将字符串拆分为交替的单词(Scala)

Question

我想将一个字符串拆分为交替的单词.总会有一个偶数.

例如

val text = "this here is a test sentence"

应该转换为包含的某些有序集合类型

"this", "is", "test"

和

"here", "a", "sentence"

我想出来了

val (l1, l2) = text.split(" ").zipWithIndex.partition(_._2 % 2 == 0) match {
  case (a,b) => (a.map(_._1), b.map(_._1))}

这给了我两个数组的正确结果.

这可以更优雅地完成吗？

Answer 1

scala> val s = "this here is a test sentence"
s: java.lang.String = this here is a test sentence

scala> val List(l1, l2) = s.split(" ").grouped(2).toList.transpose
l1: List[java.lang.String] = List(this, is, test)
l2: List[java.lang.String] = List(here, a, sentence)