斯卡拉，擴展迭代器

Question

即時通訊擴展迭代器來創建一個新的方法takeWhileInclusive，它將像takeWhile一樣運行，但包括最後一個元素。

我的問題是什麼是最好的做法來擴展迭代器返回一個新的迭代器，我想懶惰評估。從C＃背景我正常使用IEnumerable並使用yield關鍵字，但這樣的選項似乎不存在於斯卡拉。

例如我可能有

List(0,1,2,3,4,5,6,7).iterator.map(complex time consuming algorithm).takeWhileInclusive(_ < 6) 

在這種情況下， takeWhileInclusive只會有解決的值謂詞

，直到我得到比6的結果時，它會包括這第一個結果

到目前爲止我有：

object ImplicitIterator { 
    implicit def extendIterator(i : Iterator[Any]) = new IteratorExtension(i) 
} 

class IteratorExtension[T <: Any](i : Iterator[T]) { 
    def takeWhileInclusive(predicate:(T) => Boolean) = ? 
} 

Answer 1

這就是我的上級可變解決一個案例：

class InclusiveIterator[A](ia: Iterator[A]) { 
    def takeWhileInclusive(p: A => Boolean) = { 
    var done = false 
    val p2 = (a: A) => !done && { if (!p(a)) done=true; true } 
    ia.takeWhile(p2) 
    } 
} 
implicit def iterator_can_include[A](ia: Iterator[A]) = new InclusiveIterator(ia) 

Answer 2

scala> List(0,1,2,3,4,5,6,7).toStream.filter (_ < 6).take(2) 
res8: scala.collection.immutable.Stream[Int] = Stream(0, ?) 

scala> res8.toList 
res9: List[Int] = List(0, 1) 

您更新後：

scala> def timeConsumeDummy (n: Int): Int = { 
    | println ("Time flies like an arrow ...") 
    | n } 
timeConsumeDummy: (n: Int)Int 

scala> List(0,1,2,3,4,5,6,7).toStream.filter (x => timeConsumeDummy (x) < 6) 
Time flies like an arrow ... 
res14: scala.collection.immutable.Stream[Int] = Stream(0, ?) 

scala> res14.take (4).toList 
Time flies like an arrow ... 
Time flies like an arrow ... 
Time flies like an arrow ... 
res15: List[Int] = List(0, 1, 2, 3) 

timeConsumeDummy被調用4次。我錯過了什麼嗎？

Answer 3

您可以使用Iterator的span方法要做到這一點非常乾淨：

class IteratorExtension[A](i : Iterator[A]) { 
    def takeWhileInclusive(p: A => Boolean) = { 
    val (a, b) = i.span(p) 
    a ++ (if (b.hasNext) Some(b.next) else None) 
    } 
} 

object ImplicitIterator { 
    implicit def extendIterator[A](i : Iterator[A]) = new IteratorExtension(i) 
} 

import ImplicitIterator._ 

現在(0 until 10).toIterator.takeWhileInclusive(_ < 4).toList給List(0, 1, 2, 3, 4)，例如。

Answer 4

class IteratorExtension[T](i : Iterator[T]) { 
    def takeWhileInclusive(predicate:(T) => Boolean) = new Iterator[T] { 
    val it = i 
    var isLastRead = false 

    def hasNext = it.hasNext && !isLastRead 
    def next = { 
     val res = it.next 
     isLastRead = !predicate(res) 
     res 
    } 
    } 
} 

而你的隱含有一個錯誤。這是固定的：

object ImplicitIterator { 
    implicit def extendIterator[T](i : Iterator[T]) = new IteratorExtension(i) 
} 

Answer 5

以下要求scalaz對一個元組得到fold(A, B)

scala> implicit def Iterator_Is_TWI[A](itr: Iterator[A]) = new { 
    | def takeWhileIncl(p: A => Boolean) 
    | = itr span p fold (_ ++ _.toStream.headOption) 
    | } 
Iterator_Is_TWI: [A](itr: Iterator[A])java.lang.Object{def takeWhileIncl(p: A => Boolean): Iterator[A]} 

這裏它在工作：

scala> List(1, 2, 3, 4, 5).iterator takeWhileIncl (_ < 4) 
res0: Iterator[Int] = non-empty iterator 

scala> res0.toList 
res1: List[Int] = List(1, 2, 3, 4) 

你可以滾你自己折了一對這樣的：

scala> implicit def Pair_Is_Foldable[A, B](pair: (A, B)) = new { 
    | def fold[C](f: (A, B) => C): C = f.tupled(pair) 
    | } 
Pair_Is_Foldable: [A, B](pair: (A, B))java.lang.Object{def fold[C](f: (A, B) => C): C}