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I have been trying to compress a String. Given a String like this:
AAABBCAADEEFF, I would need to compress it like 3A2B1C2A1D2E2F
I was able to come up with a tail recursive implementation:
#scala.annotation.tailrec
def compress(str: List[Char], current: Seq[Char], acc: Map[Int, String]): String = str match {
case Nil =>
if (current.nonEmpty)
s"${acc.values.mkString("")}${current.length}${current.head}"
else
s"${acc.values.mkString("")}"
case List(x) if current.contains(x) =>
val newMap = acc ++ Map(acc.keys.toList.last + 1 -> s"${current.length + 1}${current.head}")
compress(List.empty[Char], Seq.empty[Char], newMap)
case x :: xs if current.isEmpty =>
compress(xs, Seq(x), acc)
case x :: xs if !current.contains(x) =>
if (acc.nonEmpty) {
val newMap = acc ++ Map(acc.keys.toList.last + 1 -> s"${current.length}${current.head}")
compress(xs, Seq(x), newMap)
} else {
compress(xs, Seq(x), acc ++ Map(1 -> s"${current.length}${current.head}"))
}
case x :: xs =>
compress(xs, current :+ x, acc)
}
// Produces 2F3A2B1C2A instead of 3A2B1C2A1D2E2F
compress("AAABBCAADEEFF".toList, Seq.empty[Char], Map.empty[Int, String])
It fails however for the given case! Not sure what edge scenario I'm missing! Any help?
So what I'm actually doing is, going over the sequence of characters, collecting identical ones into a new Sequence and as long as the new character in the original String input (the first param in the compress method) is found in the current (the second parameter in the compress method), I keep collecting it.
As soon as it is not the case, I empty the current sequence, count and push the collected elements into the Map! It fails for some edge cases that I'm not able to make out!
I came up with this solution:
def compress(word: List[Char]): List[(Char, Int)] =
word.map((_, 1)).foldRight(Nil: List[(Char, Int)])((e, acc) =>
acc match {
case Nil => List(e)
case ((c, i)::rest) => if (c == e._1) (c, i + 1)::rest else e::acc
})
Basically, it's a map followed by a right fold.
Took inspiration from the #nicodp code
def encode(word: String): String =
word.foldLeft(List.empty[(Char, Int)]) { (acc, e) =>
acc match {
case Nil => (e, 1) :: Nil
case ((lastChar, lastCharCount) :: xs) if lastChar == e => (lastChar, lastCharCount + 1) :: xs
case xs => (e, 1) :: xs
}
}.reverse.map { case (a, num) => s"$num$a" }.foldLeft("")(_ ++ _)
First our intermediate result will be List[(Char, Int)]. List of tuples of chars each char will be accompanied by its count.
Now lets start going through the list one char at once using the Great! foldLeft
We will accumulate the result in the acc variable and e represents the current element.
acc is of type List[(Char, Int)] and e is of type Char
Now when we start, we are at first char of the list. Right now the acc is empty list. So, we attach first tuple to the front of the list acc
with count one.
when acc is Nil do (e, 1) :: Nil or (e, 1) :: acc note: acc is Nil
Now front of the list is the node we are interested in.
Lets go to the second element. Now acc has one element which is the first element with count one.
Now, we compare the current element with the front element of the list
if it matches, increment the count and put the (element, incrementedCount) in the front of the list in place of old tuple.
if current element does not match the last element, that means we have
new element. So, we attach new element with count 1 to the front of the list and so on.
then to convert the List[(Char, Int)] to required string representation.
Note: We are using front element of the list which is accessible in O(1) (constant time complexity) has buffer and increasing the count in case same element is found.
Scala REPL
scala> :paste
// Entering paste mode (ctrl-D to finish)
def encode(word: String): String =
word.foldLeft(List.empty[(Char, Int)]) { (acc, e) =>
acc match {
case Nil => (e, 1) :: Nil
case ((lastChar, lastCharCount) :: xs) if lastChar == e => (lastChar, lastCharCount + 1) :: xs
case xs => (e, 1) :: xs
}
}.reverse.map { case (a, num) => s"$num$a" }.foldLeft("")(_ ++ _)
// Exiting paste mode, now interpreting.
