Scala将字符串拆分为元组

Question

我想在有4个元素的空白上拆分一个字符串:

1 1 4.57 0.83

我试图转换成List [(String,String,Point)],这样前两个分裂是列表中的前两个元素,最后两个是Point.我正在做以下但它似乎不起作用:

Source.fromFile(filename).getLines.map(string => { 
            val split = string.split(" ")
            (split(0), split(1), split(2))
        }).map{t => List(t._1, t._2, t._3)}.toIterator

Answer 1

这个怎么样:

scala> case class Point(x: Double, y: Double)
defined class Point

scala> s43.split("\\s+") match { case Array(i, j, x, y) => (i.toInt, j.toInt, Point(x.toDouble, y.toDouble)) }
res00: (Int, Int, Point) = (1,1,Point(4.57,0.83))

Answer 2

您可以使用模式匹配从数组中提取所需内容:

    case class Point(pts: Seq[Double])
    val lines = List("1 1 4.34 2.34")

    val coords = lines.collect(_.split("\\s+") match {
      case Array(s1, s2, points @ _*) => (s1, s2, Point(points.map(_.toDouble)))
    })