列表模式匹配添加基于案例对象的过滤答案

【问题标题】：List pattern matching add filtering based on case object列表模式匹配添加基于案例对象的过滤
【发布时间】：2021-01-17 19:24:42
【问题描述】：

我有一个男性和女性居民的列表要迭代。

如何在列表模式匹配中添加基于性别的过滤？（这样countOldManFloor 只有在居民性别为Male 时才会返回1，因此countOldManFloor(inhabitantsFemale) 将返回0）

import scala.annotation.tailrec

trait Gender

case object Female extends Gender

case object Male extends Gender

case class Inhabitant(age: Int= 50, gender: Gender)

val olderThen = 30

val inhabitantsBoth: List[Inhabitant] = List(Inhabitant(gender=Male), Inhabitant(gender=Female))
val inhabitantsFemale: List[Inhabitant] = List(Inhabitant(gender=Female), Inhabitant(gender=Female))
val inhabitantsMale: List[Inhabitant] = List(Inhabitant(gender=Male), Inhabitant(gender=Male))


@tailrec
def countOldManFloor(inhabitants: List[Inhabitant]): Int = inhabitants match {
  case inhabitant :: inhabitants  if inhabitant.age > olderThen => 1
  case inhabitant :: inhabitants => countOldManFloor(inhabitants)
  case Nil => 0
}


println(countOldManFloor(inhabitantsBoth))
println(countOldManFloor(inhabitantsMale))
println(countOldManFloor(inhabitantsFemale))

Online code

我尝试了case inhabitant: Male :: inhabitants if inhabitant.age > olderThen => 1 和= inhabitants.filter() match {}，但没有成功

【问题讨论】：

嗨@techkuz，您想通过Gender 和age 进行检查，还是两者兼而有之？我不明白。
嗨@Chema。两者，我只需要计算年龄过滤的男性
嗨@techkuz，如果我没记错的话，结果将是30岁以上的男性。

标签： scala functional-programming pattern-matching tail-recursion

【解决方案1】：

您可以在模式中匹配模式。在这种情况下，Male 模式，在 Inhabitant() 模式内，在 List 的 :: 模式内。

@tailrec
def countOldManFloor(inhabitants : List[Inhabitant]
                    ,ageLimit    : Int
                    ,acc         : Int = 0): Int = inhabitants match {
  case Inhabitant(age,Male) :: tl if age > ageLimit => 
                  countOldManFloor(tl, ageLimit, acc + 1)
  case _ :: tl => countOldManFloor(tl, ageLimit, acc)
  case Nil     => acc
}

countOldManFloor(inhabitantsBoth, olderThan)    // 1
countOldManFloor(inhabitantsMale, olderThan)    // 2
countOldManFloor(inhabitantsFemale, olderThan)  // 0

请注意，我将olderThan 设为传递参数。在定义空间之外引用变量的方法是代码异味。

【讨论】：

【解决方案2】：

我知道您需要的是 30 岁以上的男性计数器，所以我添加了一个性别条件检查。

  def countOldManFloor(inhabitants: List[Inhabitant]): Int = {
    def checkGender(inhabitant: Gender): Boolean = inhabitant match {
      case Male => true
      case _ => false
    }

    @tailrec
    def loop(lst: List[Inhabitant], cont: Int): Int = {
      lst match {
        case Nil => cont
        case (h :: tail) if h.age > olderThen && checkGender(h.gender) => loop(tail, cont + 1)
        case _ => loop(lst.tail, cont)
      }
    }
    loop(inhabitants, 0)
  }

【讨论】：

【解决方案3】：

您的方法不起作用，因为您没有添加任何内容，只能返回 1 和 0。如果您不关心尾递归，这可能会起作用：

def countOldManFloor(inhabitants: List[Inhabitant]): Int = inhabitants match {
  case Inhabitant(age, Male) :: inhabitants if age > olderThan => countOldManFloor(inhabitants) + 1
  case inhabitant :: inhabitants => countOldManFloor(inhabitants)
  case Nil => 0
}

Scastie

我认为您可以只使用count 而不是任何递归来做到这一点：

def countOldManFloor(inhabitants: List[Inhabitant]): Int = 
  inhabitants.count(inhabitant => inhabitant.age > olderThan && inhabitant.gender == Male)

Scastie

【讨论】：