Merge arrays with condition

Question

I would like to merge two arrays with specific condition and update objects that they are containing.

First my struct that is in arrays:

struct Item          


        

Answer 1

The map function cannot directly mutate its elements. And since you're using structs (passed by value), it wouldn't work anyway, because the version you see in $0 would be a different instance than the one in the array. To use map correctly, I'd use a closure like this:

fisrtArray = zip(fisrtArray, secondArray).map() {
   return Item(id: $0.id, name: $1.name, value: $0.value)
}

This produces the result you're expecting.

Now, if your structs were objects (value types instead of reference types), you could use forEach and do the $0.name = $1.name in there.

Answer 2

The following code does work independently by the order of the elements inside the 2 arrays

firstArray = firstArray.map { (item) -> Item in
    guard
        let index = secondArray.index(where: { $0.id == item.id })
        else { return item }
    var item = item
    item.name = secondArray[index].name
    return item
}

"[Item(id: 1, name: "Bogdan", value: 3), Item(id: 2, name: "Max", value: 5)]\n"

Update

The following version uses the first(where: method as suggested by Martin R.

firstArray = firstArray.map { item -> Item in
    guard let secondElm = secondArray.first(where: { $0.id == item.id }) else { return item }
    var item = item
    item.name = secondElm.name
    return item
}

Answer 3

A solution for your specific problem above would be:

struct Item {
    var id: Int
    var name: String
}

let first = Item(id: 1, name: "Oleg")
let second = Item(id: 2, name: "Olexander")

let firstInSecond = Item(id: 1, name: "Bogdan")
let secondInSecond = Item(id: 2, name: "Max")

let ret = zip([first, second], [firstInSecond, secondInSecond]).map({
    return $0.id == $1.id ? $1 : $0
})

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=> But it requires that there are as many items in the first as in the second array - and that they have both the same ids in the same order...