如何从具有特定 id 的数组中获取对象与另一个数组编号进行比较

Question

I need to compare id objects array from a firstArray and array numbers (secondArray) and return a new array with objects from the first array which id number exists in the second array.

So at the end, I want a new array with objects with id 39 and 41.

Actually I find something like this:

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

const arr1 =
"blocks": [
    {
      "id": 1,
      "functions": [ 0, 1 ]
    },
    {
      "id": 39,
      "functions": [ 0, 1, 3, 4 ]
    },
    {
      "id": 41,
      "functions": [ 0, 1 ]
    }
]

const arr2 = [39, 41]

Answer 1

You can create a Map to see whether there is an item of Map exists in filtering array. Getting an item from Map method is O(1) :

 const blocks = [ { "id": 1, "functions": [ 0, 1 ] }, { "id": 39, "functions": [ 0, 1, 3, 4 ] }, { "id": 41, "functions": [ 0, 1 ] } ]; const arr2 = [39, 41]; const arr2Maps = new Map(arr2.map(a=>[a, a])); const result = blocks.filter(o => arr2Maps.get(o.id)); console.log(result)

In addition, you can use filter and some methods. However, some method has O(n) :

 const blocks = [ { "id": 1, "functions": [ 0, 1 ] }, { "id": 39, "functions": [ 0, 1, 3, 4 ] }, { "id": 41, "functions": [ 0, 1 ] } ]; const arr2 = [39, 41] const result = blocks.filter(o => arr2.some(a=> a ==o.id )); console.log(result)

Answer 2

You can use includes() function during filtering. Includes() works like in array function.

 const arr1 = [ { "id": 1, "functions": [ 0, 1 ] }, { "id": 39, "functions": [ 0, 1, 3, 4 ] }, { "id": 41, "functions": [ 0, 1 ] } ] const arr2 = [39, 41] const result = arr1.filter(o => arr2.includes(o.id)); console.log(result)