Generic type inference in Scala - scala

I've written the following code, which is actually a dumb merge-sort implementation in scala:
import scala.collection.immutable.List
object MergeSort {
def sort[T,E]( comparator: (E,E) => Int ) (l: List[T]): List[T] = {
def merge[T](first: List[T], second: List[T]): List[T] = (first, second) match {
case (_, List()) => first
case (List(), _) => second
case (f::restFirst, s::restSecond) if comparator(f.asInstanceOf[E],s.asInstanceOf[E]) < 0 => f :: merge(restFirst, second)
case (f::restFirst, s::restSecond) => s :: merge(first, restSecond)
}
l match {
case List() => return l
case List(x) => return l
case _ => {
val (first, second) = l.splitAt( l.length / 2 )
merge( sort(comparator)(first), sort(comparator)(second) )
}
}
}
}
This is instead of the following, more elegant, solution:
import scala.collection.immutable.List
object MergeSort {
def sort[T]( comparator: (T,T) => Int ) (l: List[T]): List[T] = {
def merge[T](first: List[T], second: List[T]): List[T] = (first, second) match {
case (_, List()) => first
case (List(), _) => second
case (f::restFirst, s::restSecond) if comparator(f,s) < 0 => f :: merge(restFirst, second)
case (f::restFirst, s::restSecond) => s :: merge(first, restSecond)
}
l match {
case List() => return l
case List(x) => return l
case _ => {
val (first, second) = l.splitAt( l.length / 2 )
merge( sort(comparator)(first), sort(comparator)(second) )
}
}
}
}
Which doesn't compile, giving me the following error message:
MergeSort.scala:10: type mismatch;
[error] found : f.type (with underlying type T)
[error] required: T
[error] case (f::restFirst, s::restSecond) if comparator(f,s) < 0 => f :: merge(restFirst, second)
Why is the explicit cast necessary since the underlying type is T ?

This is one of the most annoying Scala gotchas I can think of (maybe after semicolon inference-related issues with operators). You're three characters from the correct answer.
The problem is the type parameter on merge. It introduces a new T that shadows the T type parameter on sort. The compiler therefore doesn't know that comparator can be applied to instances of that new T. You can boss it around with a cast, which is why your first version works, but otherwise it sees that T as a blank slate.
Just write def merge(first: List[T], ... and you'll be fine.

