I'm trying to recursively iterate through a list in Scala using pattern matching. I cannot use any list functions, or while/for loops. What I need to do is iterate through the list, and remove an element if it matches to be '4'. I'm new to Scala and I cannot find the answer in the textbook I have nor on google. Everyone else uses the filter method, or some other list method.
Here's what I tried to do (which is wrong)
def removeFours(lst: List[Int]): List[Int] = {
val newLst = lst
lst match {
case Nil => Nil
case a if a == 4 => newLst -= 0
case n => removeFours(newLst)
}
newLst
}
See if this works for you.
def removeFours(lst: List[Int], acc: List[Int] = List.empty): List[Int] = {
lst match {
case Nil => acc.reverse
case 4 :: t => removeFours( t, acc )
case h :: t => removeFours( t, h :: acc )
}
}
Usage:
scala> removeFours( List(3,7,4,9,2,4,1) )
res84: List[Int] = List(3, 7, 9, 2, 1)
Using an inner function and pattern matching to de-structure the list. If the head in the list is 4, then do not add it to the accumulator. If it is, append it to the accumulator.
def removeFours(lst: List[Int]): List[Int] = {
def loop(lst: List[Int], acc: List[Int]): List[Int] = lst match {
case Nil => acc
case h :: t =>
if (h == 4) {
loop(t, acc)
}else{
loop(t, acc :+ h)
}
}
loop(lst, List())
}
The preferred way to do this is with guards in the pattern match but the if else statement may look more familiar if you're just getting started with scala.
def removeFours(lst: List[Int]): List[Int] = {
def loop(lst: List[Int], acc: List[Int]): List[Int] = lst match {
case Nil => acc
case h :: t if (h == 4) => loop(t, acc)
case h :: t => loop(t, acc :+ h)
}
loop(lst, List())
}
I am not sure about the execution time. I am also new to scala but I am taking bollean approach to filter any list.
object Main extends App {
//fun that will remove 4
def rm_4(lst: List[Int]) : List[Int] = {
val a = lst.filter(kill_4)
a
}
// boolean fun for conditions
def kill_4(n: Int) : Boolean = {
if (n ==4) false
else true
}
println(rm_4(List(1,2,4,5,4))) // outpur List(1,2,5)
}
Related
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
}
how to sum of elements in Scala List[List[Int]]
example :
val mylist = List(List(1,2,3),List(2,3,4))
output :
List(List(6),List(9))
If you want the result list to be of type List[List[Int]] and not List[Int], you can achieve it by doing this:
mylist.map(l => l.foldLeft(List(0))((acc, el) => List(acc(0) + el)))
Since mylist.map(_.sum) will result in an List[Int] you can also do it this way:
mylist.map(_.sum).map(List(_))
And if you really want to achieve it without the scala collection library, you can do it with recursion:
object Appz extends App {
val mylist: List[List[Int]] = List(List(2,2,3),List(2,3,4))
def sumSubLists(ll: List[List[Int]]): List[Int] = {
#annotation.tailrec
def sumList(l: List[Int], acc: Int = 0): Int = l match {
case List() => acc
case h :: t => sumList(t, acc + h)
}
#annotation.tailrec
def sumSubListsInner(ll: List[List[Int]], acc: List[Int]): List[Int] =
ll match {
case List() => acc
case h :: t => sumSubListsInner(t, acc ++ List(sumList(h)))
}
sumSubListsInner(ll, List())
}
println(sumSubLists(mylist))
}
I want to remove duplicates from the list recursively using pattern matching with Scala
here is my input
val xs = List(1,2,3,4,6,3,2,7,9,4)
Tried code:
def removeDuplicates(xs : List[Int]) : List[Int] = xs match {
case Nil =>Nil
case x::ys => {
if(ys.contains(x)){
removeDuplicates(ys)
} else {
}
/// ???
}
}
I was stuck at the question mark, how to appened my result to the mutable list and return it.
Thank you.
You're close:
def removeDuplicates(xs : List[Int]) : List[Int] = xs match {
case Nil => Nil
case x::ys => if (ys.contains (x)) removeDuplicates (ys) else
x :: removeDuplicates (ys)
}
scala> removeDuplicates (List (1,2,3,4,6,3,2,7,9,4))
res143: List[Int] = List(1, 6, 3, 2, 7, 9, 4)
While this is a brief solution, it isn't tail recursive and therefore vulnerable for stackoverflows - whereas Jean Logearts solution solves the problem.
Here is an alternative solution with an inner function, tailrecursive too:
def removeDuplicates (xsOuter : List[Int]) : List[Int] = {
#annotation.tailrec
def removeDuplicates (xs: List[Int], collected: List[Int]) : List[Int] = xs match {
case Nil => collected
case x :: ys => if (collected.contains (x)) removeDuplicates (ys, collected) else
removeDuplicates (ys, x :: collected)
}
removeDuplicates (xsOuter, Nil)
}
scala> removeDuplicates (List (1,2,3,4,6,3,2,7,9,4))
res151: List[Int] = List(9, 7, 6, 4, 3, 2, 1)
with a bias to the first elements, but with the result reversed, which can be easily corrected by returning collected.reverse in the Nil case, if it is important.
The outer function serves the job to provide a simple, one-argument interface to the user, so he doesn't need to provide an empty List.
Note, that the solution is crying for a type annotation, since it doesn't depend at all on the List elements being of type Int:
scala> def removeDuplicates [A] (xsOuter : List[A]) : List[A] = {
|
| #annotation.tailrec
| def removeDuplicates (xs: List[A], collected: List[A]) : List[A] = xs match {
| case Nil => collected
| case x :: ys => if (collected.contains (x)) removeDuplicates (ys, collected) else
| removeDuplicates (ys, x :: collected)
| }
|
| removeDuplicates (xsOuter, Nil)
| }
removeDuplicates: [A](xsOuter: List[A])List[A]
scala> removeDuplicates (List (1,2,3,4,6,3,2,7,9,4))
res152: List[Int] = List(9, 7, 6, 4, 3, 2, 1)
You need to keep track of the current state: already seen elements using a set (for fast lookup), and the new list being constructed:
#tailrec
def removeDuplicatesRec(
remaining: List[Int],
seen: Set[Int],
acc: List[Int]
): List[Int] = remaining match {
case Nil => acc
case head :: tail =>
if (!seen.contains(head)) removeDuplicatesRec(tail, seen + head, acc :+ head)
else removeDuplicatesRec(tail, seen, acc)
}
def removeDuplicates(xs: List[Int]): List[Int] =
removeDuplicatesRec(xs, Set.empty, List.empty)
Here's a classic approach using an inner function and tail recursion. The tail recursion may not be necessary for small lists but it is easier for me to reason about.
def removeDuplicates(xs : List[Int]) : List[Int] = {
#scala.annotation.tailrec
def accumulator(xs: List[Int], acc: List[Int]):List[Int] = xs match {
case Nil => acc
case h::t if(!acc.contains(h)) => accumulator(t, h :: acc)
case h::t if(acc.contains(h)) => accumulator(t, acc)
}
accumulator(xs, List[Int]())
}
scala> removeDuplicates(List(1,2,3,4,6,3,2,7,9,4))
res16: List[Int] = List(9, 7, 6, 4, 3, 2, 1)
Of course, distinct is the preferred way to do this but this is a good exercise. distinct can be used to verify your solution though.
scala> List(1,2,3,4,6,3,2,7,9,4).distinct == removeDuplicates(List(1,2,3,4,6,3,2,7,9,4)).sorted
res21: Boolean = true
I want to write a function which will add an element n to a list n times for all elements in an input list.
For example:
L = List(2,4,1)
Output should be:
List(2,2,4,4,4,4,1)
I would like to do it with tail recursion. So far, I have written this:
def repeat(numbers: List[Int]): List[Int] = {
def repeat_acc(numbers: List[Int], acc: List[Int], number_acc: Int): List[Int] = {
if (numbers.length == 0)
return acc
else if (number_acc == 0)
repeat_acc(numbers.tail, acc, numbers.head)
else
repeat_acc(numbers, acc :+ numbers.head, (number_acc-1))
}
repeat_acc(numbers, List(), 0)
}
The problem is that it leaves out the first element of the list. For this output will be:
(4, 4, 1, 1, 1, 1)
I know why it happens, but I cannot fix it. I have tried many other ways, but for me it seems that tail recursion do not work here. Some always goes wrong, and I get the wrong result.
Thanks for any advice.
I know you want to do a personalized tail recursive call, but I would recommend the following instead:
def repeat(numbers: List[Int]): List[Int] = {
numbers.flatMap(n => List.fill(n)(n))
}
The inner function takes the value n and repeats it n times in a list, and then this function is flat mapped onto the original list (a regular map would turn List(1,2,3) into List(List(1), List(2, 2), List(3, 3, 3)), so we use flat map). This has the benefit of 'doing it the Scala way' with built in collections functionality.
This problem gets a lot simpler if you add another parameter to your inner recursive method, representing the current number being added. You should also familiarize yourself with match statements, since they are really powerful in Scala and can help express exactly this type of logic. Nested if/else statements and early return statements are considered unidiomatic. Try:
def repeat(numbers: List[Int]) = {
def repeatAcc(acc: List[Int], curr: Int, rem: Int, numbers:List[Int]): List[Int] =
(numbers, rem) match {
case (Nil, 0) => acc
case (hd::tl, 0) => repeatAcc(acc, hd, hd, tl)
case (_, n) => repeatAcc(acc :+ curr, curr, n - 1, numbers)
}
repeatAcc(List.empty[Int], 0, 0, numbers)
}
You might also try to use some standard Scala methods like List.fill, which can be used in conjunction with the tail recursion as follows:
def repeat(numbers: List[Int]) = {
def repeatAcc(acc: List[Int], numbers:List[Int]): List[Int] = numbers match {
case hd::tl => repeatAcc(acc ++ List.fill(hd)(hd), tl)
case Nil => acc
}
repeatAcc(List.empty[Int], numbers)
}
Finally, I understand you're trying to learn about these core concepts, but it should be mentioned that this is really simple with Scala built ins:
(List.empty[Int] /: numbers) { case (soFar, next) => soFar ++ List.fill(next)(next) }
numbers.flatMap(x => List.fill(x)(x))
Try this, it works
import scala.annotation.tailrec
#tailrec
def f(l: List[Int], res: List[Int] = Nil): List[Int] = {
#tailrec
def g(n: Int, acc: List[Int]): List[Int] = {
if (n == 1) acc
else g(n - 1, acc :+ acc.head)
}
if (l == Nil) res else f(l.tail, res ++ g(l.head, List(l.head)))
}
scala> f(List(1,2,3))
res0: List[Int] = List(1, 2, 2, 3, 3, 3)
Ok, I find the solution. Looks bad, but works. I'm too tired to figure out something better.
Here is my code:
def repeat(numbers: List[Int]): List[Int] = {
def repeat_acc(numbers: List[Int], acc: List[Int], number_acc: Int): List[Int] = {
if (numbers.length == 1 && number_acc == 0)
return acc
else if (acc.length == 0)
repeat_acc(numbers, acc :+ numbers.head, (number_acc-1))
else if (number_acc == 0 && numbers.tail != Nil ){
repeat_acc(numbers.tail, acc, (numbers.tail.head))
}
else
repeat_acc(numbers, acc :+ numbers.head, (number_acc-1))
}
repeat_acc(numbers, List(), numbers.head)
}
simple, just use fill + flatMap
val result = L.flatMap(x => List.fill(x)(x))
for recursive solution :
scala> def repeat(numbers: List[Int]): List[Int] = {
def run(nums:List[Int]):List[Int] ={
nums match {
case Nil => List.empty[Int]
case x::Nil => List.fill(x)(x)
case x::xs => List.fill(x)(x):::run(xs)
}}
run(numbers)
}
scala> repeat(List(2,4,1))
res1: List[Int] = List(2, 2, 4, 4, 4, 4, 1)
For example, I have Seq(1,2,3) and I want to get Seq(1,0,2,0,3)
The first thing that comes to mind is:
scala> Seq(1,2,3).flatMap(e => 0 :: e :: Nil).tail
res17: Seq[Int] = List(1, 0, 2, 0, 3)
Is there any better/more elegant option?
Here is another approach:
def intersperse[E](x: E, xs:Seq[E]): Seq[E] = (x, xs) match {
case (_, Nil) => Nil
case (_, Seq(x)) => Seq(x)
case (sep, y::ys) => y+:sep+:intersperse(sep, ys)
}
which is safe over empty Seqs too.
Try for comprehension:
for(i <- list; p <- List(0, i)) yield p
However you must somehow remove the first element (it yields: 0,1,0,2,0,3), either by:
(for(i <- list; p <- List(0, i)) yield p).tail
or:
list.head :: (for(i <- list.tail; p <- List(0, i)) yield p)
def intersperse[T](xs: List[T], item: T): List[T] = xs match {
case Nil => xs
case _ :: Nil => xs
case a :: ys => a :: item :: intersperse(ys, item)
}
Can also use this extension:
implicit class SeqExtensions[A](val as: Seq[A]) extends AnyVal {
def intersperse(a: A): Seq[A] = {
val b = Seq.newBuilder[A]
val it = as.iterator
if (it.hasNext) {
b += it.next()
while(it.hasNext) {
b += a
b += it.next()
}
}
b.result()
}
}