拼合[Any] Array Swift

Question

使用这个堆栈溢出问题，我有以下代码。

let numbers = [1,[2,3]] as [Any]
var flattened = numbers.flatMap { $0 }
print(flattened) // [1, [2, 3]]

取而代之的扁平设定为[1, [2, 3]]我希望它是[1, 2, 3]。

在Swift中最简单/最干净的方法是什么？

Answer 1

extension Collection {
    func joined() -> [Any] {
        return flatMap { ($0 as? [Any])?.joined() ?? [$0] }
    }
    func flatMapped<T>(with type: T.Type? = nil) -> [T] {
        return 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(with: 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"]

Answer 2

可能有更好的方法来解决这个问题，但一种解决方案是编写您自己的 Array 扩展：

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())

输出：

[1, 2, 4, 5, 3, “嗨”]

此解决方案将处理任何数组嵌套。

Answer 3

这是 @rmaddy 的替代实现anyFlatten：

它可以最简洁地写成这样，但它相当神秘：

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

这是一个更合理的实现：

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