compare 2 arrays and leave-CodePudding

var arr1 = [11, 12, 13];
var arr2 = [
      { id: 11, name: "name 1" },
      { id: 12, name: "name 2" },
      { id: 13, name: "name 3" },
      { id: 14, name: "name 4" },
];

I have this two arrays and in the second array want to leave items which id's equal to first array values. I mean this:

  var newArray = [
          { id: 11, name: "name 1" },
          { id: 12, name: "name 2" },
          { id: 13, name: "name 3" },
    ];

What is most efficient way to do this?

CodePudding user response：

First, build an index by mapping each id to its corresponding object. The index object will have this structure:

{
  '11': { id: 11, name: 'name 1' },
  '12': { id: 12, name: 'name 2' },
  '13': { id: 13, name: 'name 3' },
  '14': { id: 14, name: 'name 4' }
}

This index can be built in a variety of ways, one of which is to use Object.fromEntries. Object.fromEntries(arr2.map(obj => [obj.id, obj])).

Then map each id in your arr1 to its corresponding object with a lookup in your index object: index[id]. That will look like arr1.map(id => index[id]).

var arr1 = [11, 12, 13];
var arr2 = [
      { id: 11, name: "name 1" },
      { id: 12, name: "name 2" },
      { id: 13, name: "name 3" },
      { id: 14, name: "name 4" },
];

const index = Object.fromEntries(arr2.map(obj => [obj.id, obj]));
const result = arr1.map(id => index[id]);
console.log(result);

This approach is efficient because it does not perform a linear search for each item's index (no .includes or .find per entry is required). Instead, it leverages the hashtable lookup of a property in an object (which is more or less constant time). Overall this is O(N) instead of O(N^2).

Beware, there may more overhead in creating such an index for small values of N. But for larger values of N this will be better, and also will be better if there are multiple variations of arr1 objects to be used with a single arr2.

CodePudding user response：

This is a simple filter accessing the item id and checking if that id is included in the first array.

var arr1 = [11, 12, 13];
var arr2 = [
      { id: 11, name: "name 1" },
      { id: 12, name: "name 2" },
      { id: 13, name: "name 3" },
      { id: 14, name: "name 4" },
];
var newArray = arr2.filter(item => arr1.includes(item.id));

console.log(newArray);

Edit: In the spirit of efficient execution, another approach:

var arr1 = [11, 12, 13];
var arr2 = [
      { id: 11, name: "name 1" },
      { id: 12, name: "name 2" },
      { id: 13, name: "name 3" },
      { id: 14, name: "name 4" },
];
var objIds = arr1.reduce((obj, item) => (obj[item] = true) && obj, {})
var newArray = arr2.filter(item => objIds[item.id]);

console.log(objIds, newArray)