Remove one element from Scala List

Question

For example, if I have a list of List(1,2,1,3,2), and I want to remove only one 1, so the I get List(2,1,3,2). If the other 1 was removed it would be fine.

My solution is:

scala> val myList = List(1,2,1,3,2)
myList: List[Int] = List(1, 2, 1, 3, 2)

scala> myList.patch(myList.indexOf(1), List(), 1)
res7: List[Int] = List(2, 1, 3, 2)

But I feel like I am missing a simpler solution, if so what am I missing?

Answer 1

surely not simpler:

def rm(xs: List[Int], value: Int): List[Int] = xs match {
  case `value` :: tail =>  tail
  case x :: tail => x :: rm(tail, value)
  case _ => Nil
}

use:

scala> val xs = List(1, 2, 1, 3)
xs: List[Int] = List(1, 2, 1, 3)

scala> rm(xs, 1)
res21: List[Int] = List(2, 1, 3)

scala> rm(rm(xs, 1), 1)
res22: List[Int] = List(2, 3)

scala> rm(xs, 2)
res23: List[Int] = List(1, 1, 3)

scala> rm(xs, 3)
res24: List[Int] = List(1, 2, 1)

Answer 2

you can zipWithIndex and filter out the index you want to drop.

scala> val myList = List(1,2,1,3,2)
myList: List[Int] = List(1, 2, 1, 3, 2)

scala> myList.zipWithIndex.filter(_._2 != 0).map(_._1)
res1: List[Int] = List(2, 1, 3, 2)

The filter + map is collect,

scala> myList.zipWithIndex.collect { case (elem, index) if index != 0 => elem }
res2: List[Int] = List(2, 1, 3, 2)

To remove first occurrence of elem, you can split at first occurance, drop the element and merge back.

list.span(_ != 1) match { case (before, atAndAfter) => before ::: atAndAfter.drop(1) }

Following is expanded answer,

val list = List(1, 2, 1, 3, 2)

//split AT first occurance
val elementToRemove = 1
val (beforeFirstOccurance, atAndAfterFirstOccurance) = list.span(_ != elementToRemove)

beforeFirstOccurance ::: atAndAfterFirstOccurance.drop(1) // shouldBe List(2, 1, 3, 2)

Resource

How to remove an item from a list in Scala having only its index?

How should I remove the first occurrence of an object from a list in Scala?