scala - 将压缩流投影到最长的流

Question

Stream 上的内置 zip 函数似乎会在输入流对中最短的位置截断。我怎样才能实现这个功能：

def firstOrLongest[T]( a : Stream[ T ], b : Stream[ T ) : Stream[ T ]
// resulting stream should have the property that:
// while there are still elements of a, return (the corresponding element of) a 
// else return (the corresponding element of) b.

Answer 1

You could use zipAll method to extend the shorter collection to the length of the longer. This method involves creation of many intermediate objects.

def firstOrLongest[T]( a : Stream[T], b : Stream[T]) : Stream[T] = {
  val oa = a.map{ e => Some(e): Option[T] }
  val ob = b.map{ e => Some(e): Option[T] }
  oa.zipAll(ob, None, None).collect{
    case (Some(e), _) => e
    case (None, Some(e)) => e
  }
}

Answer 2

Stream 类有一个追加操作符，它完全符合您的要求。

您的描述指出：

// while there are still elements of a, return (the corresponding element of) a 
// else return (the corresponding element of) b.

在 Scala 中，这只是：

a ++ b