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

256
Vistas
How to group object by key names that share a part of a name and convert it into array of objects?

I'm struggling to properly convert object into a grouped array of objects. Any suggestions how to get the expected results? The most important part is to group it by HIGH_PRICE and LOW_PRICE and create arrays of those grouped keys and values but it would be nice to additionally include name from what's left from similar part of the key for example if I have HIGH_PRICE_QQQ and LOW_PRICE_QQQ the name would be QQQ.

Example of the object to sort: 

const obj = {
  HIGH_PRICE_QQQ: "10000",
  HIGH_PRICE_WWW: "2000",
  LOW_PRICE_WWW: "200",
  HIGH_PRICE_EEE: "3000",
  LOW_PRICE_EEE: "300",
}
Expected result: 

const sorted = [
  {
    name: "QQQ",
    HIGH_PRICE_QQQ: "10000",
  },
  {
    name: "WWW",
    HIGH_PRICE_WWW: "2000",
    LOW_PRICE_WWW: "200",
  },
  {
    name: "EEE",
    HIGH_PRICE_EEE: "3000",
    LOW_PRICE_EEE: "300",
  }
]
about 4 years ago · Juan Pablo Isaza
2 Respuestas
Responde la pregunta

0

you can use reduce to group by the name. Finally to get the array format as you wanted use Object.values on the result.
Object.entries used to make obj iterable
split and pop used to extract the name which is last element of splitted array

const obj = {
  HIGH_PRICE_QQQ: "10000",
  HIGH_PRICE_WWW: "2000",
  LOW_PRICE_WWW: "200",
  HIGH_PRICE_EEE: "3000",
  LOW_PRICE_EEE: "300",
}

let x = Object.values(Object.entries(obj).reduce((acc,[k,v])=> {
    let name = k.split("_").pop()
  acc[name]= acc[name] || {name}
  acc[name][k] = v
  return acc
},{}))

console.log(x)

about 4 years ago · Juan Pablo Isaza Denunciar

0

The below may be one possible solution to achieve the desired objective.

Code Snippet

// helper method to extract the "name"
const getName = s => (
  s.includes("HIGH_PRICE_")
  ? s.split("HIGH_PRICE_").at(1)
  : s.includes("LOW_PRICE_")
    ? s.split("LOW_PRICE_").at(1)
    : s
);

// group the object by name
const groupByName = obj => (
  Object.values(            // extract only values of below result
    Object.entries(obj)     // extract key-value pairs from argument "obj"
    .reduce(                // iterate over key-values to generate intermediate result
      (fin, [k, v]) => ({   // "fin" is the accumulator / aggregator
        ...fin,
        [getName(k)]: (     // "name" is used as the key, and value is constructed
          [getName(k)] in fin
          ? {...fin[getName(k)], [k]: v}
          : { name: getName(k), [k]: v}
        )
      }),
      {}                    // "fin" is being set to empty-object at begining
    )
  )
);

const rawObj = {
  HIGH_PRICE_QQQ: "10000",
  HIGH_PRICE_WWW: "2000",
  LOW_PRICE_WWW: "200",
  HIGH_PRICE_EEE: "3000",
  LOW_PRICE_EEE: "300",
};

console.log(groupByName(rawObj));
.as-console-wrapper { max-height: 100% !important; top: 0; }

Explanation

Inline comments in the above snippet provide description of the various steps.

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