如何使用高阶函数展平选项列表？

Question

使用Scala 2.7.7:

如果我有一个选项列表,我可以使用for-understandness来展平它们:

val listOfOptions = List(None, Some("hi"), None)
listOfOptions: List[Option[java.lang.String]] = List(None, Some(hi), None)

scala> for (opt <- listOfOptions; string <- opt) yield string
res0: List[java.lang.String] = List(hi)

我不喜欢这种风格,宁愿使用HOF.这种尝试过于冗长,无法接受:

scala> listOfOptions.flatMap(opt => if (opt.isDefined) Some(opt.get) else None)
res1: List[java.lang.String] = List(hi)

直觉上我会期望以下工作,但它没有:

scala> List.flatten(listOfOptions)
<console>:6: error: type mismatch;
 found   : List[Option[java.lang.String]]
 required: List[List[?]]
       List.flatten(listOfOptions)

即使以下似乎它应该工作,但不是:

scala> listOfOptions.flatMap(_: Option[String])
<console>:6: error: type mismatch;
 found   : Option[String]
 required: (Option[java.lang.String]) => Iterable[?]
       listOfOptions.flatMap(_: Option[String])
                          ^

我能想到的最好的是:

scala> listOfOptions.flatMap(_.toList)         
res2: List[java.lang.String] = List(hi)

...但我宁愿不必将选项转换为列表.这看起来很笨拙.

有什么建议？

Answer 1

在Scala 2.8中,flatten将起作用:

Welcome to Scala version 2.8.0.RC2 (Java HotSpot(TM) 64-Bit Server VM, Java 1.6.0_20).
Type in expressions to have them evaluated.
Type :help for more information.

scala> val listOfOptions = List(None, Some("hi"), None)
listOfOptions: List[Option[java.lang.String]] = List(None, Some(hi), None)

scala> listOfOptions flatten
res0: List[java.lang.String] = List(hi)

但这在2.7.7中不起作用:

Welcome to Scala version 2.7.7.final (Java HotSpot(TM) 64-Bit Server VM, Java 1.6.0_20).

scala> val listOfOptions = List(None, Some("hi"), None)
listOfOptions: List[Option[java.lang.String]] = List(None, Some(hi), None)

scala> listOfOptions.flatten
:6: error: no implicit argument matching parameter type (Option[java.lang.String]) => Iterable[Nothing] was found.
       listOfOptions.flatten

集合库已经过重新设计,并且在2.8中有了很大的改进,所以也许你可能想尝试使用最新的Scala 2.8 RC,看看是否能让你更容易使用它.

如果你真的不想使用toList方法,我想你也可以像这样写:

scala> listOfOptions.flatMap(o => o)
res: List[java.lang.String] = List(hi)

也许不是美丽的东西,但至少这在2.7.7中起作用.

Answer 2

为了补充Arjan的答案,在Scala 2.7.7中你可以使用List#flatten,但你需要帮助输入类型inferencer:

Welcome to Scala version 2.7.7.final (Java HotSpot(TM) 64-Bit Server VM, Java 1.6.0_20).
Type in expressions to have them evaluated.
Type :help for more information.

scala> val listOfOptions = List(None, Some("hi"), None)
listOfOptions: List[Option[java.lang.String]] = List(None, Some(hi), None)

scala> listOfOptions.flatten[String]                   
res0: List[String] = List(hi)

scala> val x: List[String] = listOfOptions.flatten
x: List[String] = List(hi)