JavaScript - Find multiple objects by IDs existing in prefined list from an array of objects without using the for loop [duplicate]

Question

I have a JavaScript array of objects:

var my_array = [
  {
    "id" : 1,
    "aas_name" : "aaa"
  },
  {
    "id" : 2,
    "aas_name" : "bbb"
  },
  {
    "id" : 3,
    "aas_name" : "ccc"
  },
  ...
   ...
    ...
  ,
  {
    "id" : 10,
    "aas_name" : "jjj"
  }
]

I would like to find all objects in the my_array, each of which has the id value that exists in a pre-defined array [1, 3, 8] for instance without having to use the for... loop and ES6 arrow function style.

var result = [
  {
    "id" : 1,
    "aas_name" : "aaa"
  },
  {
    "id" : 3,
    "aas_name" : "ccc"
  },
  {
    "id" : 8,
    "aas_name" : "hhh"
  }
]

Use a .filter, var result = my_array.filter(function(item) { return [1,3,8].indexOf(item.id) != -1 }); — Maksym Petrenko
– Maksym Petrenko, Commented Aug 12, 2019 at 5:57
@MaksymPetrenko this should be posted as answer instead as comment mate :) — Code Maniac
– Code Maniac, Commented Aug 12, 2019 at 6:00

Code Maniac · Accepted Answer · 2019-08-12 06:07:27Z

6

Use a .filter:

The filter() method creates a new array with all elements that pass the test implemented by the provided function. filter

var my_array = [{"id": 1,"aas_name": "aaa"},{"id": 2,
"aas_name": "bbb"},{"id": 3,"aas_name": "ccc"},{"id": 8,"aas_name": "hhh"},{"id": 10,"aas_name": "jjj"}]

let result = my_array.filter(function(item) {
  return [1, 3, 8].indexOf(item.id) != -1;
});

console.log(result)

edited Aug 12, 2019 at 6:07

Code Maniac

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answered Aug 12, 2019 at 6:02

Maksym Petrenko

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2 Comments

Jonas Johansson Over a year ago

my_array.filter(function(item) { retrun [1,3,8].includes(item.id) }) Would work as well :)

claasic Over a year ago

For bigger data it would be more efficient to create a hashtable first and instead of indexOf do lookups on it. Instead of O(n * m) we'd have O(2 * n) (or O(3 * n)). This works well though!

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