问题
Let's say I want to flatten nested lists of the same type... For example
ListA(Element(A), Element(B), ListA(Element(C), Element(D)), ListB(Element(E),Element(F)))
ListA contains nested list of the same type (ListA(Element(C), Element(D))) so I want to substitute it with the values it contains, so the result of the upper example should look like this:
ListA(Element(A), Element(B), Element(C), Element(D), ListB(Element(E),Element(F)))
Current class hierarchy:
abstract class SpecialList() extends Exp {
val elements: List[Exp]
}
case class Element(name: String) extends Exp
case class ListA(elements: List[Exp]) extends SpecialList {
override def toString(): String = "ListA("+elements.mkString(",")+")"
}
case class ListB(elements: List[Exp]) extends SpecialList {
override def toString(): String = "ListB("+elements.mkString(",")+")"
}
object ListA{def apply(elements: Exp*):ListA = ListA(elements.toList)}
object ListB{def apply(elements: Exp*):ListB = ListB(elements.toList)}
I have made three solutions that works, but I think there have to be better way to achieve this:
First solution:
def flatten[T <: SpecialList](parentList: T): List[Exp] = {
val buf = new ListBuffer[Exp]
for (feature <- parentList.elements) feature match {
case listA:ListA if parentList.isInstanceOf[ListA] => buf ++= listA.elements
case listB:ListB if parentList.isInstanceOf[ListB] => buf ++= listB.elements
case _ => buf += feature
}
buf.toList
}
Second solution:
def flatten[T <: SpecialList](parentList: T): List[Exp] = {
val buf = new ListBuffer[Exp]
parentList match {
case listA:ListA => for (elem <- listA.elements) elem match {
case listOfTypeA:ListA => buf ++= listOfTypeA.elements
case _ => buf += elem
}
case listB:ListB => for (elem <- listB.elements) elem match {
case listOfTypeB:ListB => buf ++= listOfTypeB.elements
case _ => buf += elem
}
}
buf.toList
}
Third solution
def flatten[T <: SpecialList](parentList: T): List[Exp] = parentList.elements flatMap {
case listA:ListA if parentList.isInstanceOf[ListA] => listA.elements
case listB:ListB if parentList.isInstanceOf[ListB] => listB.elements
case other => List(other)
}
My question is whether there is any better, more generic way to achieve same functionality as in all of upper three solutions there is repetition of code?
回答1:
A true functional way. Without using a variable.
def flatten[A](list: List[A]): List[A] = list match {
case Nil => Nil
case (ls: List[A]) :: tail => flatten(ls) ::: flatten(tail)
case h :: tail => h :: flatten(tail)
}
回答2:
I prefere a recursive way. In general I would do something like this:
def flattenList[A](l: List[Any]): List[A] = {
var acc = List[A]()
l foreach ( entry => entry match {
case a: List[Any] => acc = acc ::: flattenList(a)
case b: A => acc = acc :+ b
})
acc
}
This will flatten you a List(Element(A), Element(B), List(Element(C), Element(D), List(Element(E), Element(F)))) to List(Element(A), Element(B), Element(C), Element(D), Element(E), Element(F))
回答3:
In an ideal world, where the wretched type erasure would not exist, you would do something like this:
// WON'T WORK
def flatten[T <: SpecialList](parentList: T): List[Exp] = parentList.elements flatMap {
case x:T => x.elements
case somethingElse => List(somethingElse)
}
But the best solution under the circumstances is, in my opinion, this one:
def flatten[T <: SpecialList](parentList: T): List[Exp] = parentList.elements flatMap {
case x:SpecialList if x.getClass == parentList.getClass => x.elements
case somethingElse => List(somethingElse)
}
It's a bit more generic than the one proposed in the question since you don't need to bother whether the argument is a ListA or ListB and it will also work if in the future you'll add a ListC.
However, this won't solve your more general problem for flattening at arbitrary depths, since flatten(ListA(...)) must also return a ListA(...) in the end - in the case above it returns a List which looses it's initial meaning. A solution to this problem could be:
abstract class SpecialList {
val elements: List[Exp]
def flatten: SpecialList = createAnother(elements flatMap {
case x: SpecialList => {
val flattenX = x.flatten
if (flattenX.getClass == this.getClass) flattenX.elements else List(flattenX)
}
case somethingElse => List(somethingElse)
})
// Creates another special list of the same type
def createAnother(elements: List[Exp]): SpecialList
}
case class ListA(elements: List[Exp]) extends SpecialList {
override def toString: String = "ListA("+elements.mkString(",")+")"
def createAnother(elements: List[Exp]) = ListA(elements)
}
case class ListB(elements: List[Exp]) extends SpecialList {
override def toString: String = "ListB("+elements.mkString(",")+")"
def createAnother(elements: List[Exp]) = ListB(elements)
}
The problem in this case is that the createAnother bit is pure boilerplate. On the other hand, this version maintains the generality of the above solution.
A third suggestion, which may involve refactoring your code a bit more is to drop the ListA and ListB types altogether, since it seems to me that their purpose is to provide a tag to a list of Exp. So consider this solution:
case class SpecialList(tag: Tag, elements: List[Exp]) extends Exp {
def flatten: SpecialList = {
val newElems = elements flatMap {
case x: SpecialList => {
val flattenX = x.flatten
if (flattenX.tag == this.tag) flattenX.elements else List(flattenX)
}
case somethingElse => List(somethingElse)
}
SpecialList(tag, newElems)
}
override def toString = tag.toString ++ "(" + elements.mkString(",") + ")"
}
sealed abstract class Tag {
// Syntactic sugar to maintain the notation used in the question
def apply(elements: Exp*): SpecialList = SpecialList(this, elements.toList)
}
object ListA extends Tag { override val toString = "ListA" }
object ListB extends Tag { override val toString = "ListB" }
From a syntactic point of view, it's pretty much the same, since you have
val x = ListA(Element(A), Element(B), ListA(Element(C), Element(D)), ListB(Element(E),Element(F), ListA(Element(C), ListA(Element(D)))))
x.flatten => ListA(Element(A),Element(B),Element(C),Element(D),ListB(Element(E),Element(F),ListA(Element(C),Element(D))))
This may not fit your problem, however, so sorry if I went off the rails a bit there.
来源：https://stackoverflow.com/questions/10581192/flattening-nested-lists-of-the-same-type