Suppose you are writing a tail-recursive loop function to evaluate a collection of elements according to some criterium and want to end up with the element that scores best, and its score.
Naturally you will pass the best scoring element so far, as well as its score, as parameters to the function.
But since there is no best element at the start of the recursion, what should you initially pass as parameters to the loop function? Not wanting to use null, you could use Option[T] as parameter types, but then you have to check for isEmpty at each recursion while you know that it always has a value after the initial call. Isn't there a better way?
You can use list.head as initial value and loop over list. The first evaluation will be "wasted" since you're evaluating list.head against itself but that will calculate the score for list.head and the rest of the iteration can carry on and do what you want.
How is a "best element" evaluated?
Typically that is done through a numeric value.
At the start, you typically set that value to a number. Best practices state that the number should be defined as a constant, something like the name of MIN_VALUE. That value could be zero, negative, or the minimum floating point number that is representable.
The head of the list, as #vptheron answered, seems to be the best as you don't have a good starting value.
But, rather than use a tail-recursive function...
def getBestScore(scores: List[A]): A = {
def go(as: List[A], acc: A): A = as match {
case x :: xs => go(xs, getBest(acc, x)) // getBest: ((A, A) => A)
case Nil => acc
}
go(scores, scores.head)
}
... you can use foldLeft to make it concise.
val best = scores.foldLeft(scores.head)(getBest)
Example:
scala> def getBest(x: Int, y: Int) = if(x > y) x else y
getBest: (x: Int, y: Int)Int
scala> val scores = List(1, 20, 3)
scores: List[Int] = List(1, 20, 3)
scala> scores.foldLeft(scores.head)(getBest)
res6: Int = 20
Related
This question already has answers here:
Scala - Combine two lists in an alternating fashion
(4 answers)
Closed 3 years ago.
The elements of the resulting list should alternate between the elements of the arguments. Assume that the two arguments have the same length.
USE RECURSION
My code as follows
val finalString = new ListBuffer[Int]
val buff2= new ListBuffer[Int]
def alternate(xs:List[Int], ys:List[Int]):List[Int] = {
while (xs.nonEmpty) {
finalString += xs.head + ys.head
alternate(xs.tail,ys.tail)
}
return finalString.toList
}
EXPECTED RESULT:
alternate ( List (1 , 3, 5) , List (2 , 4, 6)) = List (1 , 2, 3, 4, 6)
As far for the output, I don't get any output. The program is still running and cannot be executed.
Are there any Scala experts?
There are a few problems with the recursive solutions suggested so far (including yours, which would actually work, if you replace while with if): appending to end of list is a linear operation, making the whole thing quadratic (taking a .length of a list too, as well ас accessing elements by index), don't do that; also, if the lists are long, a recursion may require a lot of space on the stack, you should be using tail-recursion whenever possible.
Here is a solution that is free of both those problems: it builds the output backwards, by prepending elements to the list (constant time operation) rather than appending them, and reverses the result at the end. It is also tail-recursive: the recursive call is the last operation in the function, which allows the compiler to convert it into a loop, so that it will only use a single stack frame for execution regardless of the size of the lists.
#tailrec
def alternate(
a: List[Int],
b: List[Int],
result: List[Int] = Nil
): List[Int] = (a,b) match {
case (Nil, _) | (_, Nil) => result.reversed
case (ah :: at, bh :: bt) => alternate(at, bt, bh :: ah :: result)
}
(if the lists are of different lengths, the whole thing stops when the shortest one ends, and whatever is left in the longer one is thrown out. You may want to modify the first case (split it into two, perhaps) if you desire a different behavior).
BTW, your own solution is actually better than most suggested here: it is actually tail recursive (or rather can be made one if you add else after your if, which is now while), and appending to ListBuffer isn't actually as bad as to a List. But using mutable state is generally considered "code smell" in scala, and should be avoided (that's one of the main ideas behind using recursion instead of loops in the first place).
Condition xs.nonEmpty is true always so you have infinite while loop.
Maybe you meant if instead of while.
A more Scala-ish approach would be something like:
def alternate(xs: List[Int], ys: List[Int]): List[Int] = {
xs.zip(ys).flatMap{case (x, y) => List(x, y)}
}
alternate(List(1,3,5), List(2,4,6))
// List(1, 2, 3, 4, 5, 6)
A recursive solution using match
def alternate[T](a: List[T], b: List[T]): List[T] =
(a, b) match {
case (h1::t1, h2::t2) =>
h1 +: h2 +: alternate(t1, t2)
case _ =>
a ++ b
}
This could be more efficient at the cost of clarity.
