函数式编程-Scala中的跨产品

Question

我想要一个二进制运算符cross（叉积/笛卡尔积），它在Scala:中与可遍历对象一起操作

val x = Seq(1, 2)
val y = List('hello', 'world', 'bye')
val z = x cross y    # i can chain as many traversables e.g. x cross y cross w etc
assert z == ((1, 'hello'), (1, 'world'), (1, 'bye'), (2, 'hello'), (2, 'world'), (2, 'bye'))

仅在Scala中实现这一点的最佳方法是什么（即不使用类似scalaz的东西）？

Answer 1

您可以在Scala 2.10:中使用隐式类和for-理解来非常直接地实现这一点

implicit class Crossable[X](xs: Traversable[X]) {
  def cross[Y](ys: Traversable[Y]) = for { x <- xs; y <- ys } yield (x, y)
}
val xs = Seq(1, 2)
val ys = List("hello", "world", "bye")

现在：

scala> xs cross ys
res0: Traversable[(Int, String)] = List((1,hello), (1,world), ...

这在2.10之前是可能的——只是不那么简洁，因为您需要定义类和隐式转换方法。

你也可以这样写：

scala> xs cross ys cross List('a, 'b)
res2: Traversable[((Int, String), Symbol)] = List(((1,hello),'a), ...

然而，如果您希望xs cross ys cross zs返回Tuple3，则需要大量样板或像Shapeless这样的库。

Answer 2

使用交叉x_list和y_list

val cross = x_list.flatMap(x => y_list.map(y => (x, y)))

Answer 3

这里是任意数量列表的递归叉积的实现：

def crossJoin[T](list: Traversable[Traversable[T]]): Traversable[Traversable[T]] =
  list match {
    case xs :: Nil => xs map (Traversable(_))
    case x :: xs => for {
      i <- x
      j <- crossJoin(xs)
    } yield Traversable(i) ++ j
  }
crossJoin(
  List(
    List(3, "b"),
    List(1, 8),
    List(0, "f", 4.3)
  )
)
res0: Traversable[Traversable[Any]] = List(List(3, 1, 0), List(3, 1, f), List(3, 1, 4.3), List(3, 8, 0), List(3, 8, f), List(3, 8, 4.3), List(b, 1, 0), List(b, 1, f), List(b, 1, 4.3), List(b, 8, 0), List(b, 8, f), List(b, 8, 4.3))

Answer 4

猫用户的替代品：

List[List[A]]上的sequence创建交叉乘积：

import cats.implicits._
val xs = List(1, 2)
val ys = List("hello", "world", "bye")
List(xs, ys).sequence 
//List(List(1, hello), List(1, world), List(1, bye), List(2, hello), List(2, world), List(2, bye))

Answer 5

这里有一些类似于Milad的响应，但不是递归的。

def cartesianProduct[T](seqs: Seq[Seq[T]]): Seq[Seq[T]] = {
  seqs.foldLeft(Seq(Seq.empty[T]))((b, a) => b.flatMap(i => a.map(j => i ++ Seq(j))))
}

基于这篇博客文章。

Answer 6

class CartesianProduct(product: Traversable[Traversable[_ <: Any]]) {
  override def toString(): String = {
    product.toString
  }
  def *(rhs: Traversable[_ <: Any]): CartesianProduct = {
      val p = product.flatMap { lhs =>
        rhs.map { r =>
          lhs.toList :+ r
        }
      }
      new CartesianProduct(p)
  }
}
object CartesianProduct {
  def apply(traversable: Traversable[_ <: Any]): CartesianProduct = {
    new CartesianProduct(
      traversable.map { t =>
        Traversable(t)
      }
    )
  }
}
// TODO: How can this conversion be made implicit?
val x = CartesianProduct(Set(0, 1))
val y = List("Alice", "Bob")
val z = Array(Math.E, Math.PI)
println(x * y * z) // Set(List(0, Alice, 3.141592653589793), List(0, Alice, 2.718281828459045), List(0, Bob, 3.141592653589793), List(1, Alice, 2.718281828459045), List(0, Bob, 2.718281828459045), List(1, Bob, 3.141592653589793), List(1, Alice, 3.141592653589793), List(1, Bob, 2.718281828459045))
// TODO: How can this conversion be made implicit?
val s0 = CartesianProduct(Seq(0, 0))
val s1 = Seq(0, 0)
println(s0 * s1) // List(List(0, 0), List(0, 0), List(0, 0), List(0, 0))

Answer 7

与其他响应类似，只是我的方法。

def loop(lst: List[List[Int]],acc:List[Int]): List[List[Int]] = {
  lst match {
    case head :: Nil => head.map(_ :: acc)
    case head :: tail => head.flatMap(x => loop(tail,x :: acc))
    case Nil => ???
  }
}
val l1 = List(10,20,30,40)
val l2 = List(2,4,6)
val l3 = List(3,5,7,9,11)
val lst = List(l1,l2,l3)
loop(lst,List.empty[Int])

Answer 8

您可以使用application:

import cats.implicits._
val xs = Seq(1, 2)
val ys = List("hello", "world", "bye")
(xs,ys).mapN((x,y) => (x,y))

https://typelevel.org/cats/typeclasses/applicative.html