Scala method to combine each element of an iterable with each element of another?

Question

If I have this:

val a = Array(\"a \",\"b \",\"c \")
val b = Array(\"x\",\"y\")

I would like to know if such a method exists which would let

Answer 1

If you want to show off your deep knowledge of higher kinded types and category theory, you can write:

trait Applicative[App[_]] {
  def pure[A](a: A): App[A]
  def fmap[A,B](f: A => B, appA: App[A]): App[B]
  def star[A,B](appF: App[A => B], appA: App[A]): App[B]
}

object ListApplicative extends Applicative[List] {
  override def pure[A](a: A): List[A] = List(a)
  override def fmap[A,B](f: A => B, listA: List[A]): List[B] = listA.map(f)
  override def star[A,B](listF: List[A => B], listA: List[A]):List[B] = 
    for(f <- listF; a <- listA) yield f(a)
}

import ListApplicative._

def pairs[A,B](listA: List[A], listB: List[B]) = 
  star(fmap((a:A) => ((b:B) => (a,b)), listA), listB)

Other than that I would prefer pst's solution...

Answer 2

For a list of a unknown number of lists, of different length, and for maybe different types, you can use this:

def xproduct (xx: List [List[_]]) : List [List[_]] = 
  xx match {
    case aa :: bb :: Nil => 
      aa.map (a => bb.map (b => List (a, b))).flatten       
    case aa :: bb :: cc => 
      xproduct (bb :: cc).map (li => aa.map (a => a :: li)).flatten
    case _ => xx
}

You would call it

xproduct List (List ("a ", "b ", "c "), List ("x", "y"))

but can call it with Lists of different kind too:

scala>  xproduct (List (List ("Beatles", "Stones"), List (8, 9, 10), List ('$', '€')))  
res146: List[List[_]] = List(List(Beatles, 8, $), List(Stones, 8, $), List(Beatles, 8, €), List(Stones, 8, €), List(Beatles, 9, $), List(Stones, 9, $), List(Beatles, 9, €), List(Stones, 9, €), List(Beatles, 10, $), List(Stones, 10, $), List(Beatles, 10, €), List(Stones, 10, €))

Arrays have to be converted to Lists, and the result converted back to Arrays, if you can't use Lists.

update:

On the way towards a lazy collection, I made a functional mapping from an index (from 0 to combination-size - 1) to the result at that position, easily calculated with modulo and division, just a bit concentration is needed:

def combicount (xx: List [List[_]]): Int = (1 /: xx) (_ * _.length)

def combination (xx: List [List[_]], i: Int): List[_] = xx match {
    case Nil => Nil
    case x :: xs => x(i % x.length) :: combination (xs, i / x.length)
}

def xproduct (xx: List [List[_]]): List [List[_]] = 
  (0 until combicount (xx)).toList.map (i => combination (xx, i))

It's no problem to use a long instead, or even BigInt.

update 2, The iterator:

class Cartesian (val ll: List[List[_]]) extends Iterator [List[_]] {

  def combicount (): Int = (1 /: ll) (_ * _.length)

  val last = combicount - 1 
  var iter = 0
  
  override def hasNext (): Boolean = iter < last
  override def next (): List[_] = {
    val res = combination (ll, iter)
    iter += 1
    res
  }

  def combination (xx: List [List[_]], i: Int): List[_] = xx match {
      case Nil => Nil
      case x :: xs => x (i % x.length) :: combination (xs, i / x.length) 
  }
}

Answer 3

I'm using the following extensively in my code. Note that this is working for an arbitrary number of lists. It is creating an Iterator instead of a collection, so you don't have to store the potentially huge result in memory.

Any improvements are very welcome.

