Question

After reading this question, I would expect the following to work:

Seq( Seq(1,2,3) , Seq(4,5,6) ).transpose()

but alas:

error: not enough arguments for method transpose: (implicit asTraversable: 
Seq[Int] => scala.collection.GenTraversableOnce[B])Seq[Seq[B]].
Unspecified value parameter asTraversable.
          Seq( Seq(1,2,3) , Seq(4,5,6) ).transpose()

Also, I can’t seem to find any reference to transpose on the scala docs, although Seq refers it

Providing the identity, it seems to work somehow:

scala> Seq( Seq(1,2,3) , Seq(4,5,6) ).transpose( a => a)
res10: Seq[Seq[Int]] = List(List(1, 4), List(2, 5), List(3, 6))

But still returns List instead of Seq

Answer 1

Just use it without parentheses:

Seq( Seq(1,2,3) , Seq(4,5,6) ).transpose
//res0: Seq[Seq[Int]] = List(List(1, 4), List(2, 5), List(3, 6))

But still returns List instead of Seq

Well, actually List is inheritor of Seq, so after all you got a Seq (look at the left part of result).

The reason of such behaviour is that transpose actually defined as a function with argument, but since it’s argument defined as implicit you have an option to delegate work of substituting argument to scala compiler (it will perform compile-time lookup for you).

If you writing parentheses, either function has to have overloaded form with no arguments, e.g.

def transpose() = ...

or you have to write something inside them (it’s actually matter of syntax).