
 e: 2,
 f: 0,
 ...
}
about 4 years ago · Juan Pablo Isaza
3 Respuestas
Responde la pregunta

0

We can use Array.reduce(), to count letters in the sampleString.

We start by creating a letterMap to specify all valid letters to be counted.

In the reduce loop, we only increment letters that are present in the letterMap, using the (c in acc) expression.

const letters = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]

const sampleString = "a bee";

const letterMap = letters.reduce((acc, c) => { 
    acc[c] = 0; 
    return acc; 
}, {});

const result = [...sampleString].reduce((acc, c) => {
    if (c in acc) acc[c]++;
    return acc;
}, letterMap);

console.log('Result:', result)
 
.as-console-wrapper { max-height: 100% !important; top: 0; }

Here's another way, using just one loop, again using Array.reduce(), this assumes we don't wish to count whitespace:

const letters = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]

const sampleString = "a bee";

const result = [...letters, ...sampleString].reduce((acc, c) => {
    if (c in acc) { 
        acc[c]++;
    } else if (c.trim()) {
        acc[c] = 0;
    }
    return acc;
}, {});

console.log('Result:', result)
.as-console-wrapper { max-height: 100% !important; top: 0; }

about 4 years ago · Juan Pablo Isaza Denunciar

0

I like using Object.fromEntries for this:

const sampleString = "a bee";
const result = Object.fromEntries(Array.from("abcdefghijklmnopqrstuvwxyz", ch => [ch, 0]));
for (let ch of sampleString) 
    if (ch in result) result[ch]++;
console.log(result);

about 4 years ago · Juan Pablo Isaza Denunciar

0

Using Array.reduce and String.match may be an idea. So, for each letter of letters, use match (length) to determine the frequency of the letter in the given sample.

const letters = `abcdefghijklmnopqrstuvwxyz`.split(``);
const freq = (chr, sample) => (sample.match(RegExp(chr, `g`)) || []).length;
const result = letters.reduce( (acc, chr) => 
  ({...acc, [chr]: freq(chr, acc.sample)}), {sample: "a bee"});
console.log(result);

about 4 years ago · Juan Pablo Isaza Denunciar
Responde la pregunta
Encuentra empleos remotos

¡Descubre la nueva forma de encontrar empleo!

Top de empleos
Top categorías de empleo
Empresas
Publicar vacante Precios Comercial
Legal
Términos y condiciones Política de privacidad
© 2026 PeakU Inc. All Rights Reserved.
Andres GPT
Recomiéndame algunas ofertas
Necesito ayuda