encode: (word: String)String
scala> encode("AAABBCAADEEFF")
res0: String = 3A2B1C2A1D2E2F
Bit more concise with back ticks e instead of guard in pattern matching
def encode(word: String): String =
word.foldLeft(List.empty[(Char, Int)]) { (acc, e) =>
acc match {
case Nil => (e, 1) :: Nil
case ((`e`, lastCharCount) :: xs) => (e, lastCharCount + 1) :: xs
case xs => (e, 1) :: xs
}
}.reverse.map { case (a, num) => s"$num$a" }.foldLeft("")(_ ++ _)
Here's another more simplified approach based upon this answer:
class StringCompressinator {
def compress(raw: String): String = {
val split: Array[String] = raw.split("(?<=(.))(?!\\1)", 0) // creates array of the repeated chars as strings
val converted = split.map(group => {
val char = group.charAt(0) // take first char of group string
s"${group.length}${char}" // use the length as counter and prefix the return string "AAA" becomes "3A"
})
converted.mkString("") // converted is again array, join turn it into a string
}
}
import org.scalatest.FunSuite
class StringCompressinatorTest extends FunSuite {
test("testCompress") {
val compress = (new StringCompressinator).compress(_)
val input = "AAABBCAADEEFF"
assert(compress(input) == "3A2B1C2A1D2E2F")
}
}
Similar idea with slight difference :
Case class for pattern matching the head so we don't need to use if and it also helps on printing end result by overriding toString
Using capital letter for variable name when pattern matching (either that or back ticks, I don't know which I like less :P)
case class Count(c : Char, cnt : Int){
override def toString = s"$cnt$c"
}
def compressor( counts : List[Count], C : Char ) = counts match {
case Count(C, cnt) :: tail => Count(C, cnt + 1) :: tail
case _ => Count(C, 1) :: counts
}
"AAABBCAADEEFF".foldLeft(List[Count]())(compressor).reverse.mkString
//"3A2B1C2A1D2E2F"
How do I rewrite the following loop (pattern) into Scala, either using built-in higher order functions or tail recursion?
This the example of an iteration pattern where you do a computation (comparison, for example) of two list elements, but only if the second one comes after first one in the original input. Note that the +1 step is used here, but in general, it could be +n.
public List<U> mapNext(List<T> list) {
List<U> results = new ArrayList();
for (i = 0; i < list.size - 1; i++) {
for (j = i + 1; j < list.size; j++) {
results.add(doSomething(list[i], list[j]))
}
}
return results;
}
So far, I've come up with this in Scala:
def mapNext[T, U](list: List[T])(f: (T, T) => U): List[U] = {
#scala.annotation.tailrec
def loop(ix: List[T], jx: List[T], res: List[U]): List[U] = (ix, jx) match {
case (_ :: _ :: is, Nil) => loop(ix, ix.tail, res)
case (i :: _ :: is, j :: Nil) => loop(ix.tail, Nil, f(i, j) :: res)
case (i :: _ :: is, j :: js) => loop(ix, js, f(i, j) :: res)
case _ => res
}
loop(list, Nil, Nil).reverse
}
Edit:
To all contributors, I only wish I could accept every answer as solution :)
Here's my stab. I think it's pretty readable. The intuition is: for each head of the list, apply the function to the head and every other member of the tail. Then recurse on the tail of the list.
def mapNext[U, T](list: List[U], fun: (U, U) => T): List[T] = list match {
case Nil => Nil
case (first :: Nil) => Nil
case (first :: rest) => rest.map(fun(first, _: U)) ++ mapNext(rest, fun)
}
Here's a sample run
scala> mapNext(List(1, 2, 3, 4), (x: Int, y: Int) => x + y)
res6: List[Int] = List(3, 4, 5, 5, 6, 7)
This one isn't explicitly tail recursive but an accumulator could be easily added to make it.
Recursion is certainly an option, but the standard library offers some alternatives that will achieve the same iteration pattern.
Here's a very simple setup for demonstration purposes.
val lst = List("a","b","c","d")
def doSomething(a:String, b:String) = a+b
And here's one way to get at what we're after.
val resA = lst.tails.toList.init.flatMap(tl=>tl.tail.map(doSomething(tl.head,_)))
// resA: List[String] = List(ab, ac, ad, bc, bd, cd)
This works but the fact that there's a map() within a flatMap() suggests that a for comprehension might be used to pretty it up.
val resB = for {
tl <- lst.tails
if tl.nonEmpty
h = tl.head
x <- tl.tail
} yield doSomething(h, x) // resB: Iterator[String] = non-empty iterator
resB.toList // List(ab, ac, ad, bc, bd, cd)
In both cases the toList cast is used to get us back to the original collection type, which might not actually be necessary depending on what further processing of the collection is required.
