为了理解而键入不匹配:获取“具有可序列化的产品”

时间:2016-10-20 15:19:00

标签: scala types yield for-comprehension

我正在编写一个函数,它会在字符串中包含字符出现列表(List[(Char, Int)]),并生成该出现列表的所有子集。

所以,给定

List(('a', 2), ('b', 2))

它会产生

List(
  List(),
  List(('a', 1)),
  List(('a', 2)),
  List(('b', 1)),
  List(('a', 1), ('b', 1)),
  List(('a', 2), ('b', 1)),
  List(('b', 2)),
  List(('a', 1), ('b', 2)),
  List(('a', 2), ('b', 2))
)

我已经实现了这样:

type Occurrences = List[(Char, Int)]

def combinations(occurrences: Occurrences): List[Occurrences] =
  if (occurrences.isEmpty) List(List())
  else for {
    (c, n) <- occurrences
    i <- n to 1 by -1
  } yield (c, i) :: combinations(occurrences.tail)

我收到了这个错误:

type mismatch;
 found   : List[List[Product with Serializable]]
 required: List[Occurrences]
    (which expands to)  List[List[(Char, Int)]]

请帮助我理解,为什么会发生这种情况,我该如何解决?

我尝试将其重写为flatMap ...,使用Intellij的“Explain Scala代码”等。

1 个答案:

答案 0 :(得分:4)

实际上缺少括号:

type Occurrences = List[(Char, Int)]

  def combinations(occurrences: Occurrences): List[Occurrences] =
    if (occurrences.isEmpty) List(List())
    else (for {
      (c, n) <- occurrences
      i <- n to 1 by -1
    } yield (c, i)) :: combinations(occurrences.tail)

在您原始的理解代码中尝试产生(c, i) :: combinations(occurrences.tail) ListAny /另一个Tuple内有List元素。

<强>更新

需要魔术的正确方法:

type Occurrences = List[(Char, Int)]


  /**
    * Returns the list of all subsets of the occurrence list.
    *  This includes the occurrence itself, i.e. `List(('k', 1), ('o', 1))`
    *  is a subset of `List(('k', 1), ('o', 1))`.
    *  It also include the empty subset `List()`.
    *
    *  Example: the subsets of the occurrence list `List(('a', 2), ('b', 2))` are:
    *
    *    List(
    *      List(),
    *      List(('a', 1)),
    *      List(('a', 2)),
    *      List(('b', 1)),
    *      List(('a', 1), ('b', 1)),
    *      List(('a', 2), ('b', 1)),
    *      List(('b', 2)),
    *      List(('a', 1), ('b', 2)),
    *      List(('a', 2), ('b', 2))
    *    )
    *
    *  Note that the order of the occurrence list subsets does not matter -- the subsets
    *  in the example above could have been displayed in some other order.
    */
  def combinations(occurrences: Occurrences): List[Occurrences] =
    occurrences.foldRight(List[Occurrences](Nil)) {
      case ((ltr, cnt), acc) =>
        acc ::: (for {
          comb <- acc
          ltrNum <- 1 to cnt
        } yield (ltr, ltrNum) :: comb)
    }

this code的作者的所有赞誉。