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Run a function in scala with a list as input


Run a function in scala with a list as input

By : user7454494
Date : November 18 2020, 03:01 PM
I hope this helps you . Well, first if you want to have method that takes list of strings, then it should look like:
code :
def myMethod(someList: List[String]) = {...}
someList.exists(s => s.contains("programmation"))
someList.count(s => s.contains("programmation"))
def countElements(allElements: List[String], substring: String): Int = 
    allElements.count(s => s.toLowerCase.contains(substring.toLowerCase))


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Scala polymorphic function for filtering an input List of Either

Scala polymorphic function for filtering an input List of Either


By : Alfredo Monroy
Date : March 29 2020, 07:55 AM
I hope this helps you . Seeking a more elegant solution , collect is made for exactly this kind of situation:
code :
def filterMe[U,T](in: List[Either[U,T]]): List[T] = in.collect{
  case Right(r) => r
}
ins.map(DSJsonMapper.parseDsResult).collect{ case Right(r) => r }
Scala way of applying function to list and getting highest result with input

Scala way of applying function to list and getting highest result with input


By : Chris
Date : March 29 2020, 07:55 AM
To fix the issue you can do You can use map to construct a sequence of (score, split) pairs and then use maxBy to find the pair with the highest score:
code :
splitCandidates.map(c => (getSplitScore(tSeriesDistances, c), c)).maxBy(_._1)
scala: list as input and output of function

scala: list as input and output of function


By : Wayne Drury
Date : March 29 2020, 07:55 AM
Hope that helps You can't get a List[String] from a List[Any]. (Well, you can, but it's a really bad thing to do.)
Don't, don't, don't create a List[Any]. Unlike Python, Scala is a strictly typed language, which means that the compiler keeps a close watch on the type of each variable and every collection. When the compiler looses track of the List type it becomes List[Any] and you've lost all the assistance the compiler offers to help write programs that don't crash.
code :
def g(tup: (Int,Int,String,String)): List[String] =
  if (tup._1 >= 1 & tup._2 <= 100) List(tup._3, tup._4)
  else List()
val v = (1, 100, "a1", "b1")
val w = g(v)  //w: List[String] = List(a1, b1)
Scala Seq's filterNot implementation doesn't accept another List as function input parameter

Scala Seq's filterNot implementation doesn't accept another List as function input parameter


By : Dhananjai Bajpai
Date : March 29 2020, 07:55 AM
Hope that helps The difference is the apply() method for Seq and Set collections. Look at the definition of filterNot():
code :
def filterNot(p: (A) ⇒ Boolean): Seq[A]
Set(2,3,6).apply(0)  //res0: Boolean = false
Seq(2,3,6).apply(0)  //res1: Int = 2
Scala recursive function directly return the input list

Scala recursive function directly return the input list


By : user3404906
Date : March 29 2020, 07:55 AM
this one helps. The Scala definition f: , You have a mistake in your step 3.
Prior to Step 3, your call is
code :
f(xs=Nil)
case Nil => Nil
> 1 :: 2 :: Nil 
> 1 :: List(2)
> List(1,2)
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