Question

如何使此代码更具功能性，其中当前迭代的值保留到下一个（变量last），

def f(i: Int): Int = ???

var last = -1

for (v <- xs) yield {
  val start = f(v)
  val end = f(start)

  val res = if (last == start - 2) "abc" else "xyz"

  last = end
  res
}

Answer 1

我认为foldLeft应该有效。您可以传递下一次迭代中所需的值，并将结果传递给元组

   val list = (1 to 10 by 2)                       
   //> list  : scala.collection.immutable.Range = Range(1, 3, 5, 7, 9)
   val last = -1
   list.foldLeft((last, List[String]())) {
     case (r, c) => println(s"Current - $c last - ${r._1}")
     (c, r._2 :+ if (...) "abc" else "xyz")
     // c will be the value that you need to pass to the next iteration
     // r._2 will contain the list which you would have got with your for comprehension 
   }
  //> Current - 1 last - -1
  //| Current - 3 last - 1
  //| Current - 5 last - 3
  //| Current - 7 last - 5
  //| Current - 9 last - 7

Answer 2

使用scanLeft注意到这种方法（受@ {hhit方法使用foldLeft启发），此处为start和end的预处理对集合，即实例

val a = Array((1,2), (3,4), (5,6))

接下来，假设s代表start，e代表end，所以

a.scanLeft((-1,-1,-1)) { case(acc, (s,e)) => (s,e,acc._2) }.drop(1)

生成包含start，end和last的三元组，

Array((1,2,-1), (3,4,2), (5,6,4))

我们可以应用if-else表达式;共，

a.scanLeft((-1,-1,-1)) { case(acc, (s,e)) => (s,e,acc._2) }.
  drop(1).
  map(t => if (t._3 == t._1 - 2) "abc" else "xyz")

用于迭代的Scala函数代码，其值转发到下一次迭代

2 个答案: