Swift - Compare array with same array in swift and if find similar values then add or combine or sum values

This code should work:

class SomeClass: NSObject {

    var displays: Int
    var id: String

    init(_ id: String, _ displays: Int) {
        self.id = id
        self.displays = displays
    }
}


//Example of filtering

let array = [
    SomeClass("1", 1),
    SomeClass("3", 3),
    SomeClass("4", 7),
    SomeClass("8", 8),
    SomeClass("2", 3),
    SomeClass("7", 2),
    SomeClass("5", 5),
    SomeClass("1", 1),
    SomeClass("3", 4),
    SomeClass("2", 2),
    SomeClass("7", 5),
    SomeClass("6", 8)
]

var resultDictionary: [String : Int] = [:]

for element in array {
    let lastDisplays = resultDictionary[element.id] ?? 0 //if object with this id is first time counted, the resultDictionary[element.id] will return nil and then lastDisplays will be equal to 0

    resultDictionary[element.id] = lastDisplays + element.displays
}

var result = resultDictionary.map { SomeClass($0.key, $0.value) }

In the end the result array will contain SomeClass objects with counted displays.

Because dictionary have only unique keys, we can use it for counting displays sum.

Try this approach:

class SomeClass: NSObject {
    internal var displays: Int?
    internal var id: String?

    init(displays: Int?, id: String?) {
        self.displays = displays
        self.id = id
    }
}

var array = [
    SomeClass(displays: 2, id: "123"),
    SomeClass(displays: 3, id: "456"),
    SomeClass(displays: 4, id: "123"),
]

var counts: [String: Int] = [:]

for obj in array {
    if let id = obj.id, displays = obj.displays {
        let prevDisplays = counts[id] ?? 0
        counts[id] = prevDisplays + displays
    }
}

print(counts)

EDIT: If you would like to have more functional version of above, I can only come up with the following:

counts = array.reduce([String: Int]()) { dict, item in
    var dict = dict
    if let id = item.id, displays = item.displays {
        dict.updateValue((dict[id] ?? 0) + displays, forKey: id)
    }
    return dict
}