Related

Filter from Seq less/greater elements and only one equal

I want to implement method in Scala which filters from Seq elements which are for example greater than provided value and additionally returns up to one equal element. For example:
greaterOrEqual(Seq(1,2,3,3,4), 3) shouldBe Seq(3,4)
I ended up with such method:
def greaterOrEqual(
seq: ArrayBuffer[Long],
value: Long
): ArrayBuffer[Long] = {
val greater = seq.filter(_ > value)
val equal = seq.filter(_ == value)
if (equal.isEmpty) {
greater
} else {
equal.tail ++ greater
}
}
but somehow it doesn't look nice to me :) Moreover, I'd like to have generic version of this method where I'd able to use not only Long type but custom case classes.
Do you have any suggestions?
Thanks in advance.
def foo[A : Ordering[A]](seq: Seq[A], value: A) = seq.find(_ == value).toList ++ seq.filter(implicitly[Ordering[A]].gt(_,value))
Or (different style)
def foo[A](seq: Seq[A], value: A)(implicit ord: Ordering[A]) = {
import ord._
seq.find(_ == value).toList ++ seq.filter(_ > value)
}
The code below is deprecated
scala> def foo[A <% Ordered[A]](seq: Seq[A], value: A) = seq.find(_ == value).toList ++ seq.filter(_ > value)
foo: [A](seq: Seq[A], value: A)(implicit evidence$1: A => Ordered[A])List[A]
scala> foo(Seq(1,2,3,3,4,4,5),3)
res8: List[Int] = List(3, 4, 4, 5)
Here's my take on it (preserving original order).
import scala.collection.mutable.ArrayBuffer
def greaterOrEqual[A]( seq :ArrayBuffer[A], value :A
)(implicit ord :Ordering[A]
) : ArrayBuffer[A] =
seq.foldLeft((ArrayBuffer.empty[A],true)){
case (acc, x) if ord.lt(x,value) => acc
case ((acc,bool), x) if ord.gt(x,value) => (acc :+ x, bool)
case ((acc,true), x) => (acc :+ x, false)
case (acc, _) => acc
}._1
testing:
greaterOrEqual(ArrayBuffer.from("xawbaxbt"), 'b')
//res0: ArrayBuffer[Char] = ArrayBuffer(x, w, b, x, t)
This is an excellent problem for a simple tail-recursive algorithm over lists.
def greaterOrEqual[T : Ordering](elements: List[T])(value: T): List[T] = {
import Ordering.Implicits._
#annotation.tailrec
def loop(remaining: List[T], alreadyIncludedEqual: Boolean, acc: List[T]): List[T] =
remaining match {
case x :: xs =>
if (!alreadyIncludedEqual && x == value)
loop(remaining = xs, alreadyIncludedEqual = true, x :: acc)
else if (x > value)
loop(remaining = xs, alreadyIncludedEqual, x :: acc)
else
loop(remaining = xs, alreadyIncludedEqual, acc)
case Nil =>
acc.reverse
}
loop(remaining = elements, alreadyIncludedEqual = false, acc = List.empty)
}
Which you can use like this:
greaterOrEqual(List(1, 3, 2, 3, 4, 0))(3)
// val res: List[Int] = List(3, 4)
You can use the below snippet:
val list = Seq(1,2,3,3,4)
val value = 3
list.partition(_>=3)._1.toSet.toSeq
Here partition method divide the list into two list. First list which satisfy the given condition, and second list contains the remaining elements.
For generic method you can using implicit Ordering. Any type who can compare elements can be handled by greaterOrEqual method.
import scala.math.Ordering._
def greaterOrEqual[T](seq: Seq[T], value: T)(implicit ordering: Ordering[T]): Seq[T] = {
#scala.annotation.tailrec
def go(xs: List[T], value: T, acc: List[T]): List[T] = {
xs match {
case Nil => acc
case head :: rest if ordering.compare(head, value) == 0 => rest.foldLeft(head :: acc){
case (result, x) if ordering.compare(x, value) > 0 => x :: result
case (result, _) => result
}
case head :: rest if ordering.compare(head, value) > 0 => go(rest, value, head :: acc)
case _ :: rest => go(rest, value, acc)
}
}
go(seq.toList, value, List.empty[T]).reverse
}

