`removeAt` implies passing an index though ?

# announcements

[bj0]

removeAt implies passing an index though ?

[karelpeeters]

Yes, I'm looking for something like list.withoutElementAt(index).

[stephan_marshay]

Based on @adam-mcneilly’s answer:
val index = 3
val answers = listOf(“Yes”, “No”, “Maybe”)
val yesNoList = answers - answers[index]
println(yesNoList) // [Yes, No]

[karelpeeters]

Problem with that is that answers[index] isn't guaranteed to be the first occurrence of that value.

[karelpeeters]

@adam-mcneilly

[cedric]

Also, removeAt should be (almost) constant time while - is linear

[Pavlo Liapota]

This operation cannot be done in constant time as new list should be created and all elements except one should be copied to new list.
So it will always take linear time.

[Pavlo Liapota]

I would advise to use MutableList instead.

[Pavlo Liapota]

or you can create your own extension function, here are 3 different ways to implement it:

[Pavlo Liapota]

fun <T> List<T>.withoutElementAt(index: Int): List<T>
    = this.toMutableList().apply { removeAt(index) }

[Pavlo Liapota]

fun <T> List<T>.withoutElementAt1(index: Int)
        = this.asSequence()
            .withIndex()
            .filter { it.index != index}
            .map { it.value }
            .toList()

[Pavlo Liapota]

fun <T> List<T>.withoutElementAt2(index: Int): List<T> {
    val result = mutableListOf<T>()
    for (i in this.indices) {
        if (i != index) result.add(this[i])
    }
    return result
}

[Pavlo Liapota]

@karelpeeters

[Pavlo Liapota]

ops, didn’t see discussion below…

[karelpeeters]

Haha, thanks for helping anyway!

[cedric]

@Pavlo Liapota This could be made constant time with a new List representation that I described in the follow up conversation