Swift 中的函数式思维？

Question

我开始学习 Haskell 和 Swift。我想知道这是否是 "functional" 方式的正确思考方式？问题是创建卡片组：我需要遍历花色和等级——为每个花色创建具有给定花色和等级的卡片。 "imperative" 方式是：

let suits: Array<Character> = ...
let ranks: Array<Int> = ...

var cards: [Card]
for suit in suits {
    for rank in ranks {
        cards.addObject(Card(suit: suit, rank: rank))
    }
}

然后我尝试使用递归的纯函数，它有效但是，可以用更少的代码来完成吗？对我来说 Swift 中的 "functional" 可读性较差，或者我做错了......

let cards = cardsWithSuits(suits, ranks, [Card]());

func cardsWithSuits(suits: [Character], ranks: [Int], cards: [Card]) -> [Card] {
   if suits.count == 0 { return cards }
   let suit: Character = head(suits)!
   let acc = cardsWithRanks(ranks, suit, cards)

   return cardsWithSuits(drop(1, suits), ranks, acc)
}

func cardsWithRanks(ranks: [Int], suit: Character, cards: [Card]) -> [Card] {
   if ranks.count == 0 { return cards }
   let acc = cards + [Card(suit: suit, rank: head(ranks)!)]

   return cardsWithRanks(drop(1, ranks), suit, acc)
}

Answer 1

基于 Haskell 的应用概念以及 <$> 和 <*> 的用法，您可能会发现以下内容通常有用（我认为我翻译正确，尽管它基于在数组而不是序列上）：

// use <^> because <$> is already used
infix operator <^> { associativity left }
public func <^> <T, U>(left:(T)->U, right:[T]) -> [U] {
    return map(right) { return left([=10=]) }
}

public func flatten<T>(input:[[T]]) -> [T] {
    return input.reduce([], +)
}

infix operator <*> { associativity left }
public func <*> <T, U>(left:[(T)->U], right:[T]) -> [U] {
    return flatten(map(left) { (function) -> [U] in
        return map(right) { return function([=10=]) }
    })
}

然后允许您使用以下内容：

let suits : [Character] = [ "C", "D", "H", "S"]
let ranks = Array(2...14)

struct Card {
    let suit : Character
    let rank : Int

    static func build(suit:Character)(rank:Int) -> Card {
        return Card(suit: suit, rank:rank)
    }
}

Card.build <^> suits <*> ranks

Answer 2

这可能不漂亮，我不认为它是函数式编程，但它的代码更少，而且它使用了 Swift 非常棒的 map 和 reduce 函数：

struct Card {
    let suit: String
    let rank: Int
}

let cards = ["Heart", "Diamond", "Club", "Spade"].reduce([Card]()) { (cards, suit) in
    return cards + map(1...13) { rank in return Card(suit: suit, rank: rank) }
}

Answer 3

Swift 不是函数式语言。但是，您可以编写 swift 函数式代码。

func map<T: Collection, U>( _ transform: (T.Iterator.Element) -> U, _ xs: T) -> [U] {
    return xs.reduce([U](), {[=10=] + [transform()]})
}

func concatMap<A, B> (_ process: (A)->[B], _ xs: [A]) -> [B] {
    return xs.reduce([B](), {[=10=] + process()})
}

infix operator <*>
func <*><A, B>(_ xs: [A], _ ys: [B]) -> [(A, B)]{
    let transform: (A, B) -> (A, B) = {([=10=], )}
    return concatMap({x in map({transform(x, [=10=])}, ys)}, xs)
}

struct Card {
    let suit : Character
    let rank : Int

    static func build(_ sr: (s:Character, r:Int)) -> Card {
        return Card(suit: sr.0, rank: sr.1)
    }
}

func test() {
    let suits : [Character] = [ "C", "D", "H", "S"]
    let ranks = Array(1...13)
    let cards = map(Card.build, suits <*> ranks)
    print(cards)
}

如果你想学习Haskell和Swift，你可以参考https://github.com/unchartedworks/HaskellSwift

Swift 中的函数式思维？

Functional thinking in Swift?

haskell

functional-programming

swift