Question

我在Scala和Java之间遇到了编译问题。

我的Java代码需要

java.util.Map<Double, java.lang.Iterable<Foo>>

我的scala代码有一个

Map[Double, Vector[Foo]]

我收到编译错误：

error: type mismatch;
found   : scala.collection.immutable.Map[scala.Double,Vector[Foo]
required: java.util.Map[java.lang.Double,java.lang.Iterable[Foo]]

scala.collection.JavaConversions似乎不适用于嵌套集合，即使Vector可以被隐式转换为Iterable。如果没有迭代scala集合并手动进行转换，我可以做些什么来使这些类型工作？

Answer 1

应该弃用

scala.collection.JavaConversions恕我直言。您最好通过scala.collection.JavaConverters明确说明转换发生的位置和时间。在你的情况下：

import scala.collection.JavaConverters._

type Foo = Int // Just to make it compile
val scalaMap = Map(1.0 -> Vector(1, 2)) // As an example

val javaMap = scalaMap.map { 
  case (d, v) => d -> v.toIterable.asJava
}.asJava

Answer 2

我写了这个通用功能，它可以很好地满足我的需求。

def toJava(x: Any): Any = {
  import scala.collection.JavaConverters._
  x match {
    case y: scala.collection.MapLike[_, _, _] => 
      y.map { case (d, v) => toJava(d) -> toJava(v) } asJava
    case y: scala.collection.SetLike[_,_] => 
      y map { item: Any => toJava(item) } asJava
    case y: Iterable[_] => 
      y.map { item: Any => toJava(item) } asJava
    case y: Iterator[_] => 
      toJava(y.toIterable)
    case _ => 
      x
  }
}

Answer 3

这更适合我的需求：

  def toJava(m: Any): Any = {
    import java.util
    import scala.collection.JavaConverters._
    m match {
      case sm: Map[_, _] => sm.map(kv => (kv._1, toJava(kv._2))).asJava
      case sl: Iterable[_] => new util.ArrayList(sl.map( toJava ).asJava.asInstanceOf[util.Collection[_]])
      case _ => m
    }
  }

Answer 4

如果有人在spark-scala中寻找解决方案，请尝试使用

import org.apache.spark.sql.catalyst.expressions.GenericRowWithSchema

下面， y是嵌套的WrapperArray

y match {
          case x : WrappedArray[x] =>
             (x.map(f => f match {case z: GenericRowWithSchema => z.mkString(",").toString()
                                                case z:Any => z  })).asJavaCollection
          case _ => row.get(i).asInstanceOf[Object]
        }

上面的代码，有两件事， 1）如果包装器数组具有原始数据类型，则条件case_通过 2）如果包装器数组具有复杂数据类型（比如struts），则执行GenericRowWithSchema。

Answer 5

所有其他解决方案都是Any到Any，对于像Scala这样的强类型语言来说，这是非常糟糕的。
这是一个尽可能保留类型的解决方案：

trait AsJava[T, R] {
  def apply(o: T): R
}

object AsJava extends LowPriorityAsJava {
  implicit class RecursiveConverter[T](o: T) {
    def asJavaRecursive[R](implicit asJava: AsJava[T, R]): R = asJava(o)
  }

  implicit lazy val longAsJava: AsJava[Long, lang.Long] = new AsJava[Long, lang.Long] {
    def apply(o: Long): lang.Long = Long.box(o)
  }

  implicit lazy val intAsJava: AsJava[Int, lang.Integer] = new AsJava[Int, lang.Integer] {
    def apply(o: Int): lang.Integer = Int.box(o)
  }

  implicit lazy val doubleAsJava: AsJava[Double, lang.Double] = new AsJava[Double, lang.Double] {
    def apply(o: Double): lang.Double = Double.box(o)
  }

  implicit def mapAsJava[K, V, KR, VR](
      implicit
      keyAsJava: AsJava[K, KR],
      valueAsJava: AsJava[V, VR]
  ): AsJava[Map[K, V], util.Map[KR, VR]] =
    new AsJava[Map[K, V], util.Map[KR, VR]] {
      def apply(map: Map[K, V]): util.Map[KR, VR] =
        map.map { case (k, v) => (keyAsJava(k), valueAsJava(v)) }.asJava
    }

  implicit def seqAsJava[V, VR](implicit valueAsJava: AsJava[V, VR]): AsJava[Seq[V], util.List[VR]] =
    new AsJava[Seq[V], util.List[VR]] {
      def apply(seq: Seq[V]): util.List[VR] = seq.map(valueAsJava(_)).asJava
    }

  implicit def setAsJava[V, VR](implicit valueAsJava: AsJava[V, VR]): AsJava[Set[V], util.Set[VR]] =
    new AsJava[Set[V], util.Set[VR]] {
      def apply(set: Set[V]): util.Set[VR] = set.map(valueAsJava(_)).asJava
    }

  implicit lazy val anyAsJava: AsJava[Any, AnyRef] = new AsJava[Any, AnyRef] {
    def apply(o: Any): AnyRef = o match {
      case x: Map[Any, Any] => mapAsJava(anyAsJava, anyAsJava)(x)
      case x: Seq[Any]      => seqAsJava(anyAsJava)(x)
      case x: Set[Any]      => setAsJava(anyAsJava)(x)
      case x: Long          => longAsJava(x)
      case x: Int           => intAsJava(x)
      case x: Double        => doubleAsJava(x)
      case x                => x.asInstanceOf[AnyRef]
    }
  }
}

trait LowPriorityAsJava {
  implicit def otherAsJava[T]: AsJava[T, T] = new AsJava[T, T] {
    def apply(o: T): T = o
  }
}

用法：

Seq(Seq.empty[Int]).asJavaRecursive

如何将嵌套的scala集合转换为嵌套的Java集合

5 个答案: