JavaScript sort array of objects based on an array of values

Question

I need to sort my list using the selected array.

const list = [
    { name: 'BMW' },
    { name: 'AUDI' },
    { name: 'MINI' },
    { name: 'FIAT' },
    { name: 'KIA' },
]

const selected = [ 'MINI', 'KIA' ]

What is the best way to do this? I have found some examples but not too sure how to adapt this to work the way I want it to!

list.sort((a,b) => (a.last_nom > b.last_nom) ? 1 : ((b.last_nom > a.last_nom) ? -1 : 0))

Expected result:

const list = [
    { name: 'MINI' },
    { name: 'KIA' },
    { name: 'BMW' },
    { name: 'AUDI' },
    { name: 'FIAT' },
]

Use selected.indexOf(a.name) and selected.indexOf(b.name), and compare those before comparing the values of a.name and b.name — Barmar
– Barmar, Commented Aug 23, 2021 at 19:55
Does this answer your question? Sort an object array by custom order — stevecomrie
– stevecomrie, Commented Aug 23, 2021 at 19:56
@stevecomrie That's not quite the same, because the custom order there includes all the names. selected just has some ofthem. — Barmar
– Barmar, Commented Aug 23, 2021 at 19:57
Does the order within the selected and non-selected groups matter? — Barmar
– Barmar, Commented Aug 23, 2021 at 19:59

Guerric P · Accepted Answer · 2021-08-23 20:17:56Z

1

You could use Array.prototype.includes and take advantage of JavaScript's numbers coercion like this:

const list = [
  { name: "BMW" },
  { name: "AUDI" },
  { name: "MINI" },
  { name: "FIAT" },
  { name: "KIA" },
];

const selected = ["MINI", "KIA"];

const sorted = list.sort(
  (a, b) => selected.includes(b.name) - selected.includes(a.name)
);

console.log(sorted);

If the order of the elements has to remain the same and be consistent across browsers:

const list = [
  { name: "BMW" },
  { name: "AUDI" },
  { name: "MINI" },
  { name: "FIAT" },
  { name: "KIA" },
];

const selected = ["MINI", "KIA"];

const sorted = list.sort(
  (a, b) =>
    selected.includes(b.name) - selected.includes(a.name) ||
    list.indexOf(a) - list.indexOf(b)
);

console.log(sorted);

edited Aug 23, 2021 at 20:17

answered Aug 23, 2021 at 19:55

Guerric P

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5 Comments

Barmar Over a year ago

includes() just returns true or false, it doesn't really make sense to subtract them.

Guerric P Over a year ago

Still it returns a number that represents the difference as if true were 1 and false were 0

Barmar Over a year ago

I assumed that if both are in selected, they want them in the same order as in selected. This puts all the selected items first, non-selected items later, but doesn't specify the order within those groups.

Guerric P Over a year ago

Indeed, chromium happens to preserve the order but it's implementation-dependent

Martyn Ball Over a year ago

Yeah as long as the selected are at the top that's all I need :D

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