list - 考虑到字符串列表代表 Scala 中映射的键，如何将 List[String] 转换为 List[Map[String,String]] ？

Question

我有一个引用文献列表:

val references: List[String]= List("S","R")

我还有变量，它是:

val variables: Map[String,List[String]]=("S"->("a","b"),"R"->("west","east"))

references 是变量映射的键列表。

我想构造一个函数，它需要:

def expandReplacements(references:List[String],variables:Map[String,List[String]]):List[Map(String,String)]

这个函数基本上应该创建返回以下组合

List(Map("S"->"a"),("R"->"west"),Map("S"->"a"),("R"->"east"),Map("S"->"b"),("R"->"west"),Map("S"->"b"),("R"->"east"))

我尝试这样做:

val variables: Map[String,List[String]] = Map("S" -> List("a", "b"), "R" -> List("east", "central"))
val references: List[String] = List("S","R")

def expandReplacements(references: List[String]): List[Map[String, String]] =
  references match {
    case ref :: refs =>
      val variableValues =
        variables(ref)
      val x = variableValues.flatMap { variableValue =>
        val remaining = expandReplacements(refs)
        remaining.map(rem => rem + (ref -> variableValue))
      }
      x

    case Nil => List.empty
  }

Answer 1

如果您有超过 2 个推荐人，您可以执行以下操作:

def expandReplacements(references: List[String], variables :Map[String,List[String]]): List[Map[String, String]] = {
  references match {
    case Nil => List(Map.empty[String, String])
    case x :: xs =>
      variables.get(x).fold {
        expandReplacements(xs, variables)
      } { variableList =>
        for {
          variable <- variableList.map(x -> _)
          otherReplacements <- expandReplacements(xs, variables)
        } yield otherReplacements + variable
      }
  }
}

代码运行于Scastie .