scala 如何从列表[选项]中过滤无？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/10104558/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How to filter None's out of List[Option]?
提问by Ralph
If I have a List[Option[A]]in Scala, what is the idiomatic way to filter out the Nonevalues?
如果我List[Option[A]]在 Scala 中有一个，那么过滤掉None值的惯用方法是什么？
One way is to use the following:
一种方法是使用以下方法：
val someList: List[Option[String]] = List(Some("Hello"), None, Some("Goodbye"))
someList.filter(_ != None)
Is there a more "idiomatic" way? This does seem pretty simple.
有没有更“惯用”的方式？这看起来很简单。
回答by Nicolas
If you want to get rid of the options at the same time, you can use flatten:
如果你想同时摆脱这些选项，你可以使用flatten：
scala> someList.flatten
res0: List[String] = List(Hello, Goodbye)
回答by Kevin O'Riordan
someList.filter(_.isDefined)if you want to keep the result type as List[Option[A]]
someList.filter(_.isDefined)如果你想保持结果类型为 List[Option[A]]
回答by dcastro
The catslibrary also has flattenOption, which turns any F[Option[A]]into an F[A](where F[_]is a FunctorFilter)
该cats文库还具有flattenOption，这将任何F[Option[A]]进入的F[A]（其中，F[_]是FunctorFilter）
import cats.implicits._
List(Some(1), Some(2), None).flattenOption == List(1, 2)
