How to sort an array of objects in the same order as an id array

Question

I have an array of players objects like this :

    var players = [{
        id: "thisIsID1",
        name: "William",
        otherProps
    },
    {
        id: "thisIsID2",
        name: "Shakespeare",
        otherProps
    },
    {
        id: "thisIsID3",
        name: "Lola",
        otherProps
    }]

And I have and array of their ID that has been shuffled, like so :

var shuffledIDs = ["thisIsID2", "thisIsID3", "thisIsID1"]

How can I sort the players var so the objects are in the same order as the corresponding IDs of shuffledIDs ?

Edit: different names just for the sake of making players different

Aplet123 · Accepted Answer · 2020-04-13 11:08:34Z

6

If your data is short, then you can sort it with the following one-liner:

players = shuffledIDs.map(id => players.find(v => v.id == id))

Essentially, for every id in shuffledID, it finds the element in players with that id and puts it in the correct spot. However, this takes O(n^2) time so it might not scale well for larger data. If you want a faster method, you can maintain an object of IDs:

var ids = {};
players.forEach(v => ids[v.id] = v);
players = shuffledIDs.map(v => ids[v]);

answered Apr 13, 2020 at 11:08

Aplet123

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1 Comment

Joseph Allain Over a year ago

I think the latter is the most efficient method I've got ! Thanks a lot !

palaѕн · Accepted Answer · 2020-04-13 11:07:02Z

1

You can achieve it using array .find() method:

var players = [{
        id: "thisIsID1",
        name: "William"
    },
    {
        id: "thisIsID2",
        name: "Shakespeare"
    },
    {
        id: "thisIsID3",
        name: "Lola"
    }]

var shuffledIDs = ["thisIsID2", "thisIsID3", "thisIsID1"]
var result = shuffledIDs.map(x => players.find(p=>p.id === x))
console.log(result)

answered Apr 13, 2020 at 11:07

palaѕн

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Comments

Siva Kondapi Venkata · Accepted Answer · 2020-04-13 11:42:19Z

1

Create object with keys and values as index from shuffle array. Use sort method and prioritize bases above shuffled indexes. This way should even the case of duplicate data in players.

var players = [
  {
    id: "thisIsID1",
    name: "William"
  },
  {
    id: "thisIsID2",
    name: "Shakespeare"
  },
  {
    id: "thisIsID3",
    name: "Lola"
  }
];

const shuffleIds = ["thisIsID2", "thisIsID3", "thisIsID1"];

const shuf_idx = Object.fromEntries(shuffleIds.map((x, i) => [x, i]));

players.sort((a, b) => shuf_idx[a.id] - shuf_idx[b.id]);

console.log(players);

answered Apr 13, 2020 at 11:42

Siva Kondapi Venkata

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Comments

norbitrial · Accepted Answer · 2020-04-13 11:08:13Z

0

With .map() and .find() where using index of the element:

const shuffledIDs = ["thisIsID2", "thisIsID3", "thisIsID1"];
const players = [{ id: "thisIsID1", name: "William" }, { id: "thisIsID2",       name: "Shakespeare" }, { id: "thisIsID3", name: "Lola" }];

const result = players.map((e, i) => players.find(f => f.id === shuffledIDs[i]));

console.log(result);

answered Apr 13, 2020 at 11:08

norbitrial

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1 Comment

Peter J Harrison Over a year ago

If there is a value in shuffledIDs that doesn't match a value in players, this approach returns an undefined element in result array. Is there a way to prevent this?

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