Sort an array of objects based on key/value of array

Question

I need to sort an array of objects. How can I sort this array based on key arrval[index]?

arr = [
  { id: 0, name: 'Name 1', arrval: [5, 3, 1] },
  { id: 1, name: 'Name 2', arrval: [6, 4, 3] },
  { id: 2, name: 'Name 3', arrval: [3, 2, 0] },
]

So if I want to sort it by arrval[1], ascending, the result should be:

arr = [
  { id: 2, name: 'Name 3', arrval: [3, 2, 0] },
  { id: 0, name: 'Name 1', arrval: [5, 3, 1] },
  { id: 1, name: 'Name 2', arrval: [6, 4, 3] },
]

I can already sort this if I sort it by id or name but I can't seem to make it work by arrval

If you know how to sort by id or name, what prevents you doing the same for the first index of arrval? — Yoshi
– Yoshi, Commented Sep 7, 2021 at 8:30
It is exactly same as you sort with id and name . arr.sort((a, b) => a.arrval[0] - b.arrval[0]) — DecPK
– DecPK, Commented Sep 7, 2021 at 8:31

Yevhen Horbunkov · Accepted Answer · 2021-09-07 09:01:12Z

1

Simple as that

const src = [
  { id: 0, name: 'Name 1', arrval: [5, 3, 1] },
  { id: 1, name: 'Name 2', arrval: [6, 4, 3] },
  { id: 2, name: 'Name 3', arrval: [3, 2, 0] },
]

const sortByKeyIdx = ([...arr], key, idx) =>
  arr
    .sort(({[key]:a}, {[key]:b}) => 
      a[idx]-b[idx])
    
console.log(sortByKeyIdx(src, 'arrval', 1))

edited Sep 7, 2021 at 9:01

answered Sep 7, 2021 at 8:31

Yevhen Horbunkov

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Comments

Фарид Ахмедов · Accepted Answer · 2021-09-07 08:31:43Z

0

Just sort by arrval like you sort by id or name

arr.sort( (a, b) => a.arrval[0] - b.arrval[0] )

answered Sep 7, 2021 at 8:31

Фарид Ахмедов

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Comments

Karma Blackshaw · Accepted Answer · 2021-09-07 08:44:37Z

0

Just pass the arrval index and sort it like you normally would. You can try this solution:

const sort = index => {
  const arr = [
    { id: 0, name: 'Name 1', arrval: [5, 3, 1] },
    { id: 1, name: 'Name 2', arrval: [6, 4, 3] },
    { id: 2, name: 'Name 3', arrval: [3, 2, 0] }
  ]

  return arr.sort((x, y) => x.arrval[index] - y.arrval[index])
}

console.log(sort(2))

edited Sep 7, 2021 at 8:44

answered Sep 7, 2021 at 8:31

Karma Blackshaw

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Comments

Sanket Shah · Accepted Answer · 2021-09-07 08:48:30Z

0

There is two way to do this

first:

let arr = [{
    id: 0,
    name: 'Name 1',
    arrval: [5, 3, 1]
  },
  {
    id: 1,
    name: 'Name 2',
    arrval: [6, 4, 3]
  },
  {
    id: 2,
    name: 'Name 3',
    arrval: [3, 2, 0]
  },
]
let index = 1
arr.sort((a, b) => {
  if (a.arrval[index] < b.arrval[index]) return -1
  else if (a.arrval[index] < b.arrval[index]) return 1
  else return 0
})

console.log(arr)

second:

let arr = [{
    id: 0,
    name: 'Name 1',
    arrval: [5, 3, 1]
  },
  {
    id: 1,
    name: 'Name 2',
    arrval: [6, 4, 3]
  },
  {
    id: 2,
    name: 'Name 3',
    arrval: [3, 2, 0]
  },
]
let index = 1
arr.sort((a, b) => a.arrval[1] - b.arrval[1])

console.log(arr)

edited Sep 7, 2021 at 8:48

Sanket Shah

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answered Sep 7, 2021 at 8:38

user15889308

Comments

trincot · Accepted Answer · 2021-09-07 09:35:10Z

0

The accepted answer sorts by the first array value. If you need to be able to deal with ties, where then the second array value becomes decisive, ...etc, then use this:

    
const src = [
  { id: 0, name: 'Name 1', arrval: [5, 3, 3] },
  { id: 1, name: 'Name 2', arrval: [5, 3, 1] },
  { id: 2, name: 'Name 3', arrval: [5, 2, 0, 1] },
  { id: 3, name: 'Name 4', arrval: [5, 2, 0] },
]

src.sort(({arrval: a}, {arrval: b}) => {
    for (let i = 0, len = Math.min(a.length, b.length); i < len; i++) {
        if (a[i] !== b[i]) return a[i] - b[i];
    }
    return a.length - b.length;
});

console.log(src);

answered Sep 7, 2021 at 9:35

trincot

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