Flatten [Any] Array Swift

Solution 1:

extension Collection where Element == Any {
    var joined: [Any] { flatMap { ($0 as? [Any])?.joined ?? [$0] } }
    func flatMapped<T>(_ type: T.Type? = nil) -> [T] { joined.compactMap { $0 as? T } }
}

let objects: [Any] = [1,[2,3],"a",["b",["c","d"]]]
let joined = objects.joined()   // [1, 2, 3, "a", "b", "c", "d"]

let integers = objects.flatMapped(Int.self)  // [1, 2, 3]
// setting the type explicitly
let integers2: [Int] = objects.flatMapped()        // [1, 2, 3]
// or casting
let strings = objects.flatMapped() as [String]     // ["a", "b", "c", "d"]

Solution 2:

There may be a better way to solve this but one solution is to write your own extension to Array:

extension Array {
    func anyFlatten() -> [Any] {
        var res = [Any]()
        for val in self {
            if let arr = val as? [Any] {
                res.append(contentsOf: arr.anyFlatten())
            } else {
                res.append(val)
            }
        }

        return res
    }
}

let numbers = [1,[2, [4, 5] ,3], "Hi"] as [Any]
print(numbers.anyFlatten())

Output:

[1, 2, 4, 5, 3, "Hi"]

This solution will handle any nesting of arrays.

Solution 3:

Here's an alternate implementation of @rmaddy's anyFlatten:

It can be most concisely written like so, but it's quite cryptic:

extension Array {
    func anyFlatten() -> [Any] {
        return self.flatMap{ ($0 as? [Any]).map{ $0.anyFlatten() } ?? [$0] }
    }
}

Here's a more reasonable implementation:

extension Array {
    func anyFlatten() -> [Any] {
        return self.flatMap{ element -> [Any] in
            if let elementAsArray = element as? [Any] { return elementAsArray.anyFlatten() }
            else { return [element] }
        }
    }
}