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

Question

I have a list as follows:

val internalIdList: List[Int] = List()

internalIdList = List(11, 12, 13, 14, 15)

From this list would remove the third element in order to obtain:

internalIdList = List(11, 12, 14, 15)

I can not use a ListBuffer, are obliged to maintain the existing structure. How can I do?

Thanks to all

Answer 1

If you know that you will be dropping the third element (index 2), then you can simply use

val trunced = internalIdList.take(2) ++ internalIdList.drop(3)

otherwise, if you don't know beforehand what the index will be of the element to remove, you could write a function like the following:

def removeIndex(ix: Int) = if (internalIdList.size < ix) internalIdList
                           else internalIdList.take(ix) ++ internalIdList.drop(ix+1)

Answer 2

There is a .patch method on Seq, so in order to remove the third element you could simply do this:

List(11, 12, 13, 14, 15).patch(2, Nil, 1)

Which says: Starting at index 2, please remove 1 element, and replace it with Nil.

Knowing this method in depth enables you to do so much more than that. You can swap out any sublist of a list with arbitrary other.

Answer 3

An idiomatic way to do it is to zip the value with their index, filter, and then project the value again:

scala> List(11,12,13,14,15).zipWithIndex.filter(_._2 != 2).map(_._1)
res0: List[Int] = List(11, 12, 14, 15)

But you can also use splitAt:

scala> val (x,y) = List(11,12,13,14,15).splitAt(2)
x: List[Int] = List(11, 12)
y: List[Int] = List(13, 14, 15)

scala> x ++ y.tail
res5: List[Int] = List(11, 12, 14, 15)

Answer 4

If you insist on using the oldschool method, use collect:

List(1,2,3,4).zipWithIndex.collect { case (a, i) if i != 2 => a }

However, I still prefer the method in my other answer.

Answer 5

(internalIdList.indices.collect { case i if i != 3 => internalList(i) }).toList

To generalise this...

def removeIndex[A](s: Seq[A], n: Int): Seq[A] = s.indices.collect { case i if i != n => s(i) }

Although this will often return a Vector, so you would need to do

val otherList = removeIndex(internalIdList, 3).toList

If you really wanted a list back.

Shadowlands has a solution which tends to be faster for linear sequences. This one will be faster with indexed sequences.

Answer 6

A generic function that implements Nicolas' first solution:

def dropIndex[T](list: List[T], idx: Int): List[T] =
  list.zipWithIndex.filter(_._2 != idx).map(_._1)

Usage:

scala> val letters = List('a', 'b', 'c')
scala> for (i <- 0 until letters.length) println(dropIndex(letters, i))
List(b, c)
List(a, c)
List(a, b)

Answer 7

Using a for comprehension on a list xs like this,

for (i <- 0 until xs.size if i != nth-1) yield xs(i)

Also consider a set of exclusion indices, for instance val excl = Set(2,4) for excluding the second and fourth items; hence we collect those items whose indices do not belong to the exclusion set, namely

for (i <- 0 until xs.size if !excl(i)) yield xs(i)