Comeback Attempt:
After deleting my first attempt to give an answer I put some more thought into it and came up with another, at least shorter solution.
def mapNext[T, U](list: List[T])(f: (T, T) => U): List[U] = {
#tailrec
def loop(in: List[T], out: List[U]): List[U] = in match {
case Nil => out
case head :: tail => loop(tail, out ::: tail.map { f(head, _) } )
}
loop(list, Nil)
}
I would also like to recommend the enrich my library pattern for adding the mapNext function to the List api (or with some adjustments to any other collection).
object collection {
object Implicits {
implicit class RichList[A](private val underlying: List[A]) extends AnyVal {
def mapNext[U](f: (A, A) => U): List[U] = {
#tailrec
def loop(in: List[A], out: List[U]): List[U] = in match {
case Nil => out
case head :: tail => loop(tail, out ::: tail.map { f(head, _) } )
}
loop(underlying, Nil)
}
}
}
}
Then you can use the function like:
list.mapNext(doSomething)
Again, there is a downside, as concatenating lists is relatively expensive.
However, variable assignemends inside for comprehensions can be quite inefficient, too (as this improvement task for dotty Scala Wart: Convoluted de-sugaring of for-comprehensions suggests).
UPDATE
Now that I'm into this, I simply cannot let go :(
Concerning 'Note that the +1 step is used here, but in general, it could be +n.'
I extended my proposal with some parameters to cover more situations:
object collection {
object Implicits {
implicit class RichList[A](private val underlying: List[A]) extends AnyVal {
def mapNext[U](f: (A, A) => U): List[U] = {
#tailrec
def loop(in: List[A], out: List[U]): List[U] = in match {
case Nil => out
case head :: tail => loop(tail, out ::: tail.map { f(head, _) } )
}
loop(underlying, Nil)
}
def mapEvery[U](step: Int)(f: A => U) = {
#tailrec
def loop(in: List[A], out: List[U]): List[U] = {
in match {
case Nil => out.reverse
case head :: tail => loop(tail.drop(step), f(head) :: out)
}
}
loop(underlying, Nil)
}
def mapDrop[U](drop1: Int, drop2: Int, step: Int)(f: (A, A) => U): List[U] = {
#tailrec
def loop(in: List[A], out: List[U]): List[U] = in match {
case Nil => out
case head :: tail =>
loop(tail.drop(drop1), out ::: tail.drop(drop2).mapEvery(step) { f(head, _) } )
}
loop(underlying, Nil)
}
}
}
}
list // [a, b, c, d, ...]
.indices // [0, 1, 2, 3, ...]
.flatMap { i =>
elem = list(i) // Don't redo access every iteration of the below map.
list.drop(i + 1) // Take only the inputs that come after the one we're working on
.map(doSomething(elem, _))
}
// Or with a monad-comprehension
for {
index <- list.indices
thisElem = list(index)
thatElem <- list.drop(index + 1)
} yield doSomething(thisElem, thatElem)
You start, not with the list, but with its indices. Then, you use flatMap, because each index goes to a list of elements. Use drop to take only the elements after the element we're working on, and map that list to actually run the computation. Note that this has terrible time complexity, because most operations here, indices/length, flatMap, map, are O(n) in the list size, and drop and apply are O(n) in the argument.
You can get better performance if you a) stop using a linked list (List is good for LIFO, sequential access, but Vector is better in the general case), and b) make this a tiny bit uglier
val len = vector.length
(0 until len)
.flatMap { thisIdx =>
val thisElem = vector(thisIdx)
((thisIdx + 1) until len)
.map { thatIdx =>
doSomething(thisElem, vector(thatIdx))
}
}
// Or
val len = vector.length
for {
thisIdx <- 0 until len
thisElem = vector(thisIdx)
thatIdx <- (thisIdx + 1) until len
thatElem = vector(thatIdx)
} yield doSomething(thisElem, thatElem)
If you really need to, you can generalize either version of this code to all IndexedSeqs, by using some implicit CanBuildFrom parameters, but I won't cover that.
I'm trying to figure out a way to group all the objects in a list depending on an x distance between the elements.
For instance, if distance is 1 then
List(2,3,1,6,10,7,11,12,14)
would give
List(List(1,2,3), List(6,7), List(10,11,12), List(14))
I can only come up with tricky approaches and loops but I guess there must be a cleaner solution.
