Filter array of objects based on values in second array

Question

I have an array of objects that I'd like to filter to create a new array based on whether or not the value of any key matches any value in another array.

const array1 = [{name: 'pink', id: 13}, {name: 'orange', id: 17}, {name: 'red, id: 64}, {name: 'purple', id: 47}, {name: 'yellow', id: 23}, {name: 'gray', id: 2}, {name: 'black', id: 200}, {name: 'violet', id: 4}]

const array2 = ['red', 'blue', 'green', 'pink']

I've tried using a for...of loop inside of a return function but that is giving me errors.

    const array3 = array1.filter(color => {
        for (mainColor of array2){
            return color.name === mainColor 
        }
    });

This works but is clearly not the best way.

    const array3 = array1.filter(color => {
            return (color.main === 'red') || (color.main === 'blue')
        });

How can I get a third array from array1 that contains only the objects where the array1.name matches a value in array2?

Is it possible with ES6 or Lodash?

Thanks in advance!

It does not work, because you are using return inside the loop, thus iterating only over the first element before exiting. If you want to use a for loop, you have to check whether the condition is true and return true only then. — Angel Politis
– Angel Politis, Commented Mar 6, 2018 at 15:19

gurvinder372 · Accepted Answer · 2018-03-06 15:15:57Z

4

Almost there, instead of for-loop use includes and return

const array3 = array1.filter(color => {
    return array2.includes ( color.name );
});

Or

const array3 = array1.filter( color => array2.includes ( color.name ) );

edited Mar 6, 2018 at 15:15

answered Mar 6, 2018 at 15:12

gurvinder372

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Comments

trincot · Accepted Answer · 2018-03-06 15:29:25Z

3

Let me give an alternative, that has slightly more code, but is more efficient as well, as it only needs to scan array2 once:

const array1 = [{name: 'pink', id: 13}, {name: 'orange', id: 17}, {name: 'red', id: 64}, {name: 'purple', id: 47}, {name: 'yellow', id: 23}, {name: 'gray', id: 2}, {name: 'black', id: 200}, {name: 'violet', id: 4}],
      array2 = ['red', 'blue', 'green', 'pink'];

const colorSet = new Set(array2),
      array3 = array1.filter(color => colorSet.has(color.name));

console.log(array3);

edited Mar 6, 2018 at 15:29

answered Mar 6, 2018 at 15:16

trincot

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2 Comments

Ele Over a year ago

Do you think is worth it a previous process with new Set(...) just to check if a color exists in array2? Just to remove the repeated? or Set has a better algorithm?

trincot Over a year ago

The has method works in constant time, while includes or indexOf need linear time, so when that is in a loop, you get O(n²) time complexity, while with a set you'll just have O(n) both for the new Set and the loop, and O(n)+O(n) is still O(n). Of course, all this is not very relevant when we speak of an array that has a limited number of entries, not hundreds of them.

Mamun · Accepted Answer · 2018-03-06 15:14:26Z

1

Try the following with Array's includes():

const array1 = [{name: 'pink', id: 13}, {name: 'orange', id: 17}, {name: 'red', id: 64}, {name: 'purple', id: 47}, {name: 'yellow', id: 23}, {name: 'gray', id: 2}, {name: 'black', id: 200}, {name: 'violet', id: 4}]

const array2 = ['red', 'blue', 'green', 'pink'];

const array3 = array1.filter(color => array2.includes(color.name));

console.log(array3);

answered Mar 6, 2018 at 15:14

Mamun

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Comments

Alexis · Accepted Answer · 2018-03-06 15:14:05Z

The filter method already iterates on each item. You just have to return true if the element is present in the second array (by using indexOf or includes)

const array1 = [{name: 'pink', id: 13}, {name: 'orange', id: 17}, {name: 'red', id: 64}, {name: 'purple', id: 47}, {name: 'yellow', id: 23}, {name: 'gray', id: 2}, {name: 'black', id: 200}, {name: 'violet', id: 4}]

const array2 = ['red', 'blue', 'green', 'pink'];

let filteredArray = array1.filter(e => array2.indexOf(e.name) > -1);

console.log(filteredArray);

connexo · Accepted Answer · 2018-03-06 15:20:04Z

0

I know indexOf might seem a little outfashioned and ES5-ish, but it still does the job:

const array1 = [{name: 'pink', id: 13}, {name: 'orange', id: 17}, {name: 'red', id: 64}, {name: 'purple', id: 47}, {name: 'yellow', id: 23}, {name: 'gray', id: 2}, {name: 'black', id: 200}, {name: 'violet', id: 4}],
  array2 = ['red', 'blue', 'green', 'pink'];

let array3 = array1.filter(x => array2.indexOf(x.name) != -1)
console.log(array3)

edited Mar 6, 2018 at 15:20

answered Mar 6, 2018 at 15:14

connexo

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