Scala 斯卡拉猫
我知道我可以遍历Scala 斯卡拉猫,scala,traversal,scala-cats,Scala,Traversal,Scala Cats,我知道我可以遍历Lists import cats.instances.list._ import cats.syntax.traverse._ def doMagic(item: A): M[B] = ??? val list: List[A] = ??? val result: M[List[B]] = list.traverse(doMagic) 我可以将Seq来回转换为List val seq: Seq[A] = ??? val result: M[Seq[B]] = seq.toL
List
s
import cats.instances.list._
import cats.syntax.traverse._
def doMagic(item: A): M[B] = ???
val list: List[A] = ???
val result: M[List[B]] = list.traverse(doMagic)
我可以将Seq
来回转换为List
val seq: Seq[A] = ???
val result: M[Seq[B]] = seq.toList.traverse(doMagic).map(_.toSeq)
但是我也可以在没有样板的情况下遍历Seq
val seq: Seq[A] = ???
val result: M[Seq[B]] = seq.traverse(doMagic)
或者,获取Traverse[Seq]实例的简单方法是什么?Cats不为
Seq
提供类型类实例,因此除了自己实现它之外,您还需要进行转换
至于原因,有一个正在进行的讨论在一个(有点老)猫。总而言之,您对Seq
底层特性了解得不够,无法确保某些TypeClass实例的法律适用
编辑:无需担心,它现在存在,请参见链接线程如果您绝对确信从所有
Seq
到List
的转换在您的代码中总是成功的,您可以通过(伪)同构将遍历结构从List
转移到Seq
:
def traverseFromIso[F[_], Z[_]]
(forward: F ~> Z, inverse: Z ~> F)
(implicit zt: Traverse[Z])
: Traverse[F] = new Traverse[F] {
def foldLeft[A, B](fa: F[A], b: B)(f: (B, A) ⇒ B): B = zt.foldLeft(forward(fa), b)(f)
def foldRight[A, B](fa: F[A], lb: Eval[B])(f: (A, Eval[B]) => Eval[B]): Eval[B] =
zt.foldRight(forward(fa), lb)(f)
def traverse[G[_], A, B]
(fa: F[A])
(f: (A) ⇒ G[B])
(implicit appG: Applicative[G])
: G[F[B]] = {
(zt.traverse(forward(fa))(f)(appG)).map(zb => inverse(zb))
}
}
这实际上不是同构,因为从Seq
到List
的转换可能会严重失败(例如,如果序列是无限的)。它所做的只是将Seq
来回转换为List
,并将所有方法调用转发给Traverse[List]
的方法调用
现在,您可以使用此方法构建遍历[Seq]
的实例:
implicit val seqTraverse: Traverse[Seq] = traverseFromIso(
new FunctionK[Seq, List] { def apply[X](sx: Seq[X]): List[X] = sx.toList },
new FunctionK[List, Seq] { def apply[X](lx: List[X]): Seq[X] = lx }
)
完整代码段(使用scala 2.12.4和cats 1.0.1编译):
我想问一下,你到底为什么要一直将Seq
包装到List
再包装回来,而不是将所有东西都转换成List
一次,然后再使用它?在我正在做的项目中,Seq
到处都在使用,不知道为什么。也许我们可以用List
来代替。
import cats._
import cats.implicits._
import cats.arrow.FunctionK
import scala.language.higherKinds
object TraverseFromIso {
// This method can build you a `Traversable[Seq]` from
// an `Traversable[List]` and a pair of polymorphic conversion
// functions:
def traverseFromIso[F[_], Z[_]]
(forward: F ~> Z, inverse: Z ~> F)
(implicit zt: Traverse[Z])
: Traverse[F] = new Traverse[F] {
def foldLeft[A, B](fa: F[A], b: B)(f: (B, A) ⇒ B): B = zt.foldLeft(forward(fa), b)(f)
def foldRight[A, B](fa: F[A], lb: Eval[B])(f: (A, Eval[B]) => Eval[B]): Eval[B] =
zt.foldRight(forward(fa), lb)(f)
def traverse[G[_], A, B]
(fa: F[A])
(f: (A) ⇒ G[B])
(implicit appG: Applicative[G])
: G[F[B]] = {
(zt.traverse(forward(fa))(f)(appG)).map(zb => inverse(zb))
}
}
// A little demo
def main(args: Array[String]): Unit = {
// To instantiate a `Traverse[Seq]`, we have to provide
// two natural transformations (from List to Seq and back):
implicit val seqTraverse: Traverse[Seq] = traverseFromIso(
new FunctionK[Seq, List] { def apply[X](sx: Seq[X]): List[X] = sx.toList },
new FunctionK[List, Seq] { def apply[X](lx: List[X]): Seq[X] = lx }
)
// do stuff with `Traversable[Seq]` here
}
}