You may try to sort your list and then use a foldLeft on it. Basically something like that
def sort = {
val l = List(2,3,1,6,10,7,11,12,14)
val dist = 1
l.sorted.foldLeft(List(List.empty[Int]))((list, n) => {
val last = list.head
last match {
case h::q if Math.abs(last.head-n) > dist=> List(n) :: list
case _ => (n :: last ) :: list.tail
}
}
)
}
The result seems to be okay but reversed. Call "reverse" if needed, when needed, on the lists. the code becomes
val l = List(2,3,1,6,10,7,11,12,14)
val dist = 1
val res = l.sorted.foldLeft(List(List.empty[Int]))((list, n) => {
val last = list.head
last match {
case h::q if Math.abs(last.head-n) > dist=> List(n) :: (last.reverse :: list.tail)
case _ => (n :: last ) :: list.tail
}
}
).reverse
The cleanest answer would rely upon a method that probably should be called groupedWhile which would split exactly where a condition was true. If you had this method, then it would just be
def byDist(xs: List[Int], d: Int) = groupedWhile(xs.sorted)((l,r) => r - l <= d)
But we don't have groupedWhile.
So let's make one:
def groupedWhile[A](xs: List[A])(p: (A,A) => Boolean): List[List[A]] = {
val yss = List.newBuilder[List[A]]
val ys = List.newBuilder[A]
(xs.take(1) ::: xs, xs).zipped.foreach{ (l,r) =>
if (!p(l,r)) {
yss += ys.result
ys.clear
}
ys += r
}
ys.result match {
case Nil =>
case zs => yss += zs
}
yss.result.dropWhile(_.isEmpty)
}
Now that you have the generic capability, you can get the specific one easily.
I have a simple piece of Scala code. I loop sequentially through a List of Strings, and I want to count the occurrence of each String which I collect as tuples (String, Int) in the list r. The part in the main function should remain (so no groupBy or something). My question is about the update function:
right now I do a find first, and then add a new tuple to r if it doesn't exist. If it does exist, I loop through r and update the counter for the matching String.
Can the update function be modified so that it is more efficient? Can r be updated in a single iteration (adding if it doesn't exist, updating the counter if it does exist)?
Thanks
var r = List[(String, Int)]() // (string, count)
def update(s: String, l: List[(String, Int)]) : List[(String, Int)] = {
if (r.find(a => a._1 == s) == None) {
(s, 1) :: r // add a new item if it does not exist
} else {
for (b <- l) yield {
if (b._1 == s) {
(b._1, b._2 + 1) // update counter if exists
} else {
b // just yield if no match
}
}
}
}
def main(args : Array[String]) : Unit = {
val l = "A" :: "B" :: "A" :: "C" :: "A" :: "B" :: Nil
for (s <- l) r = update(s, r)
r foreach println
}
I suggest you go for the functional style and use the power of Scala's collections:
ss.groupBy(identity).mapValues(_.size)
If you don't want to use groupBy or work with lazy collections (Streams), this could be the way to go:
ss.foldLeft(Map[String, Int]())((m, s) => m + (s -> (m.getOrElse(s, 0) + 1)))
Something like this also works:
val l = List("A","B","A","C","A","B")
l.foldLeft(Map[String,Int]()) {
case (a: Map[String, Int], s: String) => {
a + (s -> (1 + a.getOrElse(s, 0)))
}
}
res3: scala.collection.immutable.Map[String,Int] = Map((A,3), (B,2), (C,1))
Your present solution is horribly slow, mainly because of the choice of r as a List. But your iteration throughout the whole list in case of update can be improved on, at least. I'd write it like this
def update(s: String, l: List[(String, Int)]) : List[(String, Int)] = {
l span (_._1 != s) match {
case (before, (`s`, count) :: after) => before ::: (s, count + 1) :: after
case _ => (s, 1) :: l
}
}
Using span avoids having to search the list twice. Also, we just append after, without having to iterate through it again.
I am trying to understand the Scala quicksort example from Wikipedia. How could the sample be disassembled step by step and what does all the syntactic sugar involved mean?
def qsort: List[Int] => List[Int] = {
case Nil => Nil
case pivot :: tail =>
val (smaller, rest) = tail.partition(_ < pivot)
qsort(smaller) ::: pivot :: qsort(rest)
}
As much as I can gather at this stage qsort is a function that takes no parameters and returns a new Function1[List[Int],List[Int]] that implements quicksort through usage of pattern matching, list manipulation and recursive calls. But I can't quite figure out where the pivot comes from, and how exactly the pattern matching syntax works in this case.
