Scala way to remove duplicate in an Array

Question

I want to learn how to write for loop functions without declear "var" at the front of the code. For example, I want to remove duplicates in an interger Array. In C++/Java, I can do like that:

int removeDuplicate(vector<int> nums)
{
     vector<int> output
     Map<int,int> counter;
     for(i = 0; i < nums.size(); i++)
     {
         if(!counter.has_key(nums[i]))
         {
             counter[nums[i]]=1;  //add new key-value pair
             output.push_back(nums[i]);
         }
     }

     return output;
}

But in scala, how to use immutable variables to complete above task.

Do NOT use scala's internal functions, such as distinct. This question is about Scala implementation.

distinct is a method, not a function, so I guess that means you are allowed to use it? Note that in your C++ example, size(), has_key(), push_back(), operator[], etc. are also "internal functions", yet you seem to have no trouble using them, so what makes those "internal functions" different from other "internal functions"? And what is an "internal function" anyway? — Jörg W Mittag
– Jörg W Mittag, Commented Sep 9, 2017 at 9:39

Nyavro · Accepted Answer · 2017-09-09 01:10:53Z

15

In Scala to avoid using vars in this case you can use recursion or foldLeft or foldRight:

def distinct[A](list:List[A]):List[A] = 
  list.foldLeft(List[A]()) {
    case (acc, item) if acc.contains(item) => acc
    case (acc, item) => item::acc
  }

answered Sep 9, 2017 at 1:10

Nyavro

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2 Comments

lznt Over a year ago

is this gonna reverse the list?

Nyavro Over a year ago

using foldLeft with appending to head reverses the list

Mahesh Chand · Accepted Answer · 2017-09-09 06:06:23Z

6

Scala provides lots of library functions. Like, there is distinct function. You can directly call it.

scala> val array = Array(1,2,3,2,1,4)
array: Array[Int] = Array(1, 2, 3, 2, 1, 4)
scala> array.distinct
res0: Array[Int] = Array(1, 2, 3, 4)

And if there is no library function, we prefer to write code using recursion mostly tail recursion so that we can maintain immutability. Or you may convert array into Set which will remove duplicates and then convert set into array again to get desired result.

scala> array.toSet.toArray
res3: Array[Int] = Array(1, 2, 3, 4)

In scala, list is preferred over array as array is mutable collection in scala and list is immutable. So prefer to use it.

answered Sep 9, 2017 at 6:06

Mahesh Chand

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1 Comment

jwvh Over a year ago

Worth noting; .toSet.toArray will not maintain element order.

Sohum Sachdev · Accepted Answer · 2017-09-09 06:56:51Z

0

There are several ways to do this in Scala (without distinct). All of these alternatives use immutable variables:

Using groupBy:

val x: Seq[Int] = Seq(1,2,3,3,4,5,6,6)
val unique: Seq[Int] = x.groupBy(identity).keys.toSeq

Using toSet:

val x: Seq[Int] = Seq(1,2,3,3,4,5,6,6)
val unique: Seq[Int] = x.toSet.toSeq

answered Sep 9, 2017 at 6:56

Sohum Sachdev

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