scala n-arity tree tail recursive evaluation

I have a Tree structure, which is more general than a binary tree structure
sealed trait Tree[+A]
case class Leaf[A](value: Terminal[A]) extends Tree[A]
case class Node[A](op: Function[A], branches: Tree[A]*) extends Tree[A]
As you see, it can have a arbitrary number of branches.
I'm trying to make an evaluation method to be tail recursive but i'm not being able to do it.
def evaluateTree[A](tree: Tree[A]): A = tree match {
case Leaf(terminal) => terminal.value
case Node(op: Function[A], args # _*) => op.operator((for (i <- args) yield evaluateTree(i)))
}
How can i save the stack manually?
If each Node can hold a different op then, no, I don't think tail recursion is possible.
If, on the other hand, you can feed all the Leaf.values to a single op then it might be possible.
def evaluateTree[A](tree: Tree[A]): A = {
#tailrec
def allValues(branches: Seq[Tree[A]], acc: Seq[A] = Seq()): Seq[A] =
if (branches.length < 1) acc
else branches.head match {
case Leaf(term) => allValues(branches.tail, term.value +: acc)
case Node(_, args: Seq[Tree[A]]) => allValues(branches.tail ++ args, acc)
}
tree match {
case Leaf(terminal) => terminal.value
case Node(op: Function[A], args: Seq[Tree[A]]) => op.operator(allValues(args))
}
}
I can't compile this as I don't have definitions for Terminal and Function, but it should be a reasonable outline of one approach to the problem.
Actually it was possible, using deep first search.
def evaluateTree[A](tree: Tree[A]): A = {
#tailrec
def evaluateWhile[C](l: List[Function[C]], arguments: List[List[C]], n_args: List[Int], f: Int => Boolean, acc: C): (List[Function[C]], List[List[C]], List[Int]) =
n_args match {
case h :: t if f(h) =>
evaluateWhile(l.tail, arguments.tail, n_args.tail, f, l.head.operator(arguments.head ::: List(acc)))
case h :: t =>
(l, (List(acc) ::: arguments.head) :: arguments.tail, List(n_args.head - 1) ::: n_args.tail)
case _ =>
(l, List(acc) :: arguments, n_args)
}
#tailrec
def DFS(toVisit: List[Tree[A]], visited: List[String] = Nil, operators: List[Function[A]] = Nil, arguments: List[List[A]] = Nil, n_args: List[Int] = Nil, debug: Int = 0): A = toVisit match {
case Leaf(id, terminal) :: tail if !visited.contains(id) => {
val (operators_to_pass, args_to_pass, n_args_to_pass) =
evaluateWhile[A](operators, arguments, n_args, x => x == 1, terminal.value)
DFS(toVisit.tail, visited ::: List(id), operators_to_pass, args_to_pass, n_args_to_pass, debug + 1)
}
case Node(id, op, args #_*) :: tail if !visited.contains(id) => {
DFS(args.toList ::: toVisit.tail, visited ::: List(id), op :: operators, List(Nil) ::: arguments, List(args.length ) ::: n_args, debug + 1)
}
case _ => arguments.flatten.head
}
DFS(List(tree))
}

Scala: Using a TypeTag to match on a Some's type

See the following code:
def createOption[T: TypeTag](referentialData: Any) : Option[T] = {
Option(referentialData) match {
case Some(camelMessage: CamelMessage) => {
Option(camelMessage.body) match {
case Some(option: T) => Some(option)
case _ => None
}
}
case _ => None
}
}
Basically I am looking to return an Option[T] if camelMessage.body is non-null and of type T.
The uses of Option(referentialData) is effectively referentialData != null
Likewise for Option(camelMessage.body)
How do I use the TypeTag to determine if camelMessage.body is of type T.
(I know this can be re-written to not use TypeTags and Options but I want to learn how to use TypeTags so please no suggestions to re-write, thanks!)
Edit
I tried a new approach as could not find a solution for the above, but could not get this one to work either:
def createOption[T](referentialData: Any) : Option[T] = {
Option(referentialData) match {
case Some(option) => Try(option.asInstanceOf[T]).toOption
case _ => None
}
}
When I invoke this using createOption[Long]("test") I was presuming to get a None back, but instead I got a Some(String)
Where am I going wrong here?
This is a duplicate of this one.
But you want to try it with ClassTag to show the limitation:
scala> def f[A: ClassTag](x: Any): Option[A] = x match {
| case y: A => println("OK"); Some(y) ; case _ => println("Nope"); None }
f: [A](x: Any)(implicit evidence$1: scala.reflect.ClassTag[A])Option[A]
scala> f[String]("foo")
OK
res0: Option[String] = Some(foo)
scala> f[Long](2L)
Nope
res1: Option[Long] = None
scala> f[java.lang.Long](new java.lang.Long(2L))
OK
res2: Option[Long] = Some(2)
scala> def f[A: TypeTag](x: Any): Option[A] = Option(x) match {
| case Some(y: A) => println("OK"); Some(y) ; case _ => println("Nope"); None }
<console>:51: warning: abstract type pattern A is unchecked since it is eliminated by erasure
case Some(y: A) => println("OK"); Some(y) ; case _ => println("Nope"); None }
^
f: [A](x: Any)(implicit evidence$1: reflect.runtime.universe.TypeTag[A])Option[A]

