如果 A 和 B 是单子，如何将 A[B[C]] 转换为 B[A[C]]? [英] How to convert A[B[C]] to B[A[C]] if A and B are monads?
问题描述
我正在寻找一个更通用的解决方案，它利用 monads(可能还有 monoids)来实现与if( xs.contains(None) ) None else Some(xs.flatten) 为 Seq[Option[A]] 类型的 xs 做.
I'm looking for a more general solution which exploits monads (and monoids possibly) to achieve the same as
if( xs.contains(None) ) None else Some(xs.flatten) does for xs of type Seq[Option[A]].
如何使用 Scalaz 做到这一点?我觉得我遗漏了一些明显的东西.
How can I do that with Scalaz? I feel like I'm missing something evident.
推荐答案
有两个 monad 是不够的(对于 M)和已经足够了(对于 N)——当然这还不够——但是如果 M 有一个 Traverse 实例并且 N 有一个 Applicative> 例如，您可以使用 sequence.例如:
Having two monads is both not enough (for M) and more than enough (for N)—which adds up to not enough, of course—but if M has a Traverse instance and N has an Applicative instance, you can use sequence. For example:
import scalaz._, Scalaz._
def foo[A](xs: List[Option[A]]): Option[List[A]] = xs.sequence
这具有您想要的语义.请注意，我使用的是 List 而不是 Seq，因为 Scalaz 7 不再为 Seq 提供必要的 Traverse 实例>(尽管您可以轻松编写自己的).
This has the semantics you want. Note that I'm using List instead of Seq, since Scalaz 7 no longer provides the necessary Traverse instance for Seq (although you could easily write your own).
如您所见，以下内容无法编译:
As you've noticed, the following won't compile:
List(Some(1), Some(45)).sequence
虽然如果你把 None 扔进去也没关系:
Although it's fine if you throw a None in there:
scala> List(Some(1), None, Some(45)).sequence
res0: Option[List[Int]] = None
这是因为 List(Some(1), Some(45)) 的推断类型将是 List[Some[Int]]，而我们不Some 有一个 Applicative 实例.
This is because the inferred type of List(Some(1), Some(45)) will be List[Some[Int]], and we don't have an Applicative instance for Some.
Scalaz 提供了一个方便的 some 方法，它的工作方式与 Some.apply 类似，但为您提供了一些已经输入为 Option 的内容，因此您可以写如下:
Scalaz provides a handy some method that works like Some.apply but gives you something that's already typed as an Option, so you can write the following:
scala> List(some(1), some(45)).sequence
res1: Option[List[Int]] = Some(List(1, 45))
无需额外输入.
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