Cross product in Scala | 易学教程

问题

I want to have a binary operator cross (cross-product/cartesian product) that operates with traversables in Scala:

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'))

What is the best way to do this in Scala only (i.e. not using something like scalaz)?

回答1:

You can do this pretty straightforwardly with an implicit class and a for-comprehension in Scala 2.10:

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")

And now:

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

This is possible before 2.10—just not quite as concise, since you'd need to define both the class and an implicit conversion method.

You can also write this:

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

If you want xs cross ys cross zs to return a Tuple3, however, you'll need either a lot of boilerplate or a library like Shapeless.

回答2:

cross x_list and y_list with:

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

回答3:

Here is the implementation of recursive cross product of arbitrary number of lists:

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))

回答4:

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))

回答5:

Here is something similar to Milad's response, but non-recursive.

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))))
}

Based off this blog post.

回答6:

Similar to other responses, just my approach.

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])

来源：https://stackoverflow.com/questions/14740199/cross-product-in-scala

标签

scala

functional-programming

cartesian-product

cross-product