如何展平一系列猫的 ValidatedNel 值 [英] How to flatten a sequence of cats' ValidatedNel values

问题描述

我需要将一系列 cats.data.ValidatedNel[E, T] 值展平为单个 ValidatedNel 值:

I need to flatten a sequence of cats.data.ValidatedNel[E, T] values to a single ValidatedNel value:

val results: Seq[cats.data.ValidatedNel[E, T]] = ???

val flattenedResult: cats.data.ValidatedNel[E, T]

我可以这样做:

import cats.std.list._, cats.syntax.cartesian._

results.reduce(_ |@| _ map { case _ => validatedValue })

但是想知道是否存在预定义的库方法.

but wonder if a pre-defined library methods exists.

推荐答案

这取决于你想如何组合它们(你的问题中的 validatedValue 是什么?)

It depends on how you want to combine them (what is validatedValue in your question ?)

import cats.data.{Validated, ValidatedNel}
import cats.implicits._

val validations1 = List(1.validNel[String], 2.valid, 3.valid)
val validations2 = List(1.validNel[String], "kaboom".invalidNel, "boom".invalidNel)

如果你想结合 T s，你可以使用 Foldable.combineAll 它使用一个 Monoid[T] :

If you want to combine the Ts, you can use Foldable.combineAll which uses a Monoid[T] :

val valSum1 = validations1.combineAll 
// Valid(6)
val valSum2 = validations2.combineAll 
// Invalid(OneAnd(kaboom,List(boom)))

如果你想得到一个ValidationNel[String, List[T]]，你可以使用Traverse.sequence:

If you want to get a ValidationNel[String, List[T]], you can use Traverse.sequence :

val valList1: ValidatedNel[String, List[Int]] = validations1.sequence
// Valid(List(1, 2, 3))
val valList2: ValidatedNel[String, List[Int]] = validations2.sequence
// Invalid(OneAnd(kaboom,List(boom)))

如果你不关心结果，似乎是这样，你可以使用Foldable.sequence_.

If you don't care about the result, which seems to be the case, you can use Foldable.sequence_.

val result1: ValidatedNel[String, Unit] = validations1.sequence_
//  Valid(())
val result2: ValidatedNel[String, Unit] = validations2.sequence_
// Invalid(OneAnd(kaboom,List(boom)))

validations1.sequence_.as(validatedValue) // as(x) is equal to map(_ => x)

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