How do I convert a scala.collection.Iterator containing thousands of objects to a scala.collection.immutable.Vector?
I do not believe that I can use _* because of the number of items.
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4 Answers
You can
Vector() ++ myIterator which gives the correct thing with the correct type. For very small vectors and iterators, in high-performance loops, you may instead wish to
val b = Vector.newBuilder[WhateverType] while (myIterator.hasNext) { b += myIterator.next } b.result which does the minimum work necessary (as far as I know) to create a vector. toIndexedSeq does essentially this, but returns a more generic type (so you're not actually guaranteed a Vector, even if it does return a Vector now.)
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You can use toIndexedSeq. It doesn't statically return a Vector, but it actually is one.
answered Mar 12, 2012 at 19:35
Jean-Philippe Pellet
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Dan Burton
Chances are OP does indeed just want some indexed sequence, and not necessarily a
Vector specifically.
Ralph
You are right. I need (near) constant-time random access and length calculations.
You can use _*, since all it does is pass a Seq with all the arguments. It will be inefficient, however, since it will first convert the iterator into a sequence, and then use that sequence to create another sequence.
answered Mar 13, 2012 at 16:39
Daniel C. Sobral
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Iterator supports .toVector now since at minimum Scala 11 (and maybe earlier).
class Thing val myIterator = Iterator(new Thing, new Thing, new Thing) myIterator.toVector produces:
res1: scala.collection.immutable.Vector[Thing] = Vector(Thing@22da2fe6, Thing@100ad67e, Thing@713a35c5) 