Scala迭代器与地图和为

Question

鉴于:

val list = List("one","two","three")     
val it = list.toIterator

我可以跑:

list map ("_" +) -> List(_one, _two, _three)
for (i <- list) yield("_" + i) -> List(_one, _two, _three)

如果我在迭代器上运行相同的操作,我得到:

it map ("_" + ) -> Iterator[java.lang.String] = empty iterator
for (i <- it) yield("_" + i) -> Iterator[java.lang.String] = empty iterator

在我运行map/for之后,我不应该再找回另一个(非空的)Iterator [String]吗？

Answer 1

scala> def ints(n: Int): Stream[Int] = n #:: ints(n + 1)
ints: (n: Int)Stream[Int]

scala> val list = List("one","two","three")
list: List[java.lang.String] = List(one, two, three)

scala> val it = list.toIterator
it: Iterator[java.lang.String] = non-empty iterator

scala> it map ("_" + )
res24: Iterator[java.lang.String] = non-empty iterator

scala> it map ("_" + )
res25: Iterator[java.lang.String] = non-empty iterator

scala> for (i <- it) yield("_" + i)
res26: Iterator[java.lang.String] = non-empty iterator

也许你用过你的迭代器？

scala> res26.foreach{println}
_one
_two
_three

scala> res26
res28: Iterator[java.lang.String] = empty iterator

由于迭代器是有状态的并且不可重置,因此一旦使用它,它就是空的并且不能再次使用.

相反,您可以使用视图:

scala> val v = list.view
v: java.lang.Object with scala.collection.SeqView[java.lang.String,List[java.lang.String]] = SeqView(one, two, three)

scala> v map ("_" + )
res29: scala.collection.SeqView[java.lang.String,Seq[_]] = SeqViewM(...)

scala> for (i <- v) yield("_" + i)
res30: scala.collection.SeqView[java.lang.String,Seq[_]] = SeqViewM(...)

scala> res29.foreach{println}
_one
_two
_three

scala> res29
res32: scala.collection.SeqView[java.lang.String,Seq[_]] = SeqViewM(...)

scala> res29.foreach{println}
_one
_two
_three