Update
This is the more efficient solution:
def alternate[T](a: List[T], b: List[T]): List[T] = {
#annotation.tailrec
def loop(a: List[T], b: List[T], res: List[T]): List[T] =
(a, b) match {
case (h1 :: t1, h2 :: t2) =>
loop(t1, t2, h2 +: h1 +: res)
case _ =>
a ++ b ++ res.reverse
}
loop(a, b, Nil)
}
This retains the original function signature but uses an inner function that is an efficient, tail-recursive implementation of the algorithm.
You're accessing variables from outside the method, which is bad. I would suggest something like the following:
object Main extends App {
val l1 = List(1, 3, 5)
val l2 = List(2, 4, 6)
def alternate[A](l1: List[A], l2: List[A]): List[A] = {
if (l1.isEmpty || l2.isEmpty) List()
else List(l1.head,l2.head) ++ alternate(l1.tail, l2.tail)
}
println(alternate(l1, l2))
}
Still recursive but without accessing state from outside the method.
Assuming both lists are of the same length, you can use a ListBuffer to build up the alternating list. alternate is a pure function:
import scala.collection.mutable.ListBuffer
object Alternate extends App {
def alternate[T](xs: List[T], ys: List[T]): List[T] = {
val buffer = new ListBuffer[T]
for ((x, y) <- xs.zip(ys)) {
buffer += x
buffer += y
}
buffer.toList
}
alternate(List(1, 3, 5), List(2, 4, 6)).foreach(println)
}
Suppose I am writing function foo: Seq[B] => Boolean like this:
case class B(xs: Seq[Int])
def foo(bs: Seq[B]): Boolean = bs.map(_.xs.size).sum > 0
The implementation above is suboptimal (since it's not necessary to loop over all bs elements to return true). How would you implement foo efficiently ?
Well, for 0 this is kind of trivial:
bs.exists(!_.xs.isEmpty)
does the job, because as soon as you find a non-empty xs, you are done.
Now, suppose that the threshold is not trivial, e.g. 42 instead of 0.
You can then take the iterator of bs, incrementally aggregate the values using scanLeft, and then check whether there exists an intermediate result that is greater than zero:
def foo(bs: Seq[Int]): Boolean = bs
.iterator
.scanLeft(0)(_ + _.xs.size)
.exists(_ > 42)
I'd like to find the indices (coordinates) of the first element whose value is 4, in a nested Vector of Int, in a functional way.
val a = Vector(Vector(1,2,3), Vector(4,5), Vector(3,8,4))
a.map(_.zipWithIndex).zipWithIndex.collect{
case (col, i) =>
col.collectFirst {
case (num, index) if num == 4 =>
(i, index)
}
}.collectFirst {
case Some(x) ⇒ x
}
It returns:
Some((0, 1))
the coordinate of the first 4 occurrence.
This solution is quite simple, but it has a performance penalty, because the nested col.collect is performed for all the elements of the top Vector, when we are only interested in the 1st match.
One possible solution is to write a guard in the pattern matching. But I don't know how to write a guard based in a slow condition, and return something that has already been calculated in the guard.
Can it be done better?
Recursive maybe?
If you insist on using Vectors, something like this will work (for a non-indexed seq, you'd need a different approach):
#tailrec
findit(
what: Int,
lists: IndexedSeq[IndexedSeq[Int]],
i: Int = 0,
j: Int = 0
): Option[(Int, Int)] =
if(i >= lists.length) None
else if(j >= lists(i).length) findit(what, lists, i+1, 0)
else if(lists(i)(j) == what) Some((i,j))
else findit(what, lists, i, j+1)
A simple thing you can to without changing the algorithm is to use Scala streams to be able to exit as soon as you find the match. Streams are lazily evaluated as opposed to sequences.
Just make a change similar to this
a.map(_.zipWithIndex.toStream).zipWithIndex.toStream.collect{ ...
In terms of algorithmic changes, if you can somehow have your data sorted (even before you start to search) then you can use Binary search instead of looking at each element.
import scala.collection.Searching._
val dummy = 123
implicit val anOrdering = new Ordering[(Int, Int, Int)]{
override def compare(x: (Int, Int, Int), y: (Int, Int, Int)): Int = Integer.compare(x._1, y._1)
}
val seqOfIntsWithPosition = a.zipWithIndex.flatMap(vectorWithIndex => vectorWithIndex._1.zipWithIndex.map(intWithIndex => (intWithIndex._1, vectorWithIndex._2, intWithIndex._2)))
val sorted: IndexedSeq[(Int, Int, Int)] = seqOfIntsWithPosition.sortBy(_._1)
val element = sorted.search((4, dummy, dummy))
This code is not very pretty or readable, I just quickly wanted to show an example of how it could be done.
