Scala 等价于 Haskell 的转置

Question

If to exactly line-by-line translate this form of Haskell's transpose function to Scala, what it would be?

Haskell's transpose:
tr               :: [[a]] -> [[a]]
tr []             = []
tr ([]   : xss)   = tr xss
tr ((x:xs) : xss) = (x : map head xss) : tr (xs : map tail xss)

My attempt for Scala does not work:

def scalaTranspose(m: List[List[Int]]): List[List[Int]] = m match {
case Nil => Nil
case Nil::xss => scalaTranspose(xss)
case (x::xs)::xss => (x:: map head xss) :: scalaTranspose(xs :: map tail xss)
}

After struggling with translating map function, this as a last line also doesn't work:

case (x::xs)::xss => (x:: xss.map(head xss)) :: scalaTranspose(xs :: xss.map(tail xss))

Answer 1

Scala is a bit different from haskell, and the pattern matching here is also a bit lacking compared to haskell. Also, since Scala has a mix of ObjectOriented components, head and tail are methods defined on List objects and not pure functions . You will have to replace map head xss by scala equivalent xss.map(_.head) .

def scalaTranspose(m: List[List[Int]]): List[List[Int]] = m match {
  case Nil => Nil
  case Nil :: xss => scalaTranspose(xss)
  case (x :: xs) :: xss => (x :: xss.map(_.head)) :: scalaTranspose(xs :: xss.map(_.tail))
}