Convert Seq to ArrayBuffer

Question

Is there any concise way to convert a Seq into ArrayBuffer in Scala?

duplicate: http://stackoverflow.com/questions/3074118/scala-how-do-i-convert-a-mapint-any-to-a-sortedmap-or-a-treemap — kiritsuku, Sep 26 '11 at 09:50

Eastsun · Answer 1 · 2013-05-28T07:52:32.343

28

scala> val seq = 1::2::3::Nil
seq: List[Int] = List(1, 2, 3)

scala> seq.toBuffer
res2: scala.collection.mutable.Buffer[Int] = ArrayBuffer(1, 2, 3)

EDIT After Scala 2.1x, there is a method .to[Coll] defined in TraversableLike, which can be used as follow:

scala> import collection.mutable
import collection.mutable

scala> seq.to[mutable.ArrayBuffer]
res1: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 2, 3)

scala> seq.to[mutable.Set]
res2: scala.collection.mutable.Set[Int] = Set(1, 2, 3)

edited May 28 '13 at 07:52

answered Sep 26 '11 at 10:42

Eastsun

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+1 `seq.toBuffer` has got to be the clearest and most memorable way – karmakaze May 28 '13 at 06:07

Philippe · Answer 2 · 2011-09-26T09:59:00.450

This will work:

ArrayBuffer(mySeq : _*)

Some explanations: this uses the apply method in the ArrayBuffer companion object. The signature of that method is

def apply [A] (elems: A*): ArrayBuffer[A]

meaning that it takes a variable number of arguments. For instance:

ArrayBuffer(1, 2, 3, 4, 5, 6, 7, 8)

is also a valid call. The ascription : _* indicates to the compiler that a Seq should be used in place of that variable number of arguments (see Section 4.6.2 in the Scala Reference).

Convert Seq to ArrayBuffer

2 Answers2