Append strings to array with Immutable

Question

Having as Input a list of String (one of them could be None). How can I return a Array of String using Immutable Object.

It's pretty easy if I use var or Mutable Object, here an example:

def getArrayString(string: String, list1: List[String], list2: Option[List[String]]): Array[String] = {
  var ret = Array[String]()
  ret = ret :+ string
  if (list1.nonEmpty) {
    for (item <- list1) {
      ret = ret :+ item
    }
  }
  if (list2.isDefined) {
    for (item <- list2.get) {
      ret = ret :+ item
    }
  }
  ret
}

Question1: What if I want to use just val object?

N.B.: if list2 is None the returning array should not have any None Object

Question2: ..and if list1 and list2 were List[CustomClass] where CustomClass is

case class CustomClass(string:String)

How would you do?

Question3: ...What if we complicate the method with...

case class CustomClass1(string1:String)

case class CustomClass2(string2:String)

obviously CustomClass1 and CustomClass2 might have some other parameters in their class that make them different from each other. The signature of the method would be then:

def getArrayString( string: String
                , list1: List[CustomClass1]
                , list2: Option[List[CustomClass2]]
              ): Array[String]`

If CustomClass is the same for both lists, you can map it just once at the end — Alvaro Carrasco
– Alvaro Carrasco, Commented Jan 10, 2017 at 18:55

Alvaro Carrasco · Accepted Answer · 2017-01-10 18:54:35Z

1

You can use :: which will prepend an element to a list, and ++ which will concatenate 2 lists:

val ret = (string :: (list1 ++ list2.getOrElse(Nil))).toArray

For the updated version:

val ret = (string :: (list1 ++ list2.getOrElse(Nil)).map(_.string)).toArray

edited Jan 10, 2017 at 18:54

answered Jan 10, 2017 at 15:56

Alvaro Carrasco

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5 Comments

tkachuko Over a year ago

In a hurry to answer first you got a compilation error :) Move brackets to include string variable.

Alvaro Carrasco Over a year ago

@tkachuko Updated

Seth Tisue Over a year ago

Nitpick: mixing :: and ++ seems a bit weird to me — I would either use :: and :::, or +: and ++. One is the List API, the other is the Seq API (which also works for Lists).

salvob Over a year ago

@SethTisue, why is it weird? is it inefficient?

Seth Tisue Over a year ago

The differently-named methods do the same thing, there's no efficiency difference. It's no big deal, it just reads a bit oddly to me to mix and match the two APIs.

jwvh · Accepted Answer · 2017-01-10 19:46:30Z

1

Every time I see Option I think "fold?".

def getArrayString( string: String
                  , list1: List[String]
                  , list2: Option[List[String]]
                  ): Array[String] =
  list2.foldLeft(string +: list1)(_++_).toArray

update (as requested)

case class CustomClass(string:String)
def getArrayString( string: String
                    , list1: List[CustomClass]
                    , list2: Option[List[CustomClass]]
                  ): Array[String] =
  string +: list2.foldLeft(list1)(_++_).map(_.string).toArray

Or the slightly more concise:

  Array(string) ++ (list1 /: list2)(_++_).map(_.string)

edited Jan 10, 2017 at 19:46

answered Jan 10, 2017 at 16:39

jwvh

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9 Comments

salvob Over a year ago

Thank you jwvh, but this doesn't work if list2 is None and it is less readable than Alvaro's answer

jwvh Over a year ago

Really? What should be returned if list2 is None?

salvob Over a year ago

It has to return an Array of String.. EDIT: I did some testing and yes. Your code return what it should :)

salvob Over a year ago

I edited the question, could you extend your answer please?

salvob Over a year ago

Thank you, really appreciate it! ... but if CustomClass (say CustomClass1 and CustomClass2, respectively for list1 and list2) is not the same for both lists, this would not work though

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