Reputation: 41
Let's say, I have a list. If I want to return something at the index of that list, I can just pass the lists' val, the index. Like so.
val list = List(1, 2, 3)
list(0) //Int = 1
But what if I want items at multiple indexes on this list? I want to be able to do this... list(0, 1)
and get a collection of the items at those indices.
This is crazily simple in Ruby. Anyone have any suggestions?
Upvotes: 4
Views: 5482
Reputation: 16324
You can flip the logic around, so that for each index, you're getting the index, you retrieve the element at that index. Since you are using the apply
method on List
, there are several shorthand expressions of this logic:
val indices = List(0, 1)
indices.map(index => list.apply(index))
indices.map(index => list(index))
indices.map(list(_))
indices.map(list)
indices map list
It's worth noting that since these are all maps
on indices
, the resulting collection will generally have the same type as indices
, and not list
:
val list = Array(1, 2, 3)
val indices = List(0, 1)
indices map list //List(1, 2), instead of Array(1, 2)
This can be an undesirable property here. One solution to this is to use breakOut
(in scala.collection
):
val newList: Array[Int] = indices.map(list)(breakOut)
You can read more about breakOut here. The following solutions also maintain the collection type of list
when possible by performing operations on list
instead of indeces
:
If you are looking for a contiguous range of the list, you might consider using slice
:
list.slice(1, 2) //List(2)
You could also use drop
and take
(and the dropRight
, takeRight
versions) to a similar effect:
list.drop(1).take(1)
For more complex versions of this type of filtering, you might be interested in the zipWithIndex
method, which would allow you to express arbitrary logic on the index:
list.zipWithIndex.collect {
case(el, index) if indices.contains(index) /* any other logic */ => el
}
Upvotes: 13
Reputation: 8996
The answer from @Ben is spot on.
However, if you want to be able to use the syntax you describe list(0, 1)
, then Scala allows you to do that through implicit conversions:
implicit class MultiIndexList[A](list: List[A]){
def apply(indices: Int *) = indices.map(list)
}
Assuming you are working on the REPL:
val list = List(1,2,3)
list(1,2)
res1: Seq[Int] = ArrayBuffer(2, 3)
list(0,2)
res2: Seq[Int] = ArrayBuffer(1, 3)
Upvotes: 5
Reputation: 369428
If you are looking for an equivalent to Ruby's Array#slice
(aka Array#[]
) method, there are two:
Array#[](index)
is, as you mentioned, List.apply(n: Int): A
Array#[](range)
is List.slice(from: Int, until: Int): List[A]
Array#[](start, length)
There is also no equivalent to Ruby's Array#values_at
.
Upvotes: 1