Return a different type according to an input parameter

Let's see an example (it's a naive example but sufficient to illustrate the problem).
def produce(l: List[Int]) : Any =
l match {
case List(x) => x
case List(x, y) => (x, y)
}
val client1 : Int = produce(List(1)).asInstanceOf[Int]
Drawback : client need to cast !
def produce2[A](l: List[Int])(f: List[Int] => A) = {
f(l)
}
val toOne = (l: List[Int]) => l.head
val toTwo = (l: List[Int]) => (l.head, l.tail.head)
val client2 : Int = produce2(List(1))(toOne)
Drawback : type safety, i.e. we can call toTwo with a singleton List.
Is there a better solution ?
If you only have two possible return values you could use Either:
def produce(l : List[Any]) : Either[Any, (Any, Any)] = l match {
case List(x) => Left(x)
case List(x, y) => Right((x, y))
}
If you don't want to create an Either, you could pass a function to transform each case:
def produce[A](l : List[Int])(sf: Int => A)(pf: (Int, Int) => A): A = l match {
case List(x) => sf(x)
case List(x, y) => pf(x, y)
}
Will this work?
def produce(l: List[Int]) = {
l match {
case List(x) => (x, None)
case List(x,y) => (x,y)
case Nil => (None, None)
}
}
or even better, to avoid match errors on lists longer than 2 elements:
def produce(l: List[Int]) =
l match {
case x :: Nil => (x, None)
case x :: xs => (x,xs.head)
case Nil => (None, None)
}

Scala's Implementation of haskell last method

I am trying to do some examples programs in scala to get more familiar with the language, For that I am trying to re-implement some of the built in methods in Haskell, Most of these methods I am sure are also implemented in Scala too, But these are just for my practice. I think I can post some of code snippets (not all of them) to get a better way of doing things and to validate my understanding of scala. So please let me know if this is not the place to do these things.
Here is my scala implementation to get the last element of any list. Is this the right way of doing things, By using Any am I loosing the type of the object containing in the list? Is this how this kind of things implemented in scala?
def getLast(xs: List[Any]): Any = xs match {
case List() => null
case x :: List() => x
case _ :: ys => getLast(ys)
}
Parameterize the type of your function and use "Nil" instead of List() like so:
def getLast[T](xs: List[T]): T = xs match {
case Nil => null.asInstanceOf[T]
case x :: Nil => x
case _ :: ys => getLast(ys)
}
Also, consider making it return an Option type:
def getLast[T](xs: List[T]): Option[T] = xs match {
case Nil => None
case x :: Nil => Some(x)
case _ :: ys => getLast(ys)
}
Usage:
val listOfInts = List(1,2,3)
assert(getLast(listOfInts).isInstanceOf[Int])
val listOfStrings = List("one","two","three")
assert(getLast(listOfStrings).isInstanceOf[String])
Firstly, avoid the null, and especially null.asInstanceOf[T]. Observe the danger with primitives:
scala> null.asInstanceOf[Int]
res19: Int = 0
scala> null.asInstanceOf[Boolean]
res20: Boolean = false
So the signature should either be List[T] => T, whereby last on an empty iterator throws an exception:
def last[T](ts: List[T]): T = ts match {
case Nil => throw new NoSuchElementException
case t :: Nil => t
case t :: ts => last(ts)
}
Or instead: List[T] => Option[T]
def lastOption[T](ts: List[T]): Option[T] = ts match {
case Nil => None
case t :: Nil => Some(t)
case t :: ts => lastOption(ts)
}
def lastOption1[T](ts: List[T]): Option[T] = ts.reverse.headOption
def lastOptionInScala28[T](ts: List[T]): Option[T] = ts.lastOption // :)