Take over positions of an array after comparison with another array?

Question

I have a function (in Javascript) that compares 2 arrays:

let diferencas = [];
    let numD = 0;
    for (let a = 0; a < dados.length; a++) {
        if (dados[a].idEmail === results[a].idEmail) {
            console.log('Repetido!')
        } else {
            numD++;
            diferencas.push(dados[a]);
        }
    }
    console.log('Diferenças encontradas: ' + numD)
    console.log(diferencas)

The 2 arrays are composed in their first 8 positions by identical objects, with the array results has only 8 positions as it comes from a query to Mongodb and the array dados has 9 positions as it comes from a query to the email provider and comes with newer emails.

My problem is: How Do I Pick Up Those Over Positions in the Array dados ??

Answer 1

A solution to the problem itself is by using the filter, thus:

let diferencas = dados.filter(x => results.indexOf(x) ==-1 ? x : false );

In which the filter part of the array dados which is the largest, and filters to see if each element exists in the results through the indexOf. If the current element of dados not exist in results is returned x and so appears in diferencas. Otherwise it returns false and excludes from diferencas

Example:

const dados = ['[email protected]','[email protected]','[email protected]','[email protected]', '[email protected]'];
const results = ['[email protected]','[email protected]','[email protected]','[email protected]'];

let diferencas = dados.filter(x => results.indexOf(x) == -1 ? x:false);

console.log('Diferenças encontradas: ' + diferencas.length);
console.log(diferencas);

If you have an object array, you can still use this solution, but you have to take an additional step to stay with the emails, which is to first map the array to an array of strings with the emails, using the map and then perform the same procedure:

const dados = [
{email : '[email protected]'},
{email : '[email protected]'},
{email : '[email protected]'},
{email : '[email protected]'},
{email : '[email protected]'}
];

const results = [
{email : '[email protected]'},
{email : '[email protected]'},
{email : '[email protected]'},
{email : '[email protected]'}
];

const dados2 = dados.map(x => x.email); //mapeamento aqui
const results2 = results.map(x => x.email);//mapeamento aqui

//o resto igual
let diferencas = dados2.filter(x => results2.indexOf(x)==-1?x:false);

console.log('Diferenças encontradas: ' + diferencas.length);
console.log(diferencas);

//Para diferenças em formato de objeto pode-se utilizar o map do diferencas para dados
let diferencasObj = diferencas.map(x => dados[dados2.indexOf(x)]);
console.log(diferencasObj);

Comparing with the solution that has this has the advantage of also working even if the order of the arrays is not equal.

Example:

const dados = ['[email protected]','[email protected]', '[email protected]','[email protected]','[email protected]']; //desordenado
const results = ['[email protected]','[email protected]','[email protected]','[email protected]',]; //desordenado

let diferencas = dados.filter(x => results.indexOf(x) == -1 ? x:false);

console.log('Diferenças encontradas: ' + diferencas.length);
console.log(diferencas);

Answer 2

Just make sure results exists if there is no dados exceeded...

var dados = [
{idEmail : '1'},
{idEmail : '2'},
{idEmail : '3'},
{idEmail : '4'},
{idEmail : '5'},
];

var results = [
{idEmail : '1'},
{idEmail : '2'},
{idEmail : '3'},
{idEmail : '4'},
];

let diferencas = [];
    let numD = 0;
    for (let a = 0; a < dados.length; a++) {
        if (results[a]){
        	if(dados[a].idEmail === results[a].idEmail) {
            console.log('Repetido!')
        }}else {
            numD++;
            diferencas.push(dados[a]);
        }
    }
    console.log('Diferenças encontradas: ' + numD);
    console.log(diferencas);