UPDATE:
Thanks everyone for the great explanations!
I just wanted to share another example of quicksort implementation which I have discovered in the Scala by Example by Martin Odersky. Although based around arrays instead of lists and less of a show-off in terms of varios Scala features I personally find it much less convoluted than its Wikipedia counterpart, and just so much more clear and to the point expression of the underlying algorithm:
def sort(xs: Array[Int]): Array[Int] = {
if (xs.length <= 1) xs
else {
val pivot = xs(xs.length / 2)
Array.concat(
sort(xs filter (pivot >)),
xs filter (pivot ==),
sort(xs filter (pivot <)))
}
}
def qsort: List[Int] => List[Int] = {
case Nil => Nil
case pivot :: tail =>
val (smaller, rest) = tail.partition(_ < pivot)
qsort(smaller) ::: pivot :: qsort(rest)
}
let's pick apart a few bits.
Naming
Operators (such as * or +) are valid candidates for method and class names in Scala (hence you can have a class called :: (or a method called :: for that matter - and indeed both exist). Scala appears to have operator-overloading but in fact it does not: it's merely that you can declare a method with the same name.
Pattern Matching
target match {
case p1 =>
case p2 =>
}
Where p1 and p2 are patterns. There are many valid patterns (you can match against Strings, types, particular instances etc). You can also match against something called an extractor. An extractor basically extracts arguments for you in the case of a match, so:
target match {
case MyExtractor(arg1, arg2, arg3) => //I can now use arg1, arg2 etc
}
In scala, if an extractor (of which a case class is an example) exists called X, then the pattern X(a, b) is equivalent to a X b. The case class :: has a constructor taking 2 arguments and putting this together we get that:
case x :: xs =>
case ::(x, xs) =>
Are equivalent. This match says "if my List is an instance of :: extract the value head into x and tail into xs". pattern-matching is also used in variable declaration. For example, if p is a pattern, this is valid:
val p = expression
This why we can declare variables like:
val x :: xs = List(1, 2, 3)
val (a, b) = xs.partition(_ % 2 == 0 ) //returns a Tuple2 which is a pattern (t1, t2)
Anonymous Functions
Secondly we have a function "literal". tail is an instance of List which has a method called partition which takes a predicate and returns two lists; one of those entries satisfying the predicate and one of those entries which did not.
val pred = (el: Int) => e < 2
Declares a function predicate which takes an Int and returns true iff the int value is less than 2. There is a shorthand for writing functions inline
tail.partition(_ < pivot) // _ is a placeholder for the parameter
tail.partition( (e: Int) => e < pivot )
These two expressions mean the same thing.
Lists
A List is a sealed abstract class with only two implementations, Nil (the empty list) and :: (also called cons), which is a non-empty list consisting of a head and a tail (which is also a list). You can now see that the pattern match is a match on whether the list is empty or not. a List can be created by cons-ing it to other lists:
val l = 1 :: 2 :: Nil
val m = List(1, 2, 3) ::: List(4, 5, 6)
The above lines are simply method calls (:: is a valid method name in scala). The only difference between these and normal method calls is that, if a method end in a colon : and is called with spaces, the order of target and parameter is reversed:
a :: b === b.::(a)
Function Types
val f: A => B
the previous line types the reference f as a function which takes an A and returns a B, so I could then do:
val a = new A
val b: B = f(a)
Hence you can see that def qsort: List[Int] => List[Int] declares a method called qsort which returns a function taking a List[Int] and returning a List[Int]. So I could obviously do:
val l = List(2, 4, 1)
val m = qsort.apply(l) //apply is to Function what run is to Runnable
val n = qsort(l) //syntactic sugar - you don't have to define apply explicitly!
Recursion
When a method call is tail recursive, Scala will optimize this into the iterator pattern. There was a msitake in my original answer because the qsort above is not tail-recursive (the tail-call is the cons operator)
def qsort: List[Int] => List[Int] = {
case Nil => Nil
case pivot :: tail =>
val (smaller, rest) = tail.partition(_ < pivot)
qsort(smaller) ::: pivot :: qsort(rest)
}
Let's rewrite that. First, replace the function literal with an instance of Function1:
def qsort: List[Int] => List[Int] = new Function1[List[Int], List[Int]] {
def apply(input: List[Int]): List[Int] = input match {
case Nil => Nil
case pivot :: tail =>
val (smaller, rest) = tail.partition(_ < pivot)
qsort(smaller) ::: pivot :: qsort(rest)
}
}
Next, I'm going to replace the pattern match with equivalent if/else statements. Note that they are equivalent, not the same. The bytecode for pattern matches are more optimized. For instance, the second if and the exception throwing below do not exist, because the compile knows the second match will always happen if the first fails.
