Find 5 largest numbers in array of objects and push it to another array

Question

I have an array of objects and I want to find the five biggest numbers in object values and to push these five objects to the new array, but also remove them from the first array. I've tried using for loop combined with push and splice methods, but the result is kind of weird. Here is an example of what I've tried

const arrOfObjs = [
    {number: 1, bla: 'sadsad'},
  {number: 2, bla: 'sadsad'},
  {number: 3, bla: 'sadsad'},
  {number: 9, bla: 'sadsad'},
  {number: 4, bla: 'sadsad'},
  {number: 10, bla: 'sadsad'},
  {number: 5, bla: 'sadsad'},
  {number: 6, bla: 'sadsad'},
  {number: 7, bla: 'sadsad'},
  {number: 8, bla: 'sadsad'},
]

let largestNumbers = [];

for(var i=0; i<5; i++) {
  largestNumbers.push(
    arrOfObjs.splice(
      arrOfObjs.indexOf(
        Math.max.apply(
          Math,arrOfObjs.map(function(o){
            return o.number;
          })
        )
      ), 1)[0])
}

console.log(largestNumbers)

Everything seems to be working instead of the final result, which should be

[{
  bla: "sadsad",
  number: 8
 }, {
  bla: "sadsad",
  number: 7
 }, {
  bla: "sadsad",
  number: 6
 }, {
  bla: "sadsad",
  number: 9
 }, {
  bla: "sadsad",
  number: 10
}]

I'm not sure why number 9 isn't returned, and why the number 5 is. What am I doing wrong? Thanks in advance!

What about sorting first and then slicing?

evolutionxbox
– evolutionxbox

2021-01-13 17:15:40 +00:00
Commented Jan 13, 2021 at 17:15 — evolutionxbox
– evolutionxbox, Commented Jan 13, 2021 at 17:15
arrOfObjs.sort((a, b) => b.number - a.number).slice(0, 5)

Naren
– Naren

2021-01-13 17:17:33 +00:00
Commented Jan 13, 2021 at 17:17 — Naren
– Naren, Commented Jan 13, 2021 at 17:17

Gabriele Petrioli · Accepted Answer · 2021-01-13 17:34:10Z

The problem is that you do .indexOf on an array with objects using an integer as the parameter.

Try .findIndex(({number})=>number === Math.max.apply ...

const arrOfObjs = [
  {number: 1, bla: 'sadsad'},
  {number: 2, bla: 'sadsad'},
  {number: 3, bla: 'sadsad'},
  {number: 9, bla: 'sadsad'},
  {number: 4, bla: 'sadsad'},
  {number: 10, bla: 'sadsad'},
  {number: 5, bla: 'sadsad'},
  {number: 6, bla: 'sadsad'},
  {number: 7, bla: 'sadsad'},
  {number: 8, bla: 'sadsad'},
]
let largestNumbers = [];

for (var i = 0; i < 5; i++) {
  largestNumbers.push(
    arrOfObjs.splice(
      arrOfObjs.findIndex(({
          number
        }) => number ===
        Math.max.apply(
          Math, arrOfObjs.map(function(o) {
            return o.number;
          })
        )
      ), 1)[0])
}

console.log(largestNumbers)

trincot · Accepted Answer · 2021-01-13 20:48:05Z

1

The error is in the indexOf part. You will not find the maximum value, because your array has objects (with number property that should be compared instead). So replace

arrOfObjs.indexOf(Math.max ........)

with:

arrOfObjs.findIndex(o => o.number === max)

... where max is the Math.max expression, which you should then do separately, like this:

for(var i=0; i<5; i++) {
   let max = Math.max.apply(
      Math,
      arrOfObjs.map(o => o.number)
   );
   largestNumbers.push(arrOfObjs.splice(
      arrOfObjs.findIndex(o => o.number === max), 1)[0]
   );
}

edited Jan 13, 2021 at 20:48

answered Jan 13, 2021 at 17:22

trincot

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1 Comment

trincot Over a year ago

Please consider this answer, moving the Math.max calculation out of the findIndex callback, as you don't want to repeat the very same call over and over again as in the accepted answer... It makes it very inefficient.

ptothep · Accepted Answer · 2021-01-13 17:19:29Z

0

I would sort it and then slice it.

const arrOfObjs = [
    {number: 1, bla: 'sadsad'},
  {number: 2, bla: 'sadsad'},
  {number: 3, bla: 'sadsad'},
  {number: 9, bla: 'sadsad'},
  {number: 4, bla: 'sadsad'},
  {number: 10, bla: 'sadsad'},
  {number: 5, bla: 'sadsad'},
  {number: 6, bla: 'sadsad'},
  {number: 7, bla: 'sadsad'},
  {number: 8, bla: 'sadsad'},
]

console.log(arrOfObjs.sort((a,b)=> a.number - b.number).slice(-5))

answered Jan 13, 2021 at 17:19

ptothep

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