如何在scala中削减for-comprehension short(突破它)？

Question

我有一段代码,代码如下:

val e2 = for (e <- elements if condition(expensiveFunction(e))) yield {
            expensiveFunction(e)
         }

条件对于少数元素是正确的,然后对所有剩余的元素变为假.

不幸的是,这不起作用(即使我忽略性能)因为我elements是一个无限的迭代器.

有没有办法在for-comprehension中使用"break",这样当某个条件成立时它会停止产生元素？否则,计算我的scala-idiomatic方法是e2什么？

Answer 1

你可以采用懒惰的做法:

val e2 = elements.toIterator                    
          .map(expensiveFunction)
          .takeWhile(result => result == true) // or just .takeWhile(identity)
// you may want to strict iterator, (e.g. by calling .toList) at the end

因此,您可以按需计算昂贵的功能,如果在某个步骤中存在错误,则不会执行不必​​要的工作.

Answer 2

scala> def compute(i: Int) = { println(s"f$i"); 10*i }
compute: (i: Int)Int

scala> for (x <- Stream range (0, 20)) yield compute(x)
f0
res0: scala.collection.immutable.Stream[Int] = Stream(0, ?)

scala> res0 takeWhile (_ < 100)
res1: scala.collection.immutable.Stream[Int] = Stream(0, ?)

scala> res1.toList
f1
f2
f3
f4
f5
f6
f7
f8
f9
f10
res2: List[Int] = List(0, 10, 20, 30, 40, 50, 60, 70, 80, 90)

编辑,另一个演示:

scala> def compute(i: Int) = { println(s"f$i"); 10*i }
compute: (i: Int)Int

scala> for (x <- Stream range (0, 20)) yield compute(x)
f0
res0: scala.collection.immutable.Stream[Int] = Stream(0, ?)

scala> res0 takeWhile (_ < 100)
res1: scala.collection.immutable.Stream[Int] = Stream(0, ?)

scala> res1.toList
f1
f2
f3
f4
f5
f6
f7
f8
f9
f10
res2: List[Int] = List(0, 10, 20, 30, 40, 50, 60, 70, 80, 90)

scala> Stream.range(0,20).map(compute).toList
f0
f1
f2
f3
f4
f5
f6
f7
f8
f9
f10
f11
f12
f13
f14
f15
f16
f17
f18
f19
res3: List[Int] = List(0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180, 190)

scala> Stream.range(0,20).map(compute).takeWhile(_ < 100).toList
f0
f1
f2
f3
f4
f5
f6
f7
f8
f9
f10
res4: List[Int] = List(0, 10, 20, 30, 40, 50, 60, 70, 80, 90)