Business
Jobs
  • About Us
  • Solutions
    • Job Postings
      Post your job and receive qualified candidates in 48h.
    • Candidate Assessments
      500+ technical and psychological tests, plus anti-fraud.
    • Headhunting
      Tailor-made executive search from start to finish.
    • Payroll + EOR
      Payroll dispersal and EOR across 15+ LATAM countries.
  • Pricing
  • Jobs

0

165
Views
Sume los valores de propiedades específicas de los objetos en la matriz si tienen el mismo nombre de propiedad y devuelven una matriz única como resultado usando el script Java

mi problema es mapear dos matrices de objetos y sumar sus propiedades específicas si tienen el mismo valor de nombre.

Por ejemplo, si me gustaría sumar x e y de todos los objetos donde el nombre de la propiedad es 'v', etc. Aquí está mi código

 var arr = [{name:'v', x:1, b:2, c:3},{name:'r', x:2, b:0, c:3},{name:'v', x:4, b:3, c:3}, {name:'v', x:1, b:1, c:3}]; let arr2 = [] let obj = {name:null, x:null, b: null, c:null} arr.map(item => { for(let i=0; i<= arr.length; i++){ if(item.name === arr[i].name){ let a = arr.reduce((a, b) => ({x: ax + bx, b: ab + bb})); obj.name = item.name, obj.x = ax, obj.b = ab, obj.c = item.c } else { obj.name = item.name, obj.x=item.x, obj.b=item.b, obj.c=item.c } arr2.push(obj) } } ) console.log(arr2)

Como resultado, me gustaría devolver una matriz como esta

 arr2 = [{name: 'v', x: 6, b: 6, c: 3}, {name: 'r', x:2, b:0, c:1}]
about 4 years ago · Juan Pablo Isaza
2 answers
Answer question

0

const sumProperties = (arr) => { const result = []; const map = new Map(); for (const obj of arr) { const key = obj.name; if (!map.has(key)) { map.set(key, obj); } else { const oldObj = map.get(key); const newObj = { ...oldObj, ...obj }; map.set(key, newObj); } } map.forEach((value) => { result.push(value); }); return result; } console.log(sumProperties(arr));
about 4 years ago · Juan Pablo Isaza Report

0

Aquí está la respuesta a mi pregunta.

 var arr = [{v:'v', x:1, b:2, c:3},{v:'r', x:2, b:0, c:3},{v:'v', x:4, b:3, c:3},{v:'v', x:10, b:3, c:3},{v:'r', x:1, b:1, c:3},{v:'v', x:11, b:2, c:3},{v:'r', x:22, b:0, c:3}]; let arr2 = [] for(let j=0; j< arr.length;j++){ for(let i=j+1; i< arr.length; i++){ let obj = {v:null, x:null, b: null, c:null} if(arr[j].v === arr[i].v ){ let a = arr.filter(({v})=> v === arr[j].v).reduce((a, b) => ({x: ax + bx, b: ab + bb})); obj.v = arr[j].v obj.x = ax obj.b = ab obj.c = arr[j].ci= arr.length console.log("DA", obj) if(!arr2.some(item=> item.v === obj.v)){ arr2.push(obj) } } else if(i+1 === arr.length){ obj.v = arr[j].v obj.x = arr[j].x obj.b = arr[j].b obj.c = arr[j].c console.log("NE", obj) if(!arr2.some(item=> item.v === obj.v)){ arr2.push(obj) } } } } console.log(arr2, "REZ")
about 4 years ago · Juan Pablo Isaza Report
Answer question
Find remote jobs

Discover the new way to find a job!

Top jobs
Top job categories
Business
Post vacancy Pricing Sales
Legal
Terms and conditions Privacy policy
© 2026 PeakU Inc. All Rights Reserved.
Andres GPT
Show me some job opportunities
There's an error!