通过Id比较两个数组的元素，并从一个数组中删除未在另一个数组中显示的元素

Question

I have two arrays of objects like this:

var arr1 = [{Id: 1, Name: "Test1"}, {Id: 2, Name: "Test2"}, {Id: 3, Name: "Test3"}, {Id: 4, Name: "Test4"}]

var arr2 = [{Id: 1, Name: "Test1"}, {Id: 3, Name: "Test3"}]

I need to compare the elements of the two arrays by Id and remove the elements from arr1 that are not presented in arr2 ( does not have element with that Id ). How can I do this ?

Answer 1

var res = arr1.filter(function(o) {
    return arr2.some(function(o2) {
        return o.Id === o2.Id;
    })
});

shim, shim, shim.

Answer 2

You can use a function that accepts any number of arrays, and returns only the items that are present in all of them.

function compare() {
    let arr = [...arguments];
    return arr.shift().filter( y => 
        arr.every( x => x.some( j => j.Id === y.Id) )
    )
}

 var arr1 = [{Id: 1, Name: "Test1"}, {Id: 2, Name: "Test2"}, {Id: 3, Name: "Test3"}, {Id: 4, Name: "Test4"}]; var arr2 = [{Id: 1, Name: "Test1"}, {Id: 3, Name: "Test3"}, {Id: 30, Name: "Test3"}]; var arr3 = [{Id: 1, Name: "Test1"}, {Id: 6, Name: "Test3"}, {Id: 30, Name: "Test3"}]; var new_arr = compare(arr1, arr2, arr3); console.log(new_arr); function compare() { let arr = [...arguments] return arr.shift().filter( y => arr.every( x => x.some( j => j.Id === y.Id) ) ) } 

Answer 3

Making use of a hash (a Set) will give a performance gain:

 var arr1 = [{Id: 1, Name: "Test1"}, {Id: 2, Name: "Test2"}, {Id: 3, Name: "Test3"}, {Id: 4, Name: "Test4"}]; var arr2 = [{Id: 1, Name: "Test1"}, {Id: 3, Name: "Test3"}]; arr1 = arr1.filter(function (el) { return this.has(el.Id); }, new Set(arr2.map(el => el.Id))); console.log(arr1); 

A new Set is created that gets the Id values from arr2 :

"1","3"

That Set is passed as the thisArg to filter , so that within the filter callback it is available as this .