如何在Scala中查找滿足兩個謂詞的兩個元素？

Question

假設有一個List[A]和兩個謂詞p1: A => Boolean和p2: A => Boolean。
我需要在列表中找到兩個要素：a1滿足p1及第一要素a2滿足p2的第一個元素（在我的情況a1 != a2）

很顯然，我可以運行find兩次，但我想這樣做的一通。你如何在Scala中通過一次？

Answer 1

因此，這裏有一個嘗試。這是相當直接推廣其採取謂詞的列表（並返回找到的元素列表）

def find2[A](xs: List[A], p1: A => Boolean, p2: A => Boolean): (Option[A], Option[A]) = { 
    def find2helper(xs: List[A], p1: A => Boolean, p2: A => Boolean, soFar: (Option[A], Option[A])): (Option[A], Option[A]) = { 
    if (xs == Nil) soFar 
    else { 
     val a1 = if (soFar._1.isDefined) soFar._1 else if (p1(xs.head)) Some(xs.head) else None 
     val a2 = if (soFar._2.isDefined) soFar._2 else if (p2(xs.head)) Some(xs.head) else None 
     if (a1.isDefined && a2.isDefined) (a1, a2) else find2helper(xs.tail, p1, p2, (a1, a2)) 
    } 
    } 
    find2helper(xs, p1, p2, (None, None)) 

} //> find2: [A](xs: List[A], p1: A => Boolean, p2: A => Boolean)(Option[A], Option[A]) 

    val foo = List(1, 2, 3, 4, 5) //> foo : List[Int] = List(1, 2, 3, 4, 5) 

    find2[Int](foo, { x: Int => x > 2 }, { x: Int => x % 2 == 0 }) 
    //> res0: (Option[Int], Option[Int]) = (Some(3),Some(2)) 
    find2[Int](foo, { x: Int => x > 2 }, { x: Int => x % 7 == 0 }) 
    //> res1: (Option[Int], Option[Int]) = (Some(3),None) 
    find2[Int](foo, { x: Int => x > 7 }, { x: Int => x % 2 == 0 }) 
    //> res2: (Option[Int], Option[Int]) = (None,Some(2)) 
    find2[Int](foo, { x: Int => x > 7 }, { x: Int => x % 7 == 0 }) 
    //> res3: (Option[Int], Option[Int]) = (None,None) 

廣義的版本（這實際上是略微清晰，我認爲）

def findN[A](xs: List[A], ps: List[A => Boolean]): List[Option[A]] = { 
    def findNhelper(xs: List[A], ps: List[A => Boolean], soFar: List[Option[A]]): List[Option[A]] = { 
    if (xs == Nil) soFar 
    else { 
     val as = ps.zip(soFar).map { 
      case (p, e) => if (e.isDefined) e else if (p(xs.head)) Some(xs.head) else None 
     } 
     if (as.forall(_.isDefined)) as else findNhelper(xs.tail, ps, as) 
    } 
    } 
    findNhelper(xs, ps, List.fill(ps.length)(None)) 

} //> findN: [A](xs: List[A], ps: List[A => Boolean])List[Option[A]] 

val foo = List(1, 2, 3, 4, 5) //> foo : List[Int] = List(1, 2, 3, 4, 5) 

findN[Int](foo, List({ x: Int => x > 2 }, { x: Int => x % 2 == 0 })) 
//> res0: List[Option[Int]] = List(Some(3), Some(2)) 
findN[Int](foo, List({ x: Int => x > 2 }, { x: Int => x % 7 == 0 })) 
//> res1: List[Option[Int]] = List(Some(3), None) 
findN[Int](foo, List({ x: Int => x > 7 }, { x: Int => x % 2 == 0 })) 
//> res2: List[Option[Int]] = List(None, Some(2)) 
findN[Int](foo, List({ x: Int => x > 7 }, { x: Int => x % 7 == 0 })) 
//> res3: List[Option[Int]] = List(None, None) 

Answer 2

scala> val l = List(1,2,3) 
l: List[Int] = List(1, 2, 3) 

scala> val p1 = {x:Int => x % 2 == 0} 
p1: Int => Boolean = <function1> 

scala> val p2 = {x:Int => x % 3 == 0} 
p2: Int => Boolean = <function1> 

scala> val pp = {x:Int => p1(x) || p2(x) } 
pp: Int => Boolean = <function1> 

scala> l.find(pp) 
res2: Option[Int] = Some(2) 

scala> l.filter(pp) 
res3: List[Int] = List(2, 3) 

Answer 3

這是否適合您？

def predFilter[A](lst: List[A], p1: A => Boolean, p2: A => Boolean): List[A] = 
    lst.filter(x => p1(x) || p2(x)) // or p1(x) && p2(x) depending on your need 

這將返回一個匹配任一謂詞的新列表。

val a = List(1,2,3,4,5) 
val b = predFilter[Int](a, _ % 2 == 0, _ % 3 == 0) // b is now List(2, 3, 4)