Scala 中惯用的 Haskell-like 迭代？

Question

在Haskell中，我可以通过调用获得无限的顺序函数应用程序列表：

iterate :: (A -> A) -> A -> [A]

假设在 scala 中我有 f(x: A): A。是否有一个函数会产生一系列顺序函数应用程序？喜欢 iter(f: A => A, x: A): Stream[A]?

Answer 1

有Iterator.iterate and Stream.iterate ():

Stream.iterate(1)(_ + 1)

或者你可以自己写一个，使用 Stream:

def iter[A](f: A => A, x: A): Stream[A] = {
  val result = f(x)
  result #:: iter(f, result)
}

（请注意，按住 Stream 会导致记忆问题 warned about in the documentation）。使用它很简单：

scala> iter[Int](_ + 1, 1).iterator.drop(100 * 100 * 100).take(10).toList
res1: List[Int] = List(1000002, 1000003, 1000004, 1000005, 1000006, 1000007, 1000008, 1000009, 1000010, 1000011)

Answer 2

是的，已经在图书馆了：Iterator.iterate

Iterator.iterate(1)(_ + 1).drop(100 * 100 * 100).take(10).toList
//> List(1000001, 1000002, 1000003, 1000004, 1000005,
         1000006, 1000007, 1000008, 1000009, 1000010)

Answer 3

与@Sean Vieira 的基本相同，具有更多类似于 Haskell 的语法：

scala> def iterate[A]: (A => A) => A => Stream[A] = {
     |     f => x => x #:: iterate(f)(f(x))
     | }
iterate: [A]=> (A => A) => (A => Stream[A])

scala> iterate[Int](_+1)(1) take 10 toList
res1: List[Int] = List(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

交换f和x的位置可以帮助编译器：

scala> def iterate[A]: A => (A => A) => Stream[A] = {
     |     x => f => x #:: iterate(f(x))(f)
     | }
iterate: [A]=> A => ((A => A) => Stream[A])

scala> iterate(1)(2+) take 10 toList //Don't need to use iterate[Int] here
res3: List[Int] = List(1, 3, 5, 7, 9, 11, 13, 15, 17, 19)

Scala 中惯用的 Haskell-like 迭代？

Idiomatic Haskell-like iterate in Scala?

haskell

scala

function-composition