Streamとのzip

Scala で (1,"hoge1"),(2,"hoge2"),(3,"hoge3"),(1,"hoge4"),(2,"hoge5")...
のように 1,2,3,1,2,3… の繰り返しと 何かを zip してくっつける

val l = List(1,2,3)
val sl = List("hoge1","hoge2","hoge3","hoge4","hoge5")

Stream.continually(l).flatten.zip(sl) toList

// res0: List[(Int, String)] = List((1,hoge1), (2,hoge2), (3,hoge3), (1,hoge4), (2,hoge5))