JS 7

• Will add smaller content to compensate

A Set() is a list-like data structure that can't be accessed through index like arrays:

//we add new elements to a set with .add()

let set= new Set()
set.add(1)
set.add( {tin: 10, ten:"dieci"})

It can't have more than one strict equal element (===):

//we can use .size property to count the elements in it

let uno= new Set()
uno.add(1)
uno.add(1)
uno.add({uno: 1})
uno.add({uno: 1)}

uno.size            //Set(3) {1, {…}, {…}}
//the 2 objects arent strict equals so it will add

We use add(), delete() and has() methods on Sets:

//.has() returns a boolean if the element is present in the Set

const rik= new Set()
rik.add( {altro: "fatto"})
rik.add("non e vero")

rik.has("NON E VERO" )                 //false
rik.has("NON E VERO".toLowerCase() )   //true
rik.has({altro: "fatto"})              //false

//.delete() will remove said element from the set
rik.delete( "non e vero" )

We can use both the forEach() and for() loop on Sets:

//we could also use rik.keys() and rik.values() but it will work the same

for(const x of rik) {
  console.log(x);
}

rik.forEach((x)=>{
  console.log(x)
})

To pass from Set to Array and vice-versa:

//By becoming a set it will also eliminate its duplicates

let sort= [1,1,1,1,2,3,4,5,5,3,2,1,2,3]
let set= new Set(sort)        //Set(5) {1, 2, 3, 4, 5}

//to then go back to arrays

[...new Set(sore)]            //[1, 2, 3, 4, 5]
Array.from( new Set(sore))    //[1, 2, 3, 4, 5]

We can use Sets to get intersections and difference sets:

//we start from 2 Sets, we need to convert one into an array to use .filter(9
//while the other will pass elements from the second that return true if on the first

const intersection = new Set([...rik].filter((x) => seconn.has(x) ));

const difference = new Set([...rik].filter((x) => !seconn.has(x)));
//opposite operation with difference