如何在scala中编写元组范围函数

Question

我想要以下功能range((1,1), (2,2))返回

Seq[(Int,Int)]((1,1),(1,2),(2,1),(2,2))

与1 to 2

的一维范围相似

该函数应该适用于任何scala元组(即Tuple2, Tuple3, Tuple4，…)并且是类型安全的。

我试过了

    def tupleRange[T <: Product](t1:T, t2:T):Seq[T] = {
        assert(t1.productArity == t2.productArity)
        def tail(t:Product):Product = sys.error("todo"); 
        def join(i:Int, p:Product):T = sys.error("todo");
        for(
v <- t1.productElement(0).asInstanceOf[Int] to t2.productElement(0).asInstanceOf[Int]; 
v2 <- tupleRange(tail(t1), tail(t2)))
            yield join(v,v2)
    }
    implicit def range[T <:Product](p1:T) = new { def to(p2:T) = tupleRange(p1,p2)}

但是我想我选错方向了。

Answer 1

我的建议与@ziggystar上面的建议相同。使用List[Int]代替Int的元组

scala> import scalaz._
import scalaz._
scala> import Scalaz._
import Scalaz._
scala> def range(xs: List[Int], ys: List[Int]): List[List[Int]] = {
     |   (xs, ys).zipped.map((x, y) => List.range(x, y + 1)).sequence
     | }
range: (xs: List[Int], ys: List[Int])List[List[Int]]
scala> range(List(1, 2, 4), List(2, 5, 6))
res29: List[List[Int]] = List(List(1, 2, 4), List(1, 2, 5), List(1, 2, 6), 
List(1, 3, 4), List(1, 3, 5), List(1, 3, 6), List(1, 4, 4), List(1, 4, 5), 
List(1, 4, 6), List(1, 5, 4), List(1, 5, 5), List(1, 5, 6), List(2, 2, 4), 
List(2, 2, 5), List(2, 2, 6), List(2, 3, 4), List(2, 3, 5), List(2, 3, 6), 
List(2, 4, 4), List(2, 4, 5), List(2, 4, 6), List(2, 5, 4), List(2, 5, 5), 
List(2, 5, 6))

这个实现假设xs和ys是有序的并且长度相同。

Answer 2

首先，考虑一下:

scala> 1 to 10
res0: scala.collection.immutable.Range.Inclusive = Range(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

对于元组来说有这样的东西是很好的，对吧?

class RangedTuple(t: Tuple2[Int, Int]) {
  def to(t2: Tuple2[Int, Int]) = {
    (t, t2) match {
      case ((a1: Int, a2: Int), (b1: Int, b2: Int)) => {
        (for {
          i <- a1 to b1
        } yield (a1 to b1).map(j => (i, j))).flatMap(k => k)
      }
    }
  }
}
implicit def t2rt(t: Tuple2[Int, Int]) = new RangedTuple(t)

显示如下内容:

scala> (1, 1) to (2, 2)
res1: scala.collection.immutable.IndexedSeq[(Int, Int)] = Vector((1,1), (1,2), (2,1), (2,2))
scala> (1, 1) to (3, 3)
res2: scala.collection.immutable.IndexedSeq[(Int, Int)] = Vector((1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3))

这对你有用吗?

Answer 3

对于每个元组，您需要一个不同的版本(但是您可以使用一个预处理器来生成每个版本)。这是我的实现(这是懒惰的):

def range2( range: Range ): Seq[(Int,Int)] = range.toStream.map( i => (i,i) )

你可以这样使用:

scala> range2( 1 to 10 )
res3: Seq[(Int, Int)] = Stream((1,1), ?)

Answer 4

简单的方式，大量的剪切和粘贴，重载每个元组的方法:

def range(r: (Int, Int), s: (Int, Int)) = 
  for { p1 <- r._1 to s._1
        p2 <- r._2 to s._2 } yield (p1, p2)
def range(r: (Int, Int, Int), s: (Int, Int, Int)) = 
  for { p1 <- r._1 to s._1
        p2 <- r._2 to s._2 
        p3 <- r._3 to s._3 } yield (p1, p2, p3)
def range(r: (Int, Int, Int, Int), s: (Int, Int, Int, Int)) =    
  for // etc up to 22

另外:

def range(p1: Product, p2: Product) = {
  def toList(t: Product): List[Int] = 
    t.productIterator.toList.map(_.asInstanceOf[Int])
  def toProduct(lst: List[Int]) = lst.size match {
    case 1 => Tuple1(lst(0))
    case 2 => Tuple2(lst(0), lst(1))
    case 3 => Tuple3(lst(0), lst(1), lst(2))
    //etc up to 22
  }
  def go(xs: List[Int], ys: List[Int]): List[List[Int]] = {
    if(xs.size == 1 || ys.size == 1) (xs.head to ys.head).toList.map(List(_))
    else (xs.head to ys.head).toList.flatMap(i => go(xs.tail, ys.tail).map(i :: _))
  }
  go(toList(p1), toList(p2)) map toProduct
} 

似乎有效:

scala> range((1,2,4), (2,5,6))
res66: List[Product with Serializable] = List((1,2,4), (1,2,5), (1,2,6), 
(1,3,4), (1,3,5), (1,3,6), (1,4,4), (1,4,5), (1,4,6), (1,5,4), (1,5,5),
(1,5,6), (2,2,4), (2,2,5), (2,2,6), (2,3,4), (2,3,5), (2,3,6), (2,4,4), 
(2,4,5), (2,4,6), (2,5,4), (2,5,5), (2,5,6))

你的基本问题是，因为Scala是静态类型的，这个方法需要有一个返回类型，所以你永远不可能有一个方法同时返回Seq[(Int, Int)]和Seq[(Int, Int, Int)]以及元组的所有其他属性。您所能做的最好的事情是使用覆盖所有输出的最接近的类型，在本例中是Product with Serializable。你当然可以对结果进行强制转换，例如res0.map(_.asInstanceOf[(Int, Int, Int)]) .

像第一个例子那样重载方法允许您为每个元素使用不同的返回类型，因此您不需要进行任何强制类型转换。

如何使用Iterator，并使用两个Seq而不是两个元组进行初始化?

这是一个类Cartesian，它扩展了Iterator。

def rangeIterator (froms: Seq[Int], tos: Seq[Int]) = {
  def range (froms: Seq[Int], tos: Seq[Int]) : Seq[Seq[Int]] = 
    if (froms.isEmpty) Nil else 
    Seq (froms.head to tos.head) ++ range (froms.tail, tos.tail) 
  new Cartesian (range (froms, tos))
}  

用法:

scala> val test = rangeIterator (Seq(1, 1), Seq(2, 2))
test: Cartesian = non-empty iterator    
scala> test.toList 
res38: List[Seq[_]] = List(List(1, 1), List(2, 1), List(1, 2), List(2, 2))
scala> val test = rangeIterator (Seq(1, 0, 9), Seq(2, 2, 11))
test: Cartesian = non-empty iterator
scala> test.toList 
res43: List[Seq[_]] = List(List(1, 0, 9), List(2, 0, 9), List(1, 1, 9), List(2, 1, 9), List(1, 2, 9), List(2, 2, 9), List(1, 0, 10), List(2, 0, 10), List(1, 1, 10), List(2, 1, 10), List(1, 2, 10), List(2, 2, 10), List(1, 0, 11), List(2, 0, 11), List(1, 1, 11), List(2, 1, 11), List(1, 2, 11), List(2, 2, 11))