将一种类型的同构 HList 映射到不同类型的异构 HList [英] Map a homogenous HList of one type into a hetergenous HList of different types

问题描述

我有一个 HList 字符串:

I have an HList of strings:

val strings = "The Lorax" :: "Dr. Suess" :: HNil

我是另一个特殊类型的 HList:

I another HList of special types:

case class Title(title: String, words: List[String])
case class Author(firstName: String, lastName: String)
val book = Title("The Hobbit", List("Hobbit")) :: Author("J.R.R.", "Tolkien") :: HNil

我想把strings"，我的字符串HList，变成一个混合类型的HList，对应于book"列表.如果我有一种方法可以从字符串 -> 标题，以及一种方法可以从字符串 -> 作者，我觉得这应该非常简单，基本上可以将字符串"作为书"列表的实例-type 使用 shapeless，但我似乎想不出办法.

I want to turn "strings", my HList of strings, into an HList of mixed types, corresponding to the "book" list. If I have a method to go from a string -> Title, and a method to go from a string -> Author, I feel like this should be very straight forward to essentially get "strings" as an instance of "book"-list-type using shapeless, but I can't seem to figure out a way.

编辑

我的用例涉及处理作为案例类开始的 HList.我使用 shapeless 是因为我希望能够以相同的方式转换和修改不同案例类的数据，而不必对案例类的形状进行硬编码，我只想知道它们的值的类型.所以理想情况下，这种方法也适用于从看起来像这样的字符串列表:

My use case for this involves working on HLists that started as case classes. I'm using shapeless because I want to be able to transform and modify the data of different case classes in the same way, without having to hard-code knowledge about the shape of the case classes, I only want to have to know about the types of their values. So ideally this method would work also for going from a list of strings that looks like this:

val strings2 = "Leonardo" :: "April O'Neil" :: "The Art of Pizza" :: HNil
val book2 = Author("Michaelangelo") :: Author("Donatello") :: Title("Slicing and Dicing"), List("Slicing", "Dicing") :: HList 

所以我总是有一个它需要采用的格式的例子，但我不想将作者"的数量和书籍"的数量硬编码到翻译功能列表中.我希望能够说a，a，b"应该看起来像A，A，B"，这是从a -> A"开始的方法，这是从b ->"开始的方法B"，但我希望能够使用相同的代码从b，a，b"到B，A，B"，因为我有两个列表.

So I will always have an example of the format it needs to be in, but I don't want to have to hardcode the amount of "authors" and the amount of "books" into a list of translation functions. I want to be able to say "a,a,b" should look like "A, A, B", and here is a method go from "a -> A" and here is a method to go from "b -> B", but I want to be able to use the same code to go from "b, a, b" to "B, A, B", given that I have both lists.

推荐答案

您可以使用 zipApply 很好地做到这一点，它将函数 hlist 的每个元素应用到另一个 hlist 中的相应元素:

You can do this pretty nicely with zipApply, which applies each element of an hlist of functions to the corresponding element in another hlist:

case class Title(title: String, words: List[String])
case class Author(firstName: String, lastName: String)

// For the sake of example:
def parseTitle(s: String): Title = Title(s, s.split(' ').toList)
def parseAuthor(s: String): Author =
  Author(s.takeWhile(_ != ' '), s.dropWhile(_ != ' ').tail)

import shapeless._

val funcs = parseTitle _ :: parseAuthor _ :: HNil
val strings = "The Lorax" :: "Dr. Suess" :: HNil

val book = funcs.zipApply(strings)

然后:

scala> println(book)
Title(The Lorax,List(The, Lorax)) :: Author(Dr.,Suess) :: HNil

如果您需要更通用，您可以使用 ZipApply 类型类，而不是简单地在具有具体类型的 hlist 上调用 zipApply.

If you need this to be more generic, you can use the ZipApply type class instead of simply calling zipApply on hlists with concrete types.

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