scala - 如何使方法返回与输入相同的泛型？

Question

我想拆分由逗号分隔的字符串，并将结果用作 Seq或 Set :

def splitByComma(commaDelimited: String): Array[String]
  = commaDelimited.trim.split(',')

def splitByCommaAsSet(commaDelimited: String): Set[String]
  = splitByComma(commaDelimited).toSet

def splitByCommaAsSeq(commaDelimited: String): Seq[String]
  = splitByComma(commaDelimited).toSeq

val foods = "sushi,taco,burrito"
val foodSet = splitByCommaAsSet(foods)
// foodSet: scala.collection.immutable.Set[String] = Set(sushi, taco, burrito)
val foodSeq = splitByCommaAsSeq(foods)
// foodSeq: Seq[String] = List(sushi, taco, burrito)

然而，这是相当重复的。理想情况下，我想要类似 genericSplitByComma[Set](foods) 的东西。只返回 Set当我通过Set in，并返回 Seq当我通过Seq .

Answer 1

@KrzysztofAtłasik 的回答非常适合 斯卡拉 2.12 .
这是2.13的解决方案. (不完全确定这是否是最好的方法)。

import scala.collection.Factory
import scala.language.higherKinds

def splitByComma[C[_]](commaDelimited: String)(implicit f: Factory[String, C[String]]): C[String] =
  f.fromSpecific(commaDelimited.split(","))
  // Or, as Dmytro stated, which I have to agree looks better.
  commaDelimited.split(",").to(f)

您可以像这样使用:

splitByComma[Array]("hello,world!")
// res: Array[String] = Array(hello, world!)

splitByComma[Set]("hello,world!")
// res: Set[String] = Set(hello, world!)

splitByComma[List]("hello,world!")
// res: List[String] = List(hello, world!)