I have a homework assignment that's having us use a list and splitting it into two parts with the elements in the first part are no greater than p, and the elements in the second part are greater than p. so it's like a quick sort, except we can't use any sorting. I really need some tips on how to go about this. I know I'm using cases, but I'm not familiar with how the list class works in scala. below is what I have so far, but in not sure how to go about the splitting of the 2 lists.
using
def split(p:Int, xs:List[Int]): List[Int] = {
xs match {
case Nil => (Nil, Nil)
case head :: tail =>
}
First off, you want split to return a pair of lists, so the return type needs to be (List[Int], List[Int]). However, working with pairs and lists together can often mean decomposing return values frequently. You may want to have an auxiliary function do the heavy lifting for you.
For instance, your auxiliary function might be given two lists, initially empty, and build up the contents until the first list is empty. The result would then be the pair of lists.
The next thing you have to decide in your recursive function design is, "What is the key decision?" In your case, it is "value no greater than p". That leads to the following code:
def split(p:Int, xs: List[Int]): (List[Int], List[Int]) = {
def splitAux(r: List[Int], ngt: List[Int], gt: List[Int]): (List[Int], List[Int]) =
r match {
case Nil => (ngt, gt)
case head :: rest if head <= p =>
splitAux(rest, head :: ngt, gt)
case head :: rest if head > p =>
splitAux(rest, ngt, head :: gt)
}
val (ngt, gt) = splitAux(xs, List(), List())
(ngt.reverse, gt.reverse)
}
The reversing step isn't strictly necessary, but probably is least surprising. Similarly, the second guard predicate makes explicit the path being taken.
However, there is a much simpler way: use builtin functionality.
def split(p:Int, xs: List[Int]): (List[Int], List[Int]) = {
(xs.filter(_ <= p), xs.filter(_ > p))
}
filter extracts only those items meeting the criterion. This solution walks the list twice, but since you have the reverse step in the previous solution, you are doing that anyway.
I'm making my way through "Programming in Scala" and wrote a quick implementation of the selection sort algorithm. However, since I'm still a bit green in functional programming, I'm having trouble translating to a more Scala-ish style. For the Scala programmers out there, how can I do this using Lists and vals rather than falling back into my imperative ways?
http://gist.github.com/225870
As starblue already said, you need a function that calculates the minimum of a list and returns the list with that element removed. Here is my tail recursive implementation of something similar (as I believe foldl is tail recursive in the standard library), and I tried to make it as functional as possible :). It returns a list that contains all the elements of the original list (but kindof reversed - see the explanation below) with the minimum as a head.
def minimum(xs: List[Int]): List[Int] =
(List(xs.head) /: xs.tail) {
(ys, x) =>
if(x < ys.head) (x :: ys)
else (ys.head :: x :: ys.tail)
}
This basically does a fold, starting with a list containing of the first element of xs If the first element of xs is smaller than the head of that list, we pre-append it to the list ys. Otherwise, we add it to the list ys as the second element. And so on recursively, we've folded our list into a new list containing the minimum element as a head and a list containing all the elements of xs (not necessarily in the same order) with the minimum removed, as a tail. Note that this function does not remove duplicates.
After creating this helper function, it's now easy to implement selection sort.
def selectionSort(xs: List[Int]): List[Int] =
if(xs.isEmpty) List()
else {
val ys = minimum(xs)
if(ys.tail.isEmpty)
ys
else
ys.head :: selectionSort(ys.tail)
}
Unfortunately this implementation is not tail recursive, so it will blow up the stack for large lists. Anyway, you shouldn't use a O(n^2) sort for large lists, but still... it would be nice if the implementation was tail recursive. I'll try to think of something... I think it will look like the implementation of a fold.
Tail Recursive!
To make it tail recursive, I use quite a common pattern in functional programming - an accumulator. It works a bit backward, as now I need a function called maximum, which basically does the same as minimum, but with the maximum element - its implementation is exact as minimum, but using > instead of <.
def selectionSort(xs: List[Int]) = {
def selectionSortHelper(xs: List[Int], accumulator: List[Int]): List[Int] =
if(xs.isEmpty) accumulator
else {
val ys = maximum(xs)
selectionSortHelper(ys.tail, ys.head :: accumulator)
}
selectionSortHelper(xs, Nil)
}
EDIT: Changed the answer to have the helper function as a subfunction of the selection sort function.
It basically accumulates the maxima to a list, which it eventually returns as the base case. You can also see that it is tail recursive by replacing accumulator by throw new NullPointerException - and then inspect the stack trace.
