Tune Nested Loop in Scala - scala

Question

I was wondering if I can tune the following Scala code :
def removeDuplicates(listOfTuple: List[(Class1,Class2)]): List[(Class1,Class2)] = {
var listNoDuplicates: List[(Class1, Class2)] = Nil
for (outerIndex <- 0 until listOfTuple.size) {
if (outerIndex != listOfTuple.size - 1)
for (innerIndex <- outerIndex + 1 until listOfTuple.size) {
if (listOfTuple(i)._1.flag.equals(listOfTuple(j)._1.flag))
listNoDuplicates = listOfTuple(i) :: listNoDuplicates
}
}
listNoDuplicates
}

Answer 1

Usually if you have someting looking like:
var accumulator: A = new A
for( b <- collection ) {
accumulator = update(accumulator, b)
}
val result = accumulator
can be converted in something like:
val result = collection.foldLeft( new A ){ (acc,b) => update( acc, b ) }
So here we can first use a map to force the unicity of flags. Supposing the flag has a type F:
val result = listOfTuples.foldLeft( Map[F,(ClassA,ClassB)] ){
( map, tuple ) => map + ( tuple._1.flag -> tuple )
}
Then the remaining tuples can be extracted from the map and converted to a list:
val uniqList = map.values.toList
It will keep the last tuple encoutered, if you want to keep the first one, replace foldLeft by foldRight, and invert the argument of the lambda.
Example:
case class ClassA( flag: Int )
case class ClassB( value: Int )
val listOfTuples =
List( (ClassA(1),ClassB(2)), (ClassA(3),ClassB(4)), (ClassA(1),ClassB(-1)) )
val result = listOfTuples.foldRight( Map[Int,(ClassA,ClassB)]() ) {
( tuple, map ) => map + ( tuple._1.flag -> tuple )
}
val uniqList = result.values.toList
//uniqList: List((ClassA(1),ClassB(2)), (ClassA(3),ClassB(4)))
Edit: If you need to retain the order of the initial list, use instead:
val uniqList = listOfTuples.filter( result.values.toSet )

Answer 2

This compiles, but as I can't test it it's hard to say if it does "The Right Thing" (tm):
def removeDuplicates(listOfTuple: List[(Class1,Class2)]): List[(Class1,Class2)] =
(for {outerIndex <- 0 until listOfTuple.size
if outerIndex != listOfTuple.size - 1
innerIndex <- outerIndex + 1 until listOfTuple.size
if listOfTuple(i)._1.flag == listOfTuple(j)._1.flag
} yield listOfTuple(i)).reverse.toList
Note that you can use == instead of equals (use eq if you need reference equality).
BTW: https://codereview.stackexchange.com/ is better suited for this type of question.

Answer 3

Do not use index with lists (like listOfTuple(i)). Index on lists have very lousy performance. So, some ways...
The easiest:
def removeDuplicates(listOfTuple: List[(Class1,Class2)]): List[(Class1,Class2)] =
SortedSet(listOfTuple: _*)(Ordering by (_._1.flag)).toList
This will preserve the last element of the list. If you want it to preserve the first element, pass listOfTuple.reverse instead. Because of the sorting, performance is, at best, O(nlogn). So, here's a faster way, using a mutable HashSet:
def removeDuplicates(listOfTuple: List[(Class1,Class2)]): List[(Class1,Class2)] = {
// Produce a hash map to find the duplicates
import scala.collection.mutable.HashSet
val seen = HashSet[Flag]()
// now fold
listOfTuple.foldLeft(Nil: List[(Class1,Class2)]) {
case (acc, el) =>
val result = if (seen(el._1.flag)) acc else el :: acc
seen += el._1.flag
result
}.reverse
}
One can avoid using a mutable HashSet in two ways:
Make seen a var, so that it can be updated.
Pass the set along with the list being created in the fold. The case then becomes:
case ((seen, acc), el) =>