How to do some array comparison in JavaScript in an efficient way

Question

TMDB API returns an array of movie objects that look like this:

{
"vote_count": 1527,
"id": 338952,
"video": false,
"vote_average": 7,
"title": "Fantastic Beasts: The Crimes of Grindelwald",
"popularity": 272.487,
"poster_path": "/uyJgTzAsp3Za2TaPiZt2yaKYRIR.jpg",
"original_language": "en",
"original_title": "Fantastic Beasts: The Crimes of Grindelwald",
"genre_ids": [
   10751,
   14,
   12
],
"backdrop_path": "/xgbeBCjmFpRYHDF7tQ7U98EREWp.jpg",
"adult": false,
"overview": "Gellert Grindelwald has .....",
"release_date": "2018-11-14"
}

They also provide an API to return ALL available genres in an array of objects with a key and a label:

genres": [
{
"id": 28,
"name": "Action"
},
{
"id": 12,
"name": "Adventure"
},
{
"id": 16,
"name": "Animation"
}
]

What I need to do is get a list of all unique genres from the now playing API with their label values.

So my question isn't so much about how to do this, but what is the cleanest and most efficient approach.

My attempt:

let uniqueIds = new Set(), genres;

// First get all available unique genre IDs from the now playing list
for(var i = 0; i < this.state.items.length; i++){
    for(var x = 0; x < this.movies[i].genre_ids.length; x++){
        uniqueIds.add(this.movies[i].genre_ids[x])
    }
}

// build array of genre objects from unique genre IDs
genres = this.genres.filter((genre) => uniqueIds.has(genre.id));

Andy · Accepted Answer · 2018-11-30 14:38:00Z

3

1) For each object in the array grab the genre_ids

2) filter out the genre objects that have an id included in the ids array.

const api = [{"vote_count":1527,"id":338952,"video":false,"vote_average":7,"title":"Fantastic Beasts: The Crimes of Grindelwald","popularity":272.487,"poster_path":"/uyJgTzAsp3Za2TaPiZt2yaKYRIR.jpg","original_language":"en","original_title":"Fantastic Beasts: The Crimes of Grindelwald","genre_ids":[10751,14,12],"backdrop_path":"/xgbeBCjmFpRYHDF7tQ7U98EREWp.jpg","adult":false,"overview":"Gellert Grindelwald has .....","release_date":"2018-11-14"},{"vote_count":1527,"id":338952,"video":false,"vote_average":7,"title":"Fantastic Beasts: The Crimes of Grindelwald","popularity":272.487,"poster_path":"/uyJgTzAsp3Za2TaPiZt2yaKYRIR.jpg","original_language":"en","original_title":"Fantastic Beasts: The Crimes of Grindelwald","genre_ids":[10751,14,16],"backdrop_path":"/xgbeBCjmFpRYHDF7tQ7U98EREWp.jpg","adult":false,"overview":"Gellert Grindelwald has .....","release_date":"2018-11-14"}];
const genres = [{"id":28,"name":"Action"},{"id":12,"name":"Adventure"},{"id":16,"name":"Animation"}];

// [].concat(...arr) flattens consequtive arrays down
const idArr = [].concat(...api.map(obj => obj.genre_ids));
const matchingGenres = genres.filter(obj => idArr.includes(obj.id));

console.log(matchingGenres);

edited Nov 30, 2018 at 14:38

answered Nov 30, 2018 at 14:19

Andy

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6 Comments

shanahobo86 Over a year ago

Ah... very nice. this is destructuring? it's much neater and doesn't have that awful nested loop.

Andy Over a year ago

I updated the answer to account for the array of objects, rather than one object.

Robbie Milejczak Over a year ago

@shanahobo86 it's not destructuring, what's happening here is called spreading developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… you can read about destructuring here developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

Andy Over a year ago

@RobbieMilejczak, FYI the very first edit of this answer did contain destructuring. It's that the OP was referring to. But, yes, this edit does contain the spread syntax.

James Over a year ago

filter out the genre objects that don't have an id included

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