Convert a Scala list to a tuple? - list

Question

How can I convert a list with (say) 3 elements into a tuple of size 3?
For example, let's say I have val x = List(1, 2, 3) and I want to convert this into (1, 2, 3). How can I do this?

Answer 1

You can do it using scala extractors and pattern matching (link):
val x = List(1, 2, 3)
val t = x match {
case List(a, b, c) => (a, b, c)
}
Which returns a tuple
t: (Int, Int, Int) = (1,2,3)
Also, you can use a wildcard operator if not sure about a size of the List
val t = x match {
case List(a, b, c, _*) => (a, b, c)
}

Answer 2

You can't do this in a typesafe way. Why? Because in general we can't know the length of a list until runtime. But the "length" of a tuple must be encoded in its type, and hence known at compile time. For example, (1,'a',true) has the type (Int, Char, Boolean), which is sugar for Tuple3[Int, Char, Boolean]. The reason tuples have this restriction is that they need to be able to handle a non-homogeneous types.

Answer 3

an example using shapeless :
import shapeless._
import syntax.std.traversable._
val x = List(1, 2, 3)
val xHList = x.toHList[Int::Int::Int::HNil]
val t = xHList.get.tupled
Note: the compiler need some type informations to convert the List in the HList that the reason why you need to pass type informations to the toHList method

Answer 4

Shapeless 2.0 changed some syntax. Here's the updated solution using shapeless.
import shapeless._
import HList._
import syntax.std.traversable._
val x = List(1, 2, 3)
val y = x.toHList[Int::Int::Int::HNil]
val z = y.get.tupled
The main issue being that the type for .toHList has to be specified ahead of time. More generally, since tuples are limited in their arity, the design of your software might be better served by a different solution.
Still, if you are creating a list statically, consider a solution like this one, also using shapeless. Here, we create an HList directly and the type is available at compile time. Remember that an HList has features from both List and Tuple types. i.e. it can have elements with different types like a Tuple and can be mapped over among other operations like standard collections. HLists take a little while to get used to though so tread slowly if you are new.
scala> import shapeless._
import shapeless._
scala> import HList._
import HList._
scala> val hlist = "z" :: 6 :: "b" :: true :: HNil
hlist: shapeless.::[String,shapeless.::[Int,shapeless.::[String,shapeless.::[Boolean,shapeless.HNil]]]] = z :: 6 :: b :: true :: HNil
scala> val tup = hlist.tupled
tup: (String, Int, String, Boolean) = (z,6,b,true)
scala> tup
res0: (String, Int, String, Boolean) = (z,6,b,true)

Answer 5

Despite the simplicity and being not for lists of any length, it is type-safe and the answer in most cases:
val list = List('a','b')
val tuple = list(0) -> list(1)
val list = List('a','b','c')
val tuple = (list(0), list(1), list(2))
Another possibility, when you don't want to name the list nor to repeat it (I hope someone can show a way to avoid the Seq/head parts):
val tuple = Seq(List('a','b')).map(tup => tup(0) -> tup(1)).head
val tuple = Seq(List('a','b','c')).map(tup => (tup(0), tup(1), tup(2))).head

Answer 6

FWIW, I wanted a tuple to initalise a number of fields and wanted to use the syntactic sugar of tuple assignment.
EG:
val (c1, c2, c3) = listToTuple(myList)
It turns out that there is syntactic sugar for assigning the contents of a list too...
val c1 :: c2 :: c3 :: Nil = myList
So no need for tuples if you've got the same problem.

Answer 7

If you are very sure that your list.size<23 use it:
def listToTuple[A <: Object](list:List[A]):Product = {
val class = Class.forName("scala.Tuple" + list.size)
class.getConstructors.apply(0).newInstance(list:_*).asInstanceOf[Product]
}
listToTuple: [A <: java.lang.Object](list: List[A])Product
scala> listToTuple(List("Scala", "Smart"))
res15: Product = (Scala,Smart)

Answer 8

You can't do this in a type-safe way. In Scala, lists are arbitrary-length sequences of elements of some type. As far as the type system knows, x could be a list of arbitrary length.
In contrast, the arity of a tuple must be known at compile time. It would violate the safety guarantees of the type system to allow assigning x to a tuple type.
In fact, for technical reasons, Scala tuples were limited to 22 elements, but the limit no longer exists in 2.11 The case class limit has been lifted in 2.11 https://github.com/scala/scala/pull/2305
It would be possible to manually code a function that converts lists of up to 22 elements, and throws an exception for larger lists. Scala's template support, an upcoming feature, would make this more concise. But this would be an ugly hack.

Answer 9

This can also be done in shapeless with less boilerplate using Sized:
scala> import shapeless._
scala> import shapeless.syntax.sized._
scala> val x = List(1, 2, 3)
x: List[Int] = List(1, 2, 3)
scala> x.sized(3).map(_.tupled)
res1: Option[(Int, Int, Int)] = Some((1,2,3))
It's type-safe: you get None, if the tuple size is incorrect, but the tuple size must be a literal or final val (to be convertible to shapeless.Nat).

Answer 10

Using Pattern Matching:
val intTuple = List(1,2,3) match {case List(a, b, c) => (a, b, c)}

Answer 11

2015 post.
For the Tom Crockett's answer to be more clarifying, here is a real example.
At first, I got confused about it. Because I come from Python, where you can just do tuple(list(1,2,3)).
Is it short of Scala language ? (the answer is -- it's not about Scala or Python, it's about static-type and dynamic-type.)
That's causes me trying to find the crux why Scala can't do this .
The following code example implements a toTuple method, which has type-safe toTupleN and type-unsafe toTuple.
The toTuple method get the type-length information at run-time, i.e no type-length information at compile-time, so the return type is Product which is very like the Python's tuple indeed (no type at each position, and no length of types).
That way is proned to runtime error like type-mismatch or IndexOutOfBoundException. (so Python's convenient list-to-tuple is not free lunch. )
Contrarily , it is the length information user provided that makes toTupleN compile-time safe.
implicit class EnrichedWithToTuple[A](elements: Seq[A]) {
def toTuple: Product = elements.length match {
case 2 => toTuple2
case 3 => toTuple3
}
def toTuple2 = elements match {case Seq(a, b) => (a, b) }
def toTuple3 = elements match {case Seq(a, b, c) => (a, b, c) }
}
val product = List(1, 2, 3).toTuple
product.productElement(5) //runtime IndexOutOfBoundException, Bad !
val tuple = List(1, 2, 3).toTuple3
tuple._5 //compiler error, Good!

Answer 12

you can do this either
via pattern-matching (what you do not want) or
by iterating through the list and applying each element one by one.
val xs: Seq[Any] = List(1:Int, 2.0:Double, "3":String)
val t: (Int,Double,String) = xs.foldLeft((Tuple3[Int,Double,String] _).curried:Any)({
case (f,x) => f.asInstanceOf[Any=>Any](x)
}).asInstanceOf[(Int,Double,String)]

Answer 13

In scala 3, you can do something like this:
def totuple[A](as: List[A]): Tuple = as match
case Nil => EmptyTuple
case h :: t => h *: totuple(t)
but as has been said already, without giving the compiler any more hard-coded type information, you aren't going to know the length of the tuple or the types of its elements, so this is likely hardly any better than the original list.

Answer 14

as far as you have the type:
val x: List[Int] = List(1, 2, 3)
def doSomething(a:Int *)
doSomething(x:_*)