filter array of objects by another array of objects

Question

I want to filter array of objects by another array of objects.

I have 2 array of objects like this:

const array = [
    { id: 1, name: \'a1\', sub: {         


        

Answer 1

You can use Array.filter and then Array.some since the later would return boolean instead of the element like Array.find would:

const a1 = [ { id: 1, name: 'a1', sub: { id: 6, name: 'a1 sub' } }, { id: 2, name: 'a2', sub: null }, { id: 3, name: 'a3', sub: { id: 8, name: 'a3 sub' } }, { id: 4, name: 'a4', sub: null }, { id: 5, name: 'a5', sub: { id: 10, name: 'a5 sub' } }, ]; 
const a2 = [ { id: 1, name: 'a1', sub: { id: 6, name: 'a1 sub' } }, { id: 2, name: 'a2', sub: null }, { id: 5, name: 'a5', sub: { id: 10, name: 'a5 sub' } }, ];

const result = a1.filter(({id, sub}) => !a2.some(x => x.id == id) && sub)

console.log(result)

Answer 2

Ok, let's solve this step by step.

To simplify the process let's suppose that two elements can be considered equals if they both have the same id.

The first approach that I would use is to iterate the first array and, for each element, iterate the second one to check the conditions that you've defined above.

const A = [ /* ... */]
const B = [ /* ... */]

A.filter(el => {
  let existsInB = !!B.find(e => {
    return e.id === el.id
  }

  return existsInB && !!B.sub
})

If we are sure that the elements in A and in B are really the same when they have the same ID, we could skip all the A elements without the sub property to perform it up a little bit

A.filter(el => {
  if (!el.sub) return false

  let existsInB = !!B.find(e => {
    return e.id === el.id
  }

  return existsInB
})

Now, if our arrays are bigger than that, it means that we are wasting a lot of time looking for the element into B. Usually, in these cases, I transform the array where I look for into a map, like this

var BMap = {}
B.forEach(el => {
  BMap[el.id] = el
})

A.filter(el => {
  if (!el.sub) return false

  return !!BMap[el.id]
})

In this way you "waste" a little bit of time to create your map at the beginning, but then you can find your elements quicker.

From here there could be even more optimizations but I think this is enought for this question

Answer 3

You could use JSON.stringify to compare the two objects. It would be better to write a function that compares all properties on the objects recursively.

const array = [
    { id: 1, name: 'a1', sub: { id: 6, name: 'a1 sub' } },
    { id: 2, name: 'a2', sub: null },
    { id: 3, name: 'a3', sub: { id: 8, name: 'a3 sub' } },
    { id: 4, name: 'a4', sub: null },
    { id: 5, name: 'a5', sub: { id: 10, name: 'a5 sub' } },
];
const anotherArray = [
    { id: 1, name: 'a1', sub: { id: 6, name: 'a1 sub' } },
    { id: 2, name: 'a2', sub: null },
    { id: 5, name: 'a5', sub: { id: 10, name: 'a5 sub' } },
];

const notIn = (array1, array2) => array1.filter(item1 => {
    const item1Str = JSON.stringify(item1);
    return !array2.find(item2 => item1Str === JSON.stringify(item2))
  }
);

console.log(notIn(array, anotherArray));

Answer 4

Like Felix mentioned, Array#filter won't work faster than native for loop, however if you really want it as functional way, here's one possible solution:

const array = [
    { id: 1, name: 'a1', sub: { id: 6, name: 'a1 sub' } },
    { id: 2, name: 'a2', sub: null },
    { id: 3, name: 'a3', sub: { id: 8, name: 'a3 sub' } },
    { id: 4, name: 'a4', sub: null },
    { id: 5, name: 'a5', sub: { id: 10, name: 'a5 sub' } },
];

const anotherArray = [
    { id: 1, name: 'a1', sub: { id: 6, name: 'a1 sub' } },
    { id: 2, name: 'a2', sub: null },
    { id: 5, name: 'a5', sub: { id: 10, name: 'a5 sub' } },
];

const r = array.filter((elem) => !anotherArray.find(({ id }) => elem.id === id) && elem.sub);

console.log(r);