Question

我开始学习Haskell和Swift。我想知道它是否是正确的思维方式＆＃34; functional＆＃34;办法？问题是创建卡片组：我需要循环穿西装和排名 - 为每件西装创造出具有特定西装和等级的卡片。在＆＃34;命令式＆＃34;它的方式是：

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，它可以工作但是，可以用更少的代码完成吗？对我来说＆＃34;功能性＆＃34;在Swift中可读性较差，或者我可能做错了......

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

在使用Haskell的应用概念以及<$>和<*>的基础上，您可能会发现以下内容通常很有用（我认为我已经正确翻译了，尽管它已经＆＃39; s基于数组而不是序列）：

// use <^> because <$> is already used
infix operator <^> { associativity left }
public func <^> <T, U>(left:(T)->U, right:[T]) -> [U] {
    return map(right) { return left($0) }
}

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($0) }
    })
}

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

这可能不太漂亮，我不认为它是函数式编程，但它代码较少，它使用了Swift的极好的映射并减少了函数：

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不是一种功能语言。但是，您可以在函数式中编写快速代码。

func map<T: Collection, U>( _ transform: (T.Iterator.Element) -> U, _ xs: T) -> [U] {
    return xs.reduce([U](), {$0 + [transform($1)]})
}

func concatMap<A, B> (_ process: (A)->[B], _ xs: [A]) -> [B] {
    return xs.reduce([B](), {$0 + process($1)})
}

infix operator <*>
func <*><A, B>(_ xs: [A], _ ys: [B]) -> [(A, B)]{
    let transform: (A, B) -> (A, B) = {($0, $1)}
    return concatMap({x in map({transform(x, $0)}, 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)
}

如果您想学习Haskell和Swift，可以参考https://github.com/unchartedworks/HaskellSwift

Swift中的功能性思维？

3 个答案: