I wanted to write a Scala program that takes command-line args as list input and provide the output list without duplicates.
I want to know the custom implementation of this without using any libraries.
Input : 4 3 7 2 8 4 2 7 3
Output :4 3 7 2 8
val x= List(4, 3, 7, 2, 8, 4, 2, 7, 3)
x.foldLeft(List[Int]())((l,v)=> if (l.contains(v)) l else v :: l)
if you can't use contains you can do another fold
x.foldLeft(List[Int]())((l,v)=> if (l.foldLeft(false)((contains,c)=>if (c==v ) contains | true else contains | false)) l else v :: l)
Here's a way you could do this using recursion. I've tried to lay it out in a way that's easiest to explain:
import scala.annotation.tailrec
#tailrec
def getIndividuals(in: List[Int], out: List[Int] = List.empty): List[Int] = {
if(in.isEmpty) out
else if(!out.contains(in.head)) getIndividuals(in.tail, out :+ in.head)
else getIndividuals(in.tail, out)
}
val list = List(1, 2, 3, 4, 5, 4, 3, 5, 6, 0, 7)
val list2 = List(1)
val list3 = List()
val list4 = List(3, 3, 3, 3)
getIndividuals(list) // List(1, 2, 3, 4, 5, 6, 0, 7)
getIndividuals(list2) // List(1)
getIndividuals(list3) // List()
getIndividuals(list4) // List(3)
This function takes two parameters, in and out, and iterates through every element in the in List until it's empty (by calling itself with the tail of in). Once in is empty, the function outputs the out List.
If the out List doesn't contain the value of in you are currently looking at, the function calls itself with the tail of in and with that value of in added on to the end of the out List.
If out does contain the value of in you are currently looking at, it just calls itself with the tail of in and the current out List.
Note: This is an alternative to the fold method that Arnon proposed. I personally would write a function like mine and then maybe refactor it into a fold function if necessary. I don't naturally think in a functional, fold-y way so laying it out like this helps me picture what's going on as I'm trying to work out the logic.
Related
I have a list of elements List(1,2,3,4,5,6) which I hope to get a few elements from it to form a new List to List(2,4,5,6).
How should I go about it? Thanks!
scala collections can be mapped or filtered. In your case you simply can filter with the function you want.
eg. in scala REPL.
filter elements which are greater than or equals to 2.
scala> List(1,2,3,4,5,6).filter(_>=2)
res3: List[Int] = List(2, 3, 4, 5, 6)
or to filter all elements which are not 1 and 3,
scala> List(1,2,3,4,5,6).filter(element => (element!=1 && element!=3))
res6: List[Int] = List(2, 4, 5, 6)
Also read
https://twitter.github.io/scala_school/collections.html#filter
http://alvinalexander.com/scala/how-to-use-filter-method-scala-collections-cookbook
For a practical exercise I need to define a function that basically changes the index of every value in an odd index in a list, so that I would get this:
changePairs(List(1,2,3,4,5,6,7,8,9,10,11))
//> res62: List[Int] = List(2, 1, 4, 3, 6, 5, 8, 7, 10, 9, 11)
changePairs(List(2,2,30,4,50,6,7,80,9,100))
//> res63: List[Int] = List(2, 2, 4, 30, 6, 50, 80, 7, 100, 9)
So basically I need to swap the places of each odd-even pair, and in case I'm left with a single odd element at the last index (11 in the first example), I leave it as it is.
I have this but it's definitely not working, and I'm not really sure why.
def changePairs(a: List[Int]) = a.zipWithIndex.map {
case (s,i) => if (i % 2 != 0) a.patch(i,Seq(s),1); a.patch(i-2,Seq(s),0);
}
Here's one way:
def changePairs(a: List[Int]) = a.grouped(2).flatMap {
case List(a, b) => List(b, a)
case a => a
}.toList
changePairs(List(1, 2, 3, 4, 5, 6, 7)) // List(2, 1, 4, 3, 6, 5, 7)
Main idea that gets you going is once you think of grouping the list into sublists of two elements, which is what grouped(2) does. From then on it's an easy ride - describe two cases, one with two elements (in that case we flip them) and one with only one element, such as 7 in my example, in which case we just leave it be. We use flatMap to flatten the resulting list of 2-element lists into one big list, and we do .toList to get out of an iterator that we got from grouped.
EDIT:
I now saw a.grouped(2).map(_.reverse).flatten.toList in the comments. Yeah, that works too, it's the same as this but much less verbose since instead of "manually" swapping the elements, we just do a reverse on each sublist.
You could also use recursion and pattern matching. This is efficient as you are only going through the list once:
def changePairs(l: List[Int]): List[Int] = {
l match {
case a :: b :: tail => b :: a :: changePairs(tail)
case _ => Nil
}
}
For example I have following Scala list, I want get a sublist until there is a requirement can be met.
val list = Seq(1,2,3,4,5,5,4,1,2,5)
The requirement is the number is 5, so I want the result as:
Seq(1,2,3,4)
Currently I use Scala collection's indexWhere and splitAt to return:
list.splitAt(list.indexWhere(x => x == 5))
(Seq(1,2,3,4), Seq(5,5,4,1,2,5))
I am not sure there are more better ways to achieve the same with better Scala collection's method I didn't realise?
