Sorting an array of objects by values

Question

I need to sort an array of objects by its values and I use this code to do this:

function compare(a,b){return dict[a]-dict[b]} Object.keys(dict).sort(compare)

it works if each value is different, but if two objects have the same value, it leaves them in the order they appear in the array but I want them to be sorted alphabetically. I couldn't figure out a way to do that.

dict = {a:1, d:4, c:2, b:4, e:5, f:3} should be :

{a:1, c:2, f:3, b:4, d:4, e:5 }

but I'm getting this:

{a:1, c:2, f:3, d:4, b:4, e:5 }

Object.keys() returns an array, and you're sorting that array. The result will be an array, not an object. — Pointy
– Pointy, Commented Apr 21, 2020 at 13:07

adiga · Accepted Answer · 2020-04-21 13:12:08Z

1

You can change the compareFunction to use localeCompare

dict[a] - dict[b] || a.localeCompare(b)

If dict[a] - dict[b] returns 0, it checks the next condition and sorts the keys alphabetically

Here's a snippet:

const dict = {a:1, d:4, c:2, b:4, e:5, f:3}

function compare(a, b) {
  return dict[a] - dict[b] || a.localeCompare(b)
}

const sorted = Object.keys(dict)
                      .sort(compare)
                      .reduce((acc, k) => (acc[k] = dict[k], acc), {})

console.log( sorted )

answered Apr 21, 2020 at 13:12

adiga

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Comments

epascarello · Accepted Answer · 2020-04-21 13:14:24Z

1

So compare the keys if the diff is equal to zero

function compare(a, b) {
  var diff = dict[a] - dict[b]
  return diff === 0 ? a.localeCompare(b) : diff
}

const dict = {
  a: 1,
  d: 4,
  c: 2,
  b: 4,
  e: 5,
  f: 3
}

const result = Object.keys(dict).sort(compare).reduce((o, x) => ({
  ...o,
  [x]: dict[x],
}), {})

console.log(result)

answered Apr 21, 2020 at 13:14

epascarello

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