How to filter None's out of List[Option]?

☆樱花仙子☆ 提交于 2019-12-02 16:02:27

If you want to get rid of the options at the same time, you can use flatten:

scala> someList.flatten
res0: List[String] = List(Hello, Goodbye)
Kevin O'Riordan

someList.filter(_.isDefined) if you want to keep the result type as List[Option[A]]

The cats library also has flattenOption, which turns any F[Option[A]] into an F[A] (where F[_] is a FunctorFilter)

import cats.implicits._

List(Some(1), Some(2), None).flattenOption == List(1, 2)
标签
易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!