You can use takeWhile:
scala> val list = Seq(1,2,3,4,5,5,4,1,2,5)
list: Seq[Int] = List(1, 2, 3, 4, 5, 5, 4, 1, 2, 5)
scala> list.takeWhile(_ != 5)
res30: Seq[Int] = List(1, 2, 3, 4)
Use span like this,
val (l,r) = list.span(_ != 5)
which delivers
l: List(1, 2, 3, 4)
r: List(5, 5, 4, 1, 2, 5)
Alternatively, you can write
val l = list.span(_ != 5)._1
to access only the first element of the resulting tuple.
This bisects the list at the first element that does not hold the condition.
I've manually built a method that takes 2 arrays and combines them to 1 like this:
a0,a1,a2,b0,b1,a3,a4,a5,b2,b3,a6,...
So I always take 3 elements of the first array, then 2 of the second one.
As I said, I built that function manually.
Now I guess I could make this a one-liner instead with the help of zip. The problem is, that zip alone is not enough as zip builds tuples like (a0, b0).
Of course I can flatMap this, but still not what I want:
val zippedArray: List[Float] = data1.zip(data2).toList.flatMap(t => List(t._1, t._2))
That way I'd get a List(a0, b0, a1, b1,...), still not what I want.
(I'd then use toArray for the list... it's more convenient to work with a List right now)
I thought about using take and drop but they return new data-structures instead of modifying the old one, so not really what I want.
As you can imagine, I'm not really into functional programming (yet). I do use it and I see huge benefits, but some things are so different to what I'm used to.
Consider grouping array a by 3, and array b by 2, namely
val a = Array(1,2,3,4,5,6)
val b = Array(11,22,33,44)
val g = (a.grouped(3) zip b.grouped(2)).toArray
Array((Array(1, 2, 3),Array(11, 22)), (Array(4, 5, 6),Array(33, 44)))
Then
g.flatMap { case (x,y) => x ++ y }
Array(1, 2, 3, 11, 22, 4, 5, 6, 33, 44)
Very similar answer to #elm but I wanted to show that you can use more lazy approach (iterator) to avoid creating temp structures:
scala> val a = List(1,2,3,4,5,6)
a: List[Int] = List(1, 2, 3, 4, 5, 6)
scala> val b = List(11,22,33,44)
b: List[Int] = List(11, 22, 33, 44)
scala> val groupped = a.sliding(3, 3) zip b.sliding(2, 2)
groupped: Iterator[(List[Int], List[Int])] = non-empty iterator
scala> val result = groupped.flatMap { case (a, b) => a ::: b }
result: Iterator[Int] = non-empty iterator
scala> result.toList
res0: List[Int] = List(1, 2, 3, 11, 22, 4, 5, 6, 33, 44)
Note that it stays an iterator all the way until we materialize it with toList
I have some experience with R language and now I wanted to try Scala language. In R language I can assign one value to many elements of a vector, e.g.
(xs <- 1:10)
#[1] 1 2 3 4 5 6 7 8 9 10
k <- 3
xs[1:k] <- xs[k+1]
xs
# 4 4 4 4 5 6 7 8 9 10
It assigns value of k+1 element to all elements of indices from 1 to k. Is it also possible to do it without a loop in Scala (I mean Array in Scale)? I know there is slice method, but it only returns values of an Array, one cannot modify elements of the Array using this method.
What is even more, should I use Array or ArrayBuffer if I only want to change values of elements and I do not want to add/remove elements from a collection?
Check out the java.util.Arrays.fill methods.
scala> val xs = (1 to 9).toArray
xs: Array[Int] = Array(1, 2, 3, 4, 5, 6, 7, 8, 9)
scala> val k = 6
k: Int = 6
scala> java.util.Arrays.fill(xs, 0, k, xs(k))
scala> xs
res10: Array[Int] = Array(7, 7, 7, 7, 7, 7, 7, 8, 9)
For your second question, if not resizing the collection but editing the elements, stick with array. ArrayBuffer is much like the Java ArrayList, it resizes it self when it needs to, so insertion is amortized constant, not just constant.
For your first question, I'm not aware of any method in the collections library that would allow you to do this. It's obviously syntactic sugar for looping, so if you really care (do you really find yourself needing to do this often?), you can define an implicit class and yourself define a method which loops, and then use that. Write a comment if you would like to see example of such code, otherwise try doing it yourself, it's gonna be good training.
Scala has the Range class. You can convert the Range to an Array if you wish.
scala> val n = 10
n: Int = 10
scala> Range(1,n)
res22: scala.collection.immutable.Range = Range(1, 2, 3, 4, 5, 6, 7, 8, 9)
scala> res22.toArray
res23: Array[Int] = Array(1, 2, 3, 4, 5, 6, 7, 8, 9)
│
ArrrayBuffer has constant time update and would be good for updating values.