Here's a step by step sorting using an accumulator. The left hand side shows the list xs while the right hand side shows the accumulator. The maximum is indicated at each step by a star.
64* 25 12 22 11 ------- Nil
11 22 12 25* ------- 64
22* 12 11 ------- 25 64
11 12* ------- 22 25 64
11* ------- 12 22 25 64
Nil ------- 11 12 22 25 64
The following shows a step by step folding to calculate the maximum:
maximum(25 12 64 22 11)
25 :: Nil /: 12 64 22 11 -- 25 > 12, so it stays as head
25 :: 12 /: 64 22 11 -- same as above
64 :: 25 12 /: 22 11 -- 25 < 64, so the new head is 64
64 :: 22 25 12 /: 11 -- and stays so
64 :: 11 22 25 12 /: Nil -- until the end
64 11 22 25 12
You should have problems doing selection sort in functional style, as it is an in-place sort algorithm. In-place, by definition, isn't functional.
The main problem you'll face is that you can't swap elements. Here's why this is important. Suppose I have a list (a0 ... ax ... an), where ax is the minimum value. You need to get ax away, and then compose a list (a0 ... ax-1 ax+1 an). The problem is that you'll necessarily have to copy the elements a0 to ax-1, if you wish to remain purely functional. Other functional data structures, particularly trees, can have better performance than this, but the basic problem remains.
here is another implementation of selection sort (generic version).
def less[T <: Comparable[T]](i: T, j: T) = i.compareTo(j) < 0
def swap[T](xs: Array[T], i: Int, j: Int) { val tmp = xs(i); xs(i) = xs(j); xs(j) = tmp }
def selectiveSort[T <: Comparable[T]](xs: Array[T]) {
val n = xs.size
for (i <- 0 until n) {
val min = List.range(i + 1, n).foldLeft(i)((a, b) => if (less(xs(a), xs(b))) a else b)
swap(xs, i, min)
}
}
You need a helper function which does the selection. It should return the minimal element and the rest of the list with the element removed.
I think it's reasonably feasible to do a selection sort in a functional style, but as Daniel indicated, it has a good chance of performing horribly.
I just tried my hand at writing a functional bubble sort, as a slightly simpler and degenerate case of selection sort. Here's what I did, and this hints at what you could do:
define bubble(data)
if data is empty or just one element: return data;
otherwise, if the first element < the second,
return first element :: bubble(rest of data);
otherwise, return second element :: bubble(
first element :: (rest of data starting at 3rd element)).
Once that's finished recursing, the largest element is at the end of the list. Now,
define bubblesort [data]
apply bubble to data as often as there are elements in data.
When that's done, your data is indeed sorted. Yes, it's horrible, but my Clojure implementation of this pseudocode works.
Just concerning yourself with the first element or two and then leaving the rest of the work to a recursed activity is a lisp-y, functional-y way to do this kind of thing. But once you've gotten your mind accustomed to that kind of thinking, there are more sensible approaches to the problem.
I would recommend implementing a merge sort:
Break list into two sub-lists,
either by counting off half the elements into one sublist
and the rest in the other,
or by copying every other element from the original list
into either of the new lists.
Sort each of the two smaller lists (recursion here, obviously).
Assemble a new list by selecting the smaller from the front of either sub-list
until you've exhausted both sub-lists.
The recursion is in the middle of that, and I don't see a clever way of making the algorithm tail recursive. Still, I think it's O(log-2) in time and also doesn't place an exorbitant load on the stack.
Have fun, good luck!
Thanks for the hints above, they were very inspiring. Here's another functional approach to the selection sort algorithm. I tried to base it on the following idea: finding a max / min can be done quite easily by min(A)=if A=Nil ->Int.MaxValue else min(A.head, min(A.tail)). The first min is the min of a list, the second the min of two numbers. This is easy to understand, but unfortunately not tail recursive. Using the accumulator method the min definition can be transformed like this, now in correct Scala:
def min(x: Int,y: Int) = if (x<y) x else y
def min(xs: List[Int], accu: Int): Int = xs match {
case Nil => accu
case x :: ys => min(ys, min(accu, x))
}
(This is tail recursive)
Now a min version is needed which returns a list leaving out the min value. The following function returns a list whose head is the min value, the tail contains the rest of the original list:
def minl(xs: List[Int]): List[Int] = minl(xs, List(Int.MaxValue))
def minl(xs: List[Int],accu:List[Int]): List[Int] = xs match {
// accu always contains min as head
case Nil => accu take accu.length-1
case x :: ys => minl(ys,
if (x<accu.head) x::accu else accu.head :: x :: accu.tail )
}
Using this selection sort can be written tail recursively as:
def ssort(xs: List[Int], accu: List[Int]): List[Int] = minl(xs) match {
case Nil => accu
case min :: rest => ssort(rest, min::accu)
}
(reverses the order). In a test with 10000 list elements this algorithm is only about 4 times slower than the usual imperative algorithm.
