Scala-如何以功能方式在列表周围循环

Question

我需要遍历一个列表，并将列表中的上一个元素与当前元素进行比较。 从传统的编程背景开始，我使用带变量的循环来保存先前的值，但是对于Scala，我认为应该有一种更好的方法。

for (item <- dataList)
{
   // Code to compare previous list element
   prevElement = item
}

上面的代码可以正常工作并给出正确的结果，但是我想探索编写相同代码的功能方法。

Answer 1

有许多不同的解决方案。 也许最通用的是使用辅助函数：

// This example finds the minimum value (the hard way) in a list of integers. It's
// not looking at the previous element, as such, but you can use the same approach.
// I wanted to make the example something meaningful.
def findLowest(xs: List[Int]): Option[Int] = {

  @tailrec // Guarantee tail-recursive implementation for safety & performance.
  def helper(lowest: Int, rem: List[Int]): Int = {

    // If there are no elements left, return lowest found.
    if(rem.isEmpty) lowest

    // Otherwise, determine new lowest value for next iteration. (If you need the
    // previous value, you could put that here.)
    else {
      val newLow = Math.min(lowest, rem.head)
      helper(newLow, rem.tail)
    }
  }

  // Start off looking at the first member, guarding against an empty list.
  xs match {
    case x :: rem => Some(helper(x, rem))
    case _ => None
  }
}

在这种特殊情况下，可以用fold代替。

def findLowest(xs: List[Int]): Option[Int] = xs match {
  case h :: _ => Some(xs.fold(h)((b, m) => Math.min(b, m)))
  case _ => None
}

可以进一步简化为：

def findLowest(xs: List[Int]): Option[Int] = xs match {
  case h :: _ => Some(xs.foldLeft(h)(Math.min))
  case _ => None
}

在这种情况下，我们使用fold操作的zero参数存储最小值。

如果这看起来太抽象了，您可以在循环中发布更多有关您想做的事情的详细信息，我可以在回答中解决这个问题吗？

更新 ：好的，因此，看到您对日期的评论后，这里是一个更具体的版本。 在此示例中，我使用了DateRange （您需要将其替换为实际类型）。 您还必须确定需要进行哪些overlap和merge 。 但是，这是您的解决方案：

// Determine whether two date ranges overlap. Return true if so; false otherwise.
def overlap(a: DateRange, b: DateRange): Boolean = { ... }

// Merge overlapping a & b date ranges, return new range.
def merge(a: DateRange, b: DateRange): DateRange = { ... }

// Convert list of date ranges into another list, in which all consecutive,
// overlapping ranges are merged into a single range.
def mergeDateRanges(dr: List[DateRange]): List[DateRange] = {

  // Helper function. Builds a list of merged date ranges.
  def helper(prev: DateRange, ret: List[DateRange], rem: List[DateRange]):
  List[DateRange] = {

    // If there are no more date ranges to process, prepend the previous value
    // to the list that we'll return.
    if(rem.isEmpty) prev :: ret

    // Otherwise, determine if the previous value overlaps with the current
    // head value. If it does, create a new previous value that is the merger
    // of the two and leave the returned list alone; if not, prepend the
    // previous value to the returned list and make the previous value the
    // head value.
    else {
      val (newPrev, newRet) = if(overlap(prev, rem.head)) {
        (merge(prev, rem.head), ret)
      }
      else (rem.head, prev :: ret)

      // Next iteration of the helper (I missed this off, originally, sorry!)
      helper(newPrev, newRet, rem.tail)
    }
  }

  // Start things off, trapping empty list case. Because the list is generated by pre-pending elements, we have to reverse it to preserve the order.
  dr match {
    case h :: rem => helper(h, Nil, rem).reverse // <- Fixed bug here too...
    case _ => Nil
  }
}

Answer 2

如果您只想比较两个连续的元素并获得布尔结果：

val list = Seq("one", "two", "three", "one", "one")
val result = list.sliding(2).collect { case Seq(a,b) => a.equals(b)}.toList

前一个会给你这个

列表（false，false，false，true）

替代上一个方法是，您可能想要获取该值：

val result = list.sliding(2).collect { 
  case Seq(a,b) => if (a.equals(b)) Some(a) else None
  case _ => None
}.toList.filter(_.isDefined).map(_.get)

这将给出以下内容：

列表（之一）

但是，如果您只是想比较列表中的项目，则可以选择执行以下操作：

val list = Seq("one", "two", "three", "one")

for {
  item1 <- list
  item2 <- list
  _ <- Option(println(item1 + ","+ item2))
} yield ()

结果是：

one,one
one,two
one,three
one,one
two,one
two,two
two,three
two,one
three,one
three,two
three,three
three,one
one,one
one,two
one,three
one,one

Answer 3

如果您不更改列表，而只是尝试并排比较两个元素。 这可以为您提供帮助。 它是功能性的。

def cmpPrevElement[T](l: List[T]): List[(T,T)] = {
  (l.indices).sliding(2).toList.map(e => (l(e.head), l(e.tail.head)))
}