在Scala中是否有一种安全的方法来转置不等长列表的列表？

Question

鉴于以下列表:

val l = List(List(1, 2, 3), List(4, 5), List(6, 7, 8))

如果我尝试转置它,Scala将抛出以下错误:

scala> List.transpose(l)
java.util.NoSuchElementException: head of empty list
    at scala.Nil$.head(List.scala:1365)
    at scala.Nil$.head(List.scala:1362)
    at scala.List$$anonfun$transpose$1.apply(List.scala:417)
    at scala.List$$anonfun$transpose$1.apply(List.scala:417)
    at scala.List.map(List.scala:812)
    at scala.List$.transpose(List.scala:417)
    at .<init>(<console>:6)
    at .<clinit>(<console>)
    at RequestResult...

这是因为List.transpose假设等长列表,所以使用head方法:

def transpose[A](xss: List[List[A]]): List[List[A]] = {
  val buf = new ListBuffer[List[A]]
  var yss = xss
  while (!yss.head.isEmpty) {
    buf += (yss map (_.head))
    yss = (yss map (_.tail))
  }
  buf.toList
}

我想得到以下内容:

List(List(1, 4, 6), List(2, 5, 7), List(3, 8))

写我自己的版本transpose最好的方法来做到这一点？这就是我想出的:

def myTranspose[A](xss: List[List[A]]): List[List[A]] = {
  val buf = new ListBuffer[List[A]]
  var yss = xss
  while (!yss.head.isEmpty) {
    buf += (yss filter (!_.isEmpty) map (_.head))
    yss = (yss filter (!_.isEmpty) map (_.tail))
  }
  buf.toList
}

更新:我有兴趣比较这里提供的不同解决方案的速度,所以我把以下小基准放在一起:

import scala.testing.Benchmark
import scala.collection.mutable.ListBuffer

trait Transpose extends Benchmark {
  def transpose[Int](xss: List[List[Int]]): List[List[Int]] = Nil
  val list: List[List[Int]] = List(List(1,2,3), Nil, List(4,5,99,100), List(6,7,8))
  def run = {
    val l = transpose(list)
    println(l)
    l
  }
}

object PRTranspose extends Transpose {
  override def transpose[Int](xss: List[List[Int]]): List[List[Int]] = {
    val buf = new ListBuffer[List[Int]]
    var yss = xss
    while (!yss.head.isEmpty) {
      buf += (yss filter (!_.isEmpty) map (_.head))
      yss = (yss filter (!_.isEmpty) map (_.tail))
    }
    buf.toList
  }
}

object ACTranspose extends Transpose {
  override def transpose[Int](xss: List[List[Int]]): List[List[Int]] = {
    val b = new ListBuffer[List[Int]]
    var y = xss filter (!_.isEmpty)
    while (!y.isEmpty) {
      b += y map (_.head)
      y = y map (_.tail) filter (!_.isEmpty)
    }
    b.toList
  }
}

object ETranspose extends Transpose {
  override def transpose[Int](xss: List[List[Int]]): List[List[Int]] = xss.filter(!_.isEmpty) match {    
    case Nil => Nil
    case ys: List[List[Int]] => ys.map{ _.head }::transpose(ys.map{ _.tail })
  }
}

我的命令是:

scala PFTranspose 5 out.log
scala ACTranspose 5 out.log
scala ETranspose 5 out.log

我的结果是:

PRTranspose$            10              0               1               1               0
ACTranspose$            9               2               0               0               0
ETranspose$             9               3               2               3               1

Answer 1

这个怎么样:

    scala> def transpose[A](xs: List[List[A]]): List[List[A]] = xs.filter(_.nonEmpty) match {    
         |     case Nil    =>  Nil
         |     case ys: List[List[A]] => ys.map{ _.head }::transpose(ys.map{ _.tail })
         | }
    warning: there were unchecked warnings; re-run with -unchecked for details
    transpose: [A](xs: List[List[A]])List[List[A]]

    scala> val ls = List(List(1, 2, 3), List(4, 5), List(6, 7, 8))
    ls: List[List[Int]] = List(List(1, 2, 3), List(4, 5), List(6, 7, 8))

    scala> transpose(ls)
    res0: List[List[Int]] = List(List(1, 4, 6), List(2, 5, 7), List(3, 8))

    scala> val xs = List(List(1,2,3), List(4,5,99,100), List(6,7,8))
xs: List[List[Int]] = List(List(1, 2, 3), List(4, 5, 99, 100), List(6, 7, 8))

scala> transpose(xs)
res1: List[List[Int]] = List(List(1, 4, 6), List(2, 5, 7), List(3, 99, 8), List(100))