Scala—如何以函数方式循环列表

有许多不同的解决方案。也许最通用的方法是使用helper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 .

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

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 zero 论据 折叠

如果这看起来太抽象了,你能发布更多关于你想在你的循环中做什么的细节吗,我可以在我的回答中解决这个问题吗?

更新 DateRange (需要用实际类型替换)。你还必须确定 overlap 和 merge

// 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
  }
}

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

列表(假,假,假,真)

上一种方法的另一种方法是,您可能希望获得以下值:

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)

这将提供以下信息:

列表(一)

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

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

这将导致:

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

如果您没有改变列表,并且只尝试并排比较两个元素。

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