缩小/折叠在monoid列表上，但reducer返回

问题描述：

我发现自己有几次遇到减速器/像这样组合fn的情况:

Ive found myself in the situation a couple of times where i have a reducer / combine fn like so:

  def combiner(a: String, b: String): Either[String, String] = {
    (a + b).asRight[String]
  }

它是一个虚拟实现，但fn可能失败，因此它返回一个.然后，我有了要通过reduce/fold传递的值的列表.我能想到的最好的方法(假设列表的类型是一个monoid)是:

Its a dummy implementation but the fn can fail so it returns an either. I then I have a list of values I want to pass through this with reduce / fold. The best I can come up with (assuming the List's type is a monoid) is this:

  def combine(items: Vector[String]) = {

    items.foldLeft(Monoid[String].empty.asRight[String]) { case (acc, value) =>
      acc.flatMap( accStr => combiner(accStr, value))
    }
  }

它有点笨拙，并且作为一种相当通用的模式，我怀疑有更好的方法来使用猫.

Its a bit clumsy and as its a fairly generic pattern I suspect there's a better way to do it using cats.

答

您可能想看看

You might want to take a look at foldM. Your code would then look approximately like this:

Foldable[Vector].foldM(items, "")(combiner)

foldM方法具有签名

def foldM[G[_], A, B](fa: F[A], z: B)(f: (B, A) ⇒ G[B])(implicit G: Monad[G]): G[B]

因此，在您的情况下，type(-constructor)参数将统一如下:

so in your case, the type(-constructor) parameters would unify as follows:

G[X] = Either[String, ?]
A = String
B = String
F[X] = Vector[X]

G[X] = Either[String, ?]
A = String
B = String
F[X] = Vector[X]

，以使f: (A, B) => G[B]变为f: (String, String) => Either[String, String]，恰好是combiner转换为函数时的类型.

so that f: (A, B) => G[B] would become f: (String, String) => Either[String, String], which is exactly the type of combiner when it's converted into a function.

缩小/折叠在monoid列表上，但reducer返回

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