def qsort: List[Int] => List[Int] = new Function1[List[Int], List[Int]] {
def apply(input: List[Int]): List[Int] = if (input == Nil) {
Nil
} else if (input.isInstanceOf[::[_]] &&
scala.collection.immutable.::.unapply(input.asInstanceOf[::[Int]]) != None) {
val unapplyResult = scala.collection.immutable.::.unapply(input.asInstanceOf[::[Int]]).get
val pivot = unapplyResult._1
val tail = unapplyResult._2
val (smaller, rest) = tail.partition(_ < pivot)
qsort(smaller) ::: pivot :: qsort(rest)
} else {
throw new scala.MatchError(input)
}
}
Actually, val (smaller, rest) is pattern match as well, so Let's decompose it as well:
def qsort: List[Int] => List[Int] = new Function1[List[Int], List[Int]] {
def apply(input: List[Int]): List[Int] = if (input == Nil) {
Nil
} else if (input.isInstanceOf[::[_]] &&
scala.collection.immutable.::.unapply(input.asInstanceOf[::[Int]]) != None) {
val unapplyResult0 = scala.collection.immutable.::.unapply(input.asInstanceOf[::[Int]]).get
val pivot = unapplyResult0._1
val tail = unapplyResult0._2
val tmp0 = tail.partition(_ < pivot)
if (Tuple2.unapply(tmp0) == None)
throw new scala.MatchError(tmp0)
val unapplyResult1 = Tuple2.unapply(tmp0).get
val smaller = unapplyResult1._1
val rest = unapplyResult1._2
qsort(smaller) ::: pivot :: qsort(rest)
} else {
throw new scala.MatchError(input)
}
}
Obviously, this is highly unoptmized. Even worse, there are some function calls being done more than once, which doesn't happen in the original. Unfortunately, to fix that would require some structural changes to the code.
There's still some syntactic sugar here. There is an anonymous function being passed to partition, and there is the syntactic sugar for calling functions. Rewriting those yields the following:
def qsort: List[Int] => List[Int] = new Function1[List[Int], List[Int]] {
def apply(input: List[Int]): List[Int] = if (input == Nil) {
Nil
} else if (input.isInstanceOf[::[_]] &&
scala.collection.immutable.::.unapply(input.asInstanceOf[::[Int]]) != None) {
val unapplyResult0 = scala.collection.immutable.::.unapply(input.asInstanceOf[::[Int]]).get
val pivot = unapplyResult0._1
val tail = unapplyResult0._2
val func0 = new Function1[Int, Boolean] {
def apply(input: Int): Boolean = input < pivot
}
val tmp0 = tail.partition(func0)
if (Tuple2.unapply(tmp0) == None)
throw new scala.MatchError(tmp0)
val unapplyResult1 = Tuple2.unapply(tmp0).get
val smaller = unapplyResult1._1
val rest = unapplyResult1._2
qsort.apply(smaller) ::: pivot :: qsort.apply(rest)
} else {
throw new scala.MatchError(input)
}
}
For once, the extensive explanations about each syntactic sugar and how it works are being done by others. :-) I hope this complements their answers. Just as a final note, the following two lines are equivalent:
qsort(smaller) ::: pivot :: qsort(rest)
qsort(rest).::(pivot).:::(qsort(smaller))
The pivot in this pattern matching example is the first element of the list:
scala> List(1,2,3) match {
| case x :: xs => println(x)
| case _ => println("empty")
| }
1
The pattern matching is based on extractors and the cons is not part of the language. It uses the infix syntax. You can also write
scala> List(1,2,3) match {
| case ::(x,xs) => println(x)
| case _ => println("empty")
| }
1
as well. So there is a type :: that looks like the cons operator. This type defines how it is extracted:
final case class ::[B](private var hd: B, private[scala] var tl: List[B]){ ... }
It's a case class so the extractor will be generated by the Scala compiler. Like in this example class A.
case class A(x : Int, y : Int)
A(1,2) match { case x A y => printf("%s %s", x, y)}
-> 1 2
Based on this machinary patterns matching is supported for Lists, Regexp and XML.