在 Scala 中将 Map[String,List[String]] 展开为笛卡尔积答案

【问题标题】：Expand Map[String,List[String]] into Cartesian Product in Scala在 Scala 中将 Map[String,List[String]] 展开为笛卡尔积
【发布时间】：2021-06-25 12:39:30
【问题描述】：

我正在寻找基于地图类型的Expand a Set[Set[String]] into Cartesian Product in Scala 版本：

我想从：

val values = Map(
      "id" -> List("Paul","Joe"),
      "number" -> List("1","2","3"),
      "Type" -> List("A","B","C")
    )

并计算：

val final = Set(
Map("id" -> "Paul", "number" -> "1", "Type" -> "A" ),
Map("id" -> "Paul", "number" -> "1", "Type" -> "B" ),
Map("id" -> "Paul", "number" -> "1", "Type" -> "C" ),
Map("id" -> "Paul", "number" -> "1", "Type" -> "A" ),
Map("id" -> "Paul", "number" -> "1", "Type" -> "B" ),
Map("id" -> "Paul", "number" -> "1", "Type" -> "C" ),
Map("id" -> "Paul", "number" -> "2", "Type" -> "A" ),
Map("id" -> "Paul", "number" -> "2", "Type" -> "B" ),
Map("id" -> "Paul", "number" -> "2", "Type" -> "C" ),
....
Map("id" -> "Joe", "number" -> "3", "Type" -> "B" ),
Map("id" -> "Joe", "number" -> "3", "Type" -> "C" )
)

我尝试转换以下代码

def combine[A](xs: Traversable[Traversable[A]]): Seq[Seq[A]] =
     xs.foldLeft(Seq(Seq.empty[A])){
     (x, y) => for (a <- x.view; b <- y) yield a :+ b }

但我有一些问题要了解如何构建所有地图以添加到大 Set() 中

【问题讨论】：

标签： list scala combinations cartesian-product cartesian

【解决方案1】：

您可以这样做：
（但请注意，对于较大的地图，这会消耗大量内存，可能值得考虑生成一个 LazyList）

def cartesianProductMap[A, B](data: Map[A, List[B]]): List[Map[A, B]] =
  data.foldLeft(Map.empty[A, B] :: Nil) {
    case (acc, (key, values)) =>
      values.flatMap { b =>
        acc.map(_ + (key -> b))
      }
  }

代码在Scastie中运行。

顺便说一句，如果你使用 cats，你可以这样做：values.to(SortedMap).sequence

【讨论】：