一个scala中的两种方法

Question

开始我的第一个 Scala 项目：一个扑克框架。

所以我有以下class

class Card(rank1: CardRank, suit1: Suit){
val rank = rank1
val suit = suit1
}

还有一个 Utils 对象，它包含两个做几乎相同事情的方法：它们计算每个等级或花色的牌数

def getSuits(cards: List[Card]) = {

def getSuits(cards: List[Card], suits: Map[Suit, Int]): (Map[Suit, Int]) = {
  if (cards.isEmpty)
    return suits

  val suit = cards.head.suit
  val value = if (suits.contains(suit)) suits(suit) + 1 else 1
  getSuits(cards.tail, suits + (suit -> value))

}

getSuits(cards, Map[Suit, Int]())

}


def getRanks(cards: List[Card]): Map[CardRank, Int] = {

def getRanks(cards: List[Card], ranks: Map[CardRank, Int]): Map[CardRank, Int] = {
  if (cards isEmpty)
    return ranks

  val rank = cards.head.rank
  val value = if (ranks.contains(rank)) ranks(rank) + 1 else 1
  getRanks(cards.tail, ranks + (rank -> value))
}

getRanks(cards, Map[CardRank, Int]())
}

有什么办法可以 "unify" 这两种方法与 "field/method-as-parameter" 合而为一吗？

谢谢

Answer 1

是的，这需要高阶函数（即以函数为参数的函数）和类型 parameters/genericity

def groupAndCount[A,B](elements: List[A], toCount: A => B): Map[B, Int] = {
   // could be your implementation, just note key instead of suit/rank 
   // and change val suit = ... or val rank = ... 
   // to val key = toCount(card.head)
}

然后

def getSuits(cards: List[Card]) = groupAndCount(cards, {c : Card => c.suit})
def getRanks(cards: List[Card]) = groupAndCount(cards, {c: Card => c.rank})

你不需要类型参数A，你可以强制该方法只对Card起作用，但那会很遗憾。

为了加分，您可以使用两个参数列表，并且

def groupAndCount[A,B](elements: List[A])(toCount: A => B): Map[B, Int] = ...

这是带有类型推断的 Scala 的一个小特点，如果你使用两个参数列表，你将不需要在定义函数时键入卡片参数：

def getSuits(cards: List[Card]) = groupAndCount(cards)(c => c.suit)

或者只是

def getSuits(cards: List[Card] = groupAndCount(cards)(_.suit)

当然，图书馆可以帮助您实施

def groupAndCount[A,B](l: List[A])(toCount: A => B) : Map[A,B] =
   l.groupBy(toCount).map{case (k, elems) => (k, elems.length)}

虽然手工实现可能会稍微快一些。

小记，卡片应声明为case class :

case class Card(rank: CardRank, suit: Suit) 
// declaration done, nothing else needed

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