[FIXED] How to extract multiple sets within a single set to become their own individual array in Swift?

Issue

let groupA : Set<Int> = [1111,2000]
let groupB : Set<Int> = [221,122]
let groupC : Set<Int> = [300,12,23,232] 


StudentGroups = [groupA, groupB, groupC]

Say that I can't access the declared sets above, I'm trying to find a solution to extract out each sets inside StudentGroups to become their new own array.

For example: The expected extracted arrays from studentGroups should be:

groupA = [1111,2000]
groupB = [221,122]
groupC = [300,12,23, 232] 

This is what I did to test my progress:

for idx in StudentGroups.indices {
   let elem = StudentGroups[idx]
    print(elem)

//This will return  [1111,2000]
//                  [221,122]
//                  [300,12,23,232]

}
        

At the moment, it's only iterating through the sets. Unfortunately, I'm a bit clueless on how to proceed from here

Solution

try something like this:

let groupA : Set<Int> = [1111,2000]
let groupB : Set<Int> = [221,122]
let groupC : Set<Int> = [300,12,23,232]

let studentGroups = [groupA, groupB, groupC]
print("---> studentGroups: \(studentGroups)")

for idx in studentGroups.indices {
    let arr = Array(studentGroups[idx])
    print("\n---> arr \(idx): \(arr)")
    for i in arr.indices {
        print("     ---> element \(i) of arr \(idx): \(arr[i])")
    }
}

        

...a solution to extract out each sets inside StudentGroups to become their new own array..., try this, to transform an array of Sets into an array of arrays:

var newArr: [[Int]] = studentGroups.map{ Array($0) }

print("\n---> newArr: \(newArr)")
print("\n---> newArr[0][0]: \(newArr[0][0])")

Answered By - workingdog support Ukraine

Answer Checked By - Marie Seifert (PHPFixing Admin)