maybe zip and then flatmap kotlinlang #announcements

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Title

clubfan

03/30/2018, 12:41 PM

maybe zip and then flatmap?

Daniel

03/30/2018, 12:43 PM

Thank you and let's see about this with a short unit test

clubfan

03/30/2018, 12:46 PM

listOf("a", "b", "c").zip(listOf("d", "e")).flatMap(it.toList())

maybe? Haven't tested it though

Daniel

03/30/2018, 12:48 PM

Very close! This gets me to

<[a, d, b, e]>

On another note: the second list is always -1 the size of the first one, so this would be expect

list1.zip(list2).flatMap { it.toList() } + list1.takeLast(1)

would do it

clubfan

03/30/2018, 1:05 PM

fun <T> List<T>.merge(other : List<T>) : List<T> {
    val result = ArrayList<T>()
    for ( i in 0 .. maxOf(this.size, other.size)) {
        this.getOrNull(i)?.let { result.add(it) }
        other.getOrNull(i)?.let { result.add(it) }
    }
    return result
}

works, but it's not really idiomatic I think

and you can call it like this:

listOf("a", "b", "c").merge(listOf("d", "e"))

or this:

fun <T> List<T>.merge(other : List<T>) : List<T> {
    return this.mapIndexed{i, el -> other.getOrNull(i)?.let { listOf(el, it) } ?: listOf(el)}.flatten() + other.drop(this.size)

}

I'm pretty sure there's a better solution

Scala's zipAll would be great for this

ilya.gorbunov

03/30/2018, 2:57 PM

Please add your use case to this issue: https://youtrack.jetbrains.com/issue/KT-13017

Note that if the elements in lists are nullable, you won't be able to distinguish what the last

null

zipAll

means: either null last element or missing last element.

Daniel

03/30/2018, 9:44 PM

Thank you both for your replies! Currently I am settling for the .zip sulution with the awesome

require(other.size == size - 1)

function beforehand!

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