Compare values of two arrays and generate a third with the unique javascript values

Question

I am doing, with google script, a very simple system that monitors the progress of daily activities, I have a table with the requested activities, a table with the interactions and a table with the status (open or closed). I need to compare the list with the ID’s of the requested activities with the list with the ID’s of the closed activities. The remaining Ids I assume are open. Note: I am beginner in programming, can correct me the will!

function arrayabertas(){
var todos = [08072020105842, 08072020110953, 8072020104528];
var fechado = [8072020104528];
var abertos = [];

// Verifica se cada número jogado
// está na lista dos sorteados
for(var i=0; i<fechado.length; i++) {
    if(todos.indexOf(fechado[i]) > -1) {
        abertos.push(fechado[i]);
    }
}
    return(abertos);
}

Answer 1

If you’re already using compatible environments Ecmascript 2015, an alternative is to use Set, who already owns good support of current browsers.

The Set ensures that its elements will be unique (which is one of your requirements), and also has some operations that facilitate the algorithm for what you need (in addition to being more efficient, as we will see at the end).

For example, in your case, creating a Set with all values and remove the ones that are closed. What is left are the open items:

let todos = [ 08072020105842, 08072020110953, 8072020104528 ];
let fechado = [ 8072020104528 ];

let abertos = new Set(todos);
for (let e of fechado) { // remove todos os fechados
    abertos.delete(e);
}

// converter para array
abertos = Array.from(abertos);
console.log(abertos);

---

The other answers suggest using indexOf or includes, which actually work (except with includes you should check which elements nay are in fechado, that is to say, abertos = todos.filter(x => ! fechado.includes(x))), but are more inefficient because so much indexOf how much includes traverse the array todos from the start.

That is, for each element of fechado, you have to go through the array todos from the beginning, until you find the element. And when you go through the same array several times, you are creating a variation of the call Shlemiel the Painter’s Algorithm, which is extremely inefficient.

Of course, for small arrays - as is your case - the difference will be insignificant and maybe this is just micro-optimization. But for larger arrays the difference is significant.

Obviously, if this is running on older browsers that do not support Set, then the way is to use indexOf or includes even.

---

Another detail is that if you are using a function to check open items, it is interesting to receive arrays todos and fechado as parameters, because this way the function becomes more generic and reusable, being able to receive any arrays (as you did, it only works for the informed arrays). Would look like this:

function verificarAbertos(todos, fechado){
    let abertos = new Set(todos);
    for (let e of fechado) {
        abertos.delete(e);
    }
    
    // converter para array
    return Array.from(abertos);
}

let todos = [08072020105842, 08072020110953, 8072020104528];
let fechado = [8072020104528];
let abertos = verificarAbertos(todos, fechado);
console.log(abertos);

// posso passar outros arrays também
console.log(verificarAbertos([1,2,3,4,5,6], [3,4,5])); // [1, 2, 6]

Answer 2

Utilize filter with includes, where this filter will be satisfactory only when it has contained the tested value, example:

var todos = [08072020105842, 08072020110953, 8072020104528];
var fechado = [8072020104528];
var abertos = todos.filter(x => fechado.includes(x)); 

console.log(abertos);

^References:

• filter
• includes

Answer 3

Your algorithm is wrong, it checks which elements in todos are also closed, when in fact should check which are NOT closed.

function arrayabertas(){
  var todos = [08072020105842, 08072020110953, 8072020104528];
  var fechado = [8072020104528];
  var abertos = [];

  // Verifica se cada número jogado
  // está na lista dos sorteados
  for(var i=0; i<todos.length; i++) {
    // verifica qual em "todos" não está fechado
    if(fechado.indexOf(todos[i]) === -1) {
      abertos.push(todos[i]);
    }
  }
  return(abertos);
}

You can use the function filter to make the code cleaner

function arrayabertas(){
  var todos = [08072020105842, 08072020110953, 8072020104528];
  var fechado = [8072020104528];
  var abertos = [];

  abertos = todos.filter((x) => fechado.indexOf(x) === -1)

  return(abertos);
}