两个方法在一个scala中

Question

从Scala开始我的第一个项目：一个扑克框架。

所以我有以下的类

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

而且其中包含两个方法，做几乎同样的事情utils的对象：他们算牌数

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]()) 
} 

是每个等级或花色有什么办法可以通过“field/method-as-parameter”将这两种方法“统一”到一个方法中？

感谢

Answer 1

是的，这将需要高阶函数（即函数，它的功能参数）和类型参数/泛型

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}) 

你做不需要类型参数A，你可以强制该方法只在卡上工作，但这是可惜的。

对于额外的信用，你可以用两个参数列表，并有

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

是斯卡拉的类型推断有点特殊性，如果用两个参数列表做，你就不会需要键入卡参数定义函数时：

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

或只是

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

当然，库可以帮助您实施

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

虽然手工制作的实施可能会稍快。

A小调笔记，卡片应被宣布为case class：

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