So I read about right-associative operators like the cons operator in Scala. I'm wondering why they work in case statements. It seems like you can pattern match using the cons statement here?
def findKth1[A](k:Int, l:List[A]):A = (k, l) match {
case (0, h::_) => h
case(k, _::tail) if k > 0 => findKth1(k - 1, tail)
case _ => throw new NoSuchElementException
}
findKth1(2, List(3,4,5,6))
res1: Int = 5
What is the placeholder doing here? I've only seen placeholders used in functions like this: <SomeList>.map(_.doThing). Is it the same concept?
The only difference with :: and ::: is that ::: is used for 2 Lists right?
tl;dr It isn't pattern matching the operator, it's pattern matching a case class called ::
There are a couple of things happening at the same time. First of all, there may be a bit of confusion, because :: is a method on List:
val x: List[Int] = 1 :: 2 :: 3 :: Nil
But there is also a case class :: that, according to the docs is:
A non empty list characterized by a head and a tail.
More info here
Scala's case classes automatically come with extractor methods(unapply), which allow pattern matching, as in case User(name, age) => ....
You are also allowed to use the case class name in an infix position (although you shouldn't do so, except when the case class is used like an operator, such as in this case). So case head :: tail => ... is the same as case ::(head, tail) => .... More info here
When pattern matching, you can use _ to mean that there will be a value there, but you don't care about it, so you aren't providing it a name.
So the three cases you provided are roughly:
A tuple where the first value is 0, and the second value is a list with a head to be called h, and a tail which we will ignore.
A tuple with some integer to be called k, and a list with a head that we don't care about, and a tail, which we will call tail. Also, k must be greater than 0
Anything else, then throw a NoSuchElementException
From the Book programming in Scala I got the following line of code:
val second: List[ Int] => Int = { case x :: y :: _ => y }
//warning: match may not be exhaustive.
It states that this function will return the second element of a list of integers if the list is not empty or nil. Stil this part is a bit awkward to me:
case x :: y :: _
How does this ecxactly work? Does this mathches any list with at least 2 Elements and than return the second? If so can somebody still explain the syntax? I understood that :: is invoked on the right operand. So it could be written as
(_.::(y)).::(X)
Still I than don't get why this would return 2
val second: List[ Int] => Int = { case x :: y :: _ => y }
var x = List(1,2)
second(x) //returns 2
In the REPL, you can type:
scala> val list = "a" :: "b" :: Nil
list: List[String] = List(a, b)
which is to be read from right to left, and means take the end of a List (Nil), prepend String "b" and to this List ("b" :: Nil) prepend String a, a :: ("b" :: Nil) but you don't need the parens, so it can be written "a" :: "b" :: Nil.
In pattern matching you will more often see:
... list match {
case Nil => // ...
case x :: xs => // ...
}
to distinguish between empty list, and nonempty, where xs might be a rest of list, but matches Nil too, if the whole list is ("b" :: Nil) for example, then x="b" and xs=Nil.
But if list= "a" :: "b" :: Nil, then x="a" and xs=(b :: Nil).
In your example, the deconstruction is just one more step, and instead of a name like xs, the joker sign _ is used, indicating, that the name is probably not used and doesn't play a role.
The value second is of function type, it takes List[Int] and returns Int.
If the list has first element ("x"), and a second element ("y"), and whatever comes next (we don't care about it), we simply return the element "y" (which is the second element of the list).
In any other case, the function is not defined. You can check that:
scala> val second: PartialFunction[List[Int], Int] = {
| case x :: y :: _ => y
| }
second: PartialFunction[List[Int],Int] = <function1>
scala> second.isDefinedAt(List(1,2,3))
res18: Boolean = true
scala> second.isDefinedAt(List(1,2))
res19: Boolean = true
scala> second.isDefinedAt(List(0))
res20: Boolean = false
First of all. When you think about pattern matching you should think about matching a structure.
The first part of the case statement describes a structure. This structure may describe one or more things (variables) which are useful to deriving your result.
In your example, you are interested in deriving the second element of a list. A shorthand to build a list in Scala is to use :: method (also called cons). :: can also be used to describe a structure in case statement. At this time, you shouldn't think about evaluation of the :: method in first part of case. May be that's why you are saying about evaluation of _.::(y).::(x). The :: cons operator help us describe the structure of the list in terms of its elements. In this case, the first element (x) , the second element (y) and the rest of it (_ wildcard). We are interested in a structure that is a list with at least 2 elements and the third can be anything - a Nil to indicate end of list or another element - hence the wildcard.
The second part of the case statement, uses the second element to derive the result (y).
