Scala：如何简化嵌套模式匹配语句

Question

我正在用 Scala 编写一个 Hive UDF（因为我想学习 scala）。为此，我必须覆盖三个函数：evaluate、initialize 和 getDisplayString.

在初始化函数中我必须：

接收 ObjectInspector 和 return 一个 ObjectInspector
检查数组是否为空
检查数组大小是否正确
检查数组是否包含正确类型的对象

为此，我使用了模式匹配并提出了以下函数：

  override def initialize(genericInspectors: Array[ObjectInspector]): ObjectInspector = genericInspectors match {
    case null => throw new UDFArgumentException(functionNameString + ": ObjectInspector is null!")
    case _ if genericInspectors.length != 1 => throw new UDFArgumentException(functionNameString + ": requires exactly one argument.")
    case _ => {
      listInspector = genericInspectors(0) match {
        case concreteInspector: ListObjectInspector => concreteInspector
        case _ => throw new UDFArgumentException(functionNameString + ": requires an input array.")
     }
      PrimitiveObjectInspectorFactory.getPrimitiveWritableObjectInspector(listInspector.getListElementObjectInspector.asInstanceOf[PrimitiveObjectInspector].getPrimitiveCategory)
    }
  }

尽管如此，我的印象是函数可以变得更清晰，并且通常更漂亮，因为我不喜欢缩进级别太多的代码。

是否有惯用的 Scala 方法来改进上面的代码？

Answer 1

模式通常包含其他模式。这里x的类型是String.

scala> val xs: Array[Any] = Array("x")
xs: Array[Any] = Array(x)

scala> xs match {
     | case null => ???
     | case Array(x: String) => x
     | case _ => ???
     | }
res0: String = x

"any number of args" 的成语是 "sequence pattern"，它匹配任意参数：

scala> val xs: Array[Any] = Array("x")
xs: Array[Any] = Array(x)

scala> xs match { case Array(x: String) => x case Array(_*) => ??? }
res2: String = x

scala> val xs: Array[Any] = Array(42)
xs: Array[Any] = Array(42)

scala> xs match { case Array(x: String) => x case Array(_*) => ??? }
scala.NotImplementedError: an implementation is missing
  at scala.Predef$.$qmark$qmark$qmark(Predef.scala:230)
  ... 32 elided

scala> Array("x","y") match { case Array(x: String) => x case Array(_*) => ??? }
scala.NotImplementedError: an implementation is missing
  at scala.Predef$.$qmark$qmark$qmark(Predef.scala:230)
  ... 32 elided

不应将此答案解释为提倡匹配以返回类型安全。

Scala：如何简化嵌套模式匹配语句

Scala: How to simplify nested pattern matching statements

hive

scala

udf