Get unmatched objects between two object arrays in typescript

Question

I have two object arrays :

first = [
  { "id": "15", "name": "raza" },
  { "id": "1", "name": "sahir" },
  { "id": "54", "name": "ayyan" },
  { "id": "3", "name": "tahir" },
];

 second = [
  { "id": "15", "name": "razi" },
  { "id": "3", "name": "qasim" },
  { "id": "1", "name": "taha" },
];

I want to get unmatched objects from "first" array on the basis of their id, something like

const result = this.first.filter(e => this.second.some(({id}) => e.id ==id ));

It gives me matched objects but I want to get unmatched objects.

user4676340 · Accepted Answer · 2018-12-19 14:35:39Z

4

Something like this ?

first = [
  { "id": "15", "name": "raza" },
  { "id": "1", "name": "sahir" },
  { "id": "54", "name": "ayyan" },
  { "id": "3", "name": "tahir" },
];

second = [
  { "id": "15", "name": "razi" },
  { "id": "3", "name": "qasim" },
  { "id": "1", "name": "taha" },
];

unmatched = first.filter(item => !second.some(_item => _item.id === item.id));

console.log(unmatched);

answered Dec 19, 2018 at 14:35

user4676340

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Comments

Kamil Kiełczewski · Accepted Answer · 2018-12-20 14:40:23Z

0

Here is O(n+m) time complexity solution (n,m are arrays lengths):

let h={}; second.map(x=>h[x.id]=1);     // we use hash map with "second" ids
let result = first.filter(x=>!h[x.id]);

first = [
  { "id": "15", "name": "raza" },
  { "id": "1", "name": "sahir" },
  { "id": "54", "name": "ayyan" },
  { "id": "3", "name": "tahir" },
];

second = [
  { "id": "15", "name": "razi" },
  { "id": "3", "name": "qasim" },
  { "id": "1", "name": "taha" },
];

let h = {}; second.map(x=>h[x.id]=1);     // we use hash map with data2 ids
let result = first.filter(x=>!h[x.id]);

console.log(result);

answered Dec 20, 2018 at 14:40

Kamil Kiełczewski

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