More on List and Consing
List in Scala is similar to a LinkedList. You know about the first element called head and start of the rest of the list. When traversing the linked list you stop if the rest of the list is Nil. This :: cons operator helps us visualise the structure of the linked list. Although Scala compile would actually be calling :: methods evaluating from right to left as you described _.::(y).::(x)
As an aside, you might have already noticed that the Scala compiler might be complain that your match isn't exhaustive. This means that this second method would work for list of any size. Because there isn't any case statement to describe list with zero or one element. Also, as mentioned in comments of previous answers, if you aren't interested in first element you can describe it as a wildcard _.
case _ :: y :: _ => y
I hope this helped.
If you see the structure of list in scala its head::tail, first element is treated as head and all remaining ones as tail(Nil will be the last element of tail). whenever you do x::y::_, x will match the head of the list and remaining will be tail and again y will match the head of the next list(tail of first list)
eg:
val l = List(1,2,3,4,5)
you can see this list in differnt ways:
1::2::3::4::5::Nil
1::List(2,3,4,5)
1::2::List(2,3,4,5)
and so on
So try matching the pattern. In your question y will give the second element
I am reading the book programming in Scala from Martin O. and there is one example there to remove duplicates totally confused me:
def removeDuplicates[A](xs: List[A]): List[A] = {
if (xs.isEmpty) xs
else
xs.head :: removeDuplicates(
xs.tail filter (x => x != xs.head)
)
}
println(removeDuplicates[String](List("a", "a", "b", "a", "c")))
gives me:
List(a,b,c)
I know that .head will give you the very first element of the List while .tail give you the rest of the List. And I can understand that xs.tail filter (x => x != xs.head) will return a list containing the elements which don't equal to the head.
My Google search leads me to this cons operator however, I am still having a hard time mapping Martin's words to this example. And anyone help me understand how this :: works in this function?
A peculiarity in Scala is that operators ending in : (colon) are right-associative, and they are dispatched to the object on the right, with the parameter being on the left. For example: a :: list (infix notation) is equivalent to list.::(a) (method notation).
Have a look at the documentation for :: (cons). It constructs a linked list from an element and another list. Note that a :: b :: c :: Nil is equivalent to List(a, b, c), but note that the construction is happening from right to left, as Nil.::(c).::(b).::(a).
The example you gave uses recursion, which is based on a base case and an inductive case. The base case says that an empty list has no duplicates. The inductive case says that, assuming you have a removeDuplicates method which can remove all duplicates from a list, you can construct a new (sometimes larger) duplicate-free list by adding a value to the beginning, as long as you've remove that value from the remainder of the list first.
This is a very common pattern in functional programming.
Realize that removeDuplicates evaluates to a list, which the cons operator takes on its right side. The end result is a list where it's tail doesn't contain its head.
Every recurse, we add the head of the remaining list to the new list that we're constructing using the cons operator. We see if the current head exists in the rest of the list, and filter them out.
Look up what a the map method is. If you get how it works, this should click. They aren't exactly the same, but it involves building a list using the cons operator.
Given a list L I want to keep an element L(i) if it exists at least one value j > i such that L(j) is a multiple of L(i), otherwise L(i) should be discarded.
It is quite simple to do that by means of imperative programming paradigms, but I would like to do that using functional programming.
Is that possible to use the filter method? If so, how to write the condition (i.e. the parameter of the filter function) ? Otherwise, what can I do?
For example:
val l = (1 to 100)
l.tails.collect { case (head +: tail) if tail.exists(_ % head == 0) => head } .toList
tail produces an iterator that returns in each step the input minus one element, e.g.
(1 to 10).tails.foreach(println)
gives
Vector(1, 2, 3, 4)
Vector(2, 3, 4)
Vector(3, 4)
Vector(4)
Vector()
You can view these 'tails' as a head element to which you want to apply a filter and a tail in itself that is used to find out whether to keep the head.
The collect method is useful here, because it takes a partial function, so you only need to specify the cases where you actually retain a value—like filter—, while at the same time it acts like a map by letting you specify how the filtered value is to be collected.
So we can match on tails that have at least one head element and a tail of any size, and then see if in that tail there exists an element that is a multiple of the head. I use a guard here for the match case, so the match is a double filter. First, the tail must be non-empty, second there must be multiple. A multiple means that the modulus is zero. If the case matches, just return the verified head element.
Finally, since without specific type annotations the collect will just return another iterator, we turn the result into a list with toList.
A more "explicit" one - you accumulate elements in a case if tail has a multiple of head:
(1 to 10).tails.foldLeft(List[Int]())((acc, tl) => tl match {
case h +: t if (t.exists(_ % h == 0)) => h :: acc
case _ => acc
}).reverse
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)
}