Even though, when coding Scala, I'm used to prefer functional programming style (via combinators or recursion) over imperative style (via variables and iterations), THIS TIME, for this specific problem, old school imperative nested loops result in simpler and more performant code.
I don't think falling back to imperative style is a mistake for certain classes of problems, such as sorting algorithms which usually transform the input buffer in place rather than resulting to a new collection.
My solution is:
package bitspoke.algo
import scala.math.Ordered
import scala.collection.mutable.Buffer
abstract class Sorter[T <% Ordered[T]] {
// algorithm provided by subclasses
def sort(buffer : Buffer[T]) : Unit
// check if the buffer is sorted
def sorted(buffer : Buffer[T]) = buffer.isEmpty || buffer.view.zip(buffer.tail).forall { t => t._2 > t._1 }
// swap elements in buffer
def swap(buffer : Buffer[T], i:Int, j:Int) {
val temp = buffer(i)
buffer(i) = buffer(j)
buffer(j) = temp
}
}
class SelectionSorter[T <% Ordered[T]] extends Sorter[T] {
def sort(buffer : Buffer[T]) : Unit = {
for (i <- 0 until buffer.length) {
var min = i
for (j <- i until buffer.length) {
if (buffer(j) < buffer(min))
min = j
}
swap(buffer, i, min)
}
}
}
As you can see, to achieve parametric polymorphism, rather than using java.lang.Comparable, I preferred scala.math.Ordered and Scala View Bounds rather than Upper Bounds. That's certainly works thanks to Scala Implicit Conversions of primitive types to Rich Wrappers.
You can write a client program as follows:
import bitspoke.algo._
import scala.collection.mutable._
val sorter = new SelectionSorter[Int]
val buffer = ArrayBuffer(3, 0, 4, 2, 1)
sorter.sort(buffer)
assert(sorter.sorted(buffer))
A simple functional program for selection-sort in Scala
def selectionSort(list:List[Int]):List[Int] = {
#tailrec
def selectSortHelper(list:List[Int], accumList:List[Int] = List[Int]()): List[Int] = {
list match {
case Nil => accumList
case _ => {
val min = list.min
val requiredList = list.filter(_ != min)
selectSortHelper(requiredList, accumList ::: List.fill(list.length - requiredList.length)(min))
}
}
}
selectSortHelper(list)
}
You may want to try replacing your while loops with recursion, so, you have two places where you can create new recursive functions.
That would begin to get rid of some vars.
This was probably the toughest lesson for me, trying to move more toward FP.
I hesitate to show solutions here, as I think it would be better for you to try first.
But, if possible you should be using tail-recursion, to avoid problems with stack overflows (if you are sorting a very, very large list).
Here is my point of view on this problem: SelectionSort.scala
def selectionsort[A <% Ordered[A]](list: List[A]): List[A] = {
def sort(as: List[A], bs: List[A]): List[A] = as match {
case h :: t => select(h, t, Nil, bs)
case Nil => bs
}
def select(m: A, as: List[A], zs: List[A], bs: List[A]): List[A] =
as match {
case h :: t =>
if (m > h) select(m, t, h :: zs, bs)
else select(h, t, m :: zs, bs)
case Nil => sort(zs, m :: bs)
}
sort(list, Nil)
}
There are two inner functions: sort and select, which represents two loops in original algorithm. The first function sort iterates through the elements and call select for each of them. When the source list is empty it return bs list as result, which is initially Nil. The sort function tries to search for maximum (not minimum, since we build result list in reversive order) element in source list. It suppose that maximum is head by the default and then just replace it with a proper value.
This is 100% functional implementation of Selection Sort in Scala.
Here is my solution
def sort(list: List[Int]): List[Int] = {
#tailrec
def pivotCompare(p: Int, l: List[Int], accList: List[Int] = List.empty): List[Int] = {
l match {
case Nil => p +: accList
case x :: xs if p < x => pivotCompare(p, xs, accList :+ x)
case x :: xs => pivotCompare(x, xs, accList :+ p)
}
}
#tailrec
def loop(list: List[Int], accList: List[Int] = List.empty): List[Int] = {
list match {
case x :: xs =>
pivotCompare(x, xs) match {
case Nil => accList
case h :: tail => loop(tail, accList :+ h)
}
case Nil => accList
}
}
loop(list)
}