查找数组中不包含在另一个对象数组中的元素

Question

I have an array arr1 = [1,2,3,4,5] There is another array of objects arr2 = [{'id':2, 'name':'A'},{'id':4, 'name':'B'}]

I am looking for find elements in arr1 which are not in arr2 . The expected output is [1,3,5]

I tried the following but it doesn't work.

const arr = arr1.filter(i => arr2.includes(i.id));

Can you please help?

Answer 1

A solution with O(arr2.length) + O(arr1.length) complexity in Vanilla JS

var arr1= [1,2,3,4,5]; 
var arr2 = [{'id':2, 'name':'A'},{'id':4, 'name':'B'}];

var tmp = arr2.reduce(function (acc, obj) { 
    acc[obj['id']] = true; 
    return acc; 
}, {});

var result = arr1.filter(function(nr) { 
    return !tmp.hasOwnProperty(nr); 
})

Answer 2

arr2 is an array of objects, so arr2.includes(i.id) doesn't work because i (an item from arr1 ) is a number , which doesn't have an id property, and because arr2 is an array of objects.

Turn arr2 's id s into a Set first, then check whether the set contains the item being iterated over:

 const arr1 = [1,2,3,4,5]; const arr2 = [{'id':2, 'name':'A'},{'id':4, 'name':'B'}]; const ids = new Set(arr2.map(({ id }) => id)); const filtered = arr1.filter(num => !ids.has(num)); console.log(filtered);

Answer 3

You can try with Array.prototype.some() :

The some() method tests whether at least one element in the array passes the test implemented by the provided function. It returns a Boolean value.

 const arr1 = [1,2,3,4,5] const arr2 = [{'id':2, 'name':'A'},{'id':4, 'name':'B'}] const arr = arr1.filter(i => !arr2.some(j => j.id == i)); console.log(arr);

Answer 4

We can use the filter method like below to check the condition required

    var arr1 = [1, 2, 3, 4, 5]
    var arr2 = [{ 'id': 2, 'name': 'A' }, { 'id': 4, 'name': 'B' }]
    var ids = [];
    arr2.forEach(element => {
        ids.push(element['id'])
    });

    var result = arr1.filter(s => ids.indexOf(s) < 0)
    console.log(result)

Answer 5

let arr1= [1,2,3,4,5]; 
let arr2 = [{'id':2, 'name':'A'},{'id':4, 'name':'B'}]


let arr2Ids=arr2.map(item=>item.id); 
let result=arr1.filter(n => !arr2Ids.includes(n));

Answer 6

您可以在arr2上使用 find 而不是包含，因为arr2由对象组成

const arr = arr1.filter(i => !arr2.find(e => e.id===i));