compare values of an array, with the key of an array of objects

Question

As the question indicates, I need to compare the values of an array of this type [1,2,3], with the values of a key of an array of objects [{id: 1, name: 'jhon'}, {id: 2 , name: 'max'}] in this case I want to compare it with the value of the id key, I write the following example of what I need:

if I have this array of objects:

[
   {
      id: 1
      name: 'jhon'
   },
   {
     id: 2,
     name: 'max'
   },
   {
     id: 3,
     name: 'fer'
   }
]

and I have this array:

[1,2,3,4,5]

I need to compare them and obtain the values that are not present in the array of objects, in the previous example the values 4 and 5 are not present in the array of objects, so I need to obtain a new array or the same, but with the values that do not they were, [4,5]

NOTE: this should also work if I have an array, only with numbers that are not present, for example [8,9], they should return those same values.

EDIT: it is possible to obtain in a new array those values that are only present in the array of objects. For example, with the arrays of the previous code:

[{id: 1, name: 'jhon'}, {id: 2, name: 'max'}, {id: 3, name: 'fer'}]

now this array would have the following values [1,4,5] with the previous answers I get the following array [4,5] what is correct. how can I get the values [2,3] in a new array.

Answer 1

You can use filter and find

let arr1 = [{id: 1,name: 'jhon'},{id: 2,name: 'max'},{id: 3,name: 'fer'}]
let arr2 = [1, 2, 3, 4, 5];

let result = arr2.filter(o => !arr1.find(x => x.id === o));

console.log(result);

---

Update: This is basically the reverse of the first example. You filter the arr2 and check if the id exists on arr1. Then, use map to return the id

let arr1 = [{id: 1,name: 'jhon'},{id: 2,name: 'max'},{id: 3,name: 'fer'}]
let arr2 = [1,4,5];

let result = arr1.filter(o => !arr2.find(x => x === o.id)).map(o=>o.id);

console.log(result);

Doc: filter(), find()

Answer 2

Create a Set from the ids of arr1 using Array.map(). Then filter arr2 by checking if the Set has the number:

const arr1 = [{"id":1,"name":"jhon"},{"id":2,"name":"max"},{"id":3,"name":"fer"}];
const arr2 = [1,2,3,4,5];
const arr3 = [1, 5];

const difference = (arr1, arr2, key) => {
  const arr1Values = new Set(key ? arr1.map(({ [key]: v }) => v) : arr1);

  return arr2.filter((n) => !arr1Values.has(n));
};

// array of primitives and array of objects
console.log(difference(arr1, arr2, 'id'));

// two arrays of primitives 
console.log(difference(arr3, arr2));

Answer 3

You can use filter to check if id property of element in arr2 matches to the current element in arr2

const arr1 = [{"id":1,"name":"jhon"},{"id":2,"name":"max"},{"id":3,"name":"fer"}];

const arr2 = [8,9];
var diffArray = arr2.filter(x => !arr1.filter(y => y.id === x).length);
console.log(diffArray);

Answer [ 8,9 ]

Answer 4

You can use .filter() and .includes():

let arr1 = [{id: 1,name: 'jhon'},{id: 2, name: 'max'},{id: 3,name: 'fer'}],
    arr2 = [1, 2, 3, 4, 5];

let result = (
    ids => arr2.filter(n => !ids.includes(n))
)(arr1.map(({id}) => id));

console.log(result);

Description:

• Create a new array by extracting only ids from first array using .map().
• Filter elements using .filter() and .includes() to filter only those elements of second array that doesn't exists in newly created array.

Docs:

• Array.prototype.map()
• Array.prototype.filter()
• Array.prototype.includes()
• Object Destructuring
• Arrow Functions

Answer 5

You could take a set with the given ids and then delete the ones of the object. Later return the array with the leftover ids.

var data = [{ id: 1, name: 'jhon' }, { id: 2, name: 'max' }, { id: 3, name: 'fer' }],
    ids = [1, 2, 3, 4, 5],
    result = Array.from(data.reduce((s, { id }) => (s.delete(id), s), new Set(ids)));

console.log(result);

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