/**
  * An iterator, that traverses every combination of objects in a List of Lists.
  * The first Iterable will be incremented fastest. So consider the head as 
  * the "least significant" bit when counting.*/

class CombinationIterator[A](val components: List[Iterable[A]]) extends Iterator[List[A]]{
  private var state: List[BufferedIterator[A]] = components.map(_.iterator.buffered)
  private var depleted = state.exists(_.isEmpty)

  override def next(): List[A] = {
    //this function assumes, that every iterator is non-empty    
    def advance(s: List[(BufferedIterator[A],Iterable[A])]): List[(BufferedIterator[A],A)] = {
      if( s.isEmpty ){
        depleted = true
        Nil
      }
      else {
        assert(!s.head._1.isEmpty)

        //advance and return identity
        val it = s.head._1
        val next = it.next()
        if( it.hasNext){
          //we have simply incremented the head, so copy the rest
          (it,next) :: s.tail.map(t => (t._1,t._1.head))
        } else {
          //we have depleted the head iterator, reset it and increment the rest
          (s.head._2.iterator.buffered,next) :: advance(s.tail)
        }
      }
    }
    //zipping the iterables to the iterators is needed for resseting them
    val (newState, result) = advance(state.zip(components)).unzip
    
    //update state
    state = newState    
    
    result
  }

  override def hasNext = !depleted
}

So using this one, you have to write new CombinationIterator(List(a,b)) to obtain an iterator that goes through every combination.

Edit: based on user unkown's version

Note that the following version is not optimal (performance wise):

• indexed access into lists (use arrays instead)
• takeWhile evaluates after every element

.

scala> def combination(xx: List[List[_]], i: Int): List[_] = xx match {
     | case Nil => Nil
     | case x :: xs => x(i % x.length) :: combination(xs, i/x.length)
     | }
combination: (xx: List[List[_]], i: Int)List[_]

scala> def combinationIterator(ll: List[List[_]]): Iterator[List[_]] = {
     | Iterator.from(0).takeWhile(n => n < ll.map(_.length).product).map(combination(ll,_))
     | }
combinationIterator: (ll: List[List[_]])Iterator[List[_]]

scala> List(List(1,2,3),List("a","b"),List(0.1,0.2,0.3))
res0: List[List[Any]] = List(List(1, 2, 3), List(a, b), List(0.1, 0.2, 0.3))
    
scala> combinationIterator(res0)
res1: Iterator[List[_]] = non-empty iterator

scala> res1.mkString("\n")
res2: String = 
List(1, a, 0.1)
List(2, a, 0.1)
List(3, a, 0.1)
List(1, b, 0.1)
List(2, b, 0.1)
List(3, b, 0.1)
List(1, a, 0.2)
List(2, a, 0.2)
List(3, a, 0.2)
List(1, b, 0.2)
List(2, b, 0.2)
List(3, b, 0.2)
List(1, a, 0.3)
List(2, a, 0.3)
List(3, a, 0.3)
List(1, b, 0.3)
List(2, b, 0.3)
List(3, b, 0.3)

Answer 4

Here's one more that does the same thing as @ziggystar's last edit but doesn't use indexed access of lists.

def combinationIterator[A](xs: Iterable[Iterable[A]]): Iterator[List[A]] = {
  xs.foldRight(Iterator(List[A]())) { (heads, tails) =>
    tails.flatMap { tail =>
      heads.map(head => head :: tail)
    }
  }
}

And the sugary version:

def combinationIterator[A](xs: Iterable[Iterable[A]]): Iterator[List[A]] = {
  (xs :\ Iterator(List[A]())) { (heads, tails) =>
    for (tail <- tails; head <- heads) yield head :: tail
  }
}

Answer 5

I'm not sure of a "method", but this can be expressed with just a nested/compound for:

val a = Array("a ","b ","c ")
val b = Array("x","y")
for (a_ <- a; b_ <- b) yield (a_, b_)

res0: Array[(java.lang.String, java.lang.String)] = Array((a ,x), (a ,y), (b ,x), (b ,y), (c ,x), (c ,y))

Happy coding.