I want to zip even and odd elements in a list to make a list of pairs, like that:
["A", "B", "C", "D", "E", "F"] -> [("A", "B"), ("C", "D"), ("E", "F")]
What is the most concise expression to do this in elegant in functional way?
In 2.8, you'd probably use methods:
scala> val a = "ABCDEF".toList.map(_.toString)
a: List[java.lang.String] = List(A, B, C, D, E, F)
scala> a.grouped(2).partialMap{ case List(a,b) => (a,b) }.toList
res0: List[(java.lang.String, java.lang.String)] = List((A,B), (C,D), (E,F))
(This is 2.8.0 Beta1; the latest trunk has collect
in place of partialMap
.)
In 2.7--and not a bad runner-up in 2.8--you could create a recursive method as legoscia did:
def zipPairs[A](la : List[A]): List[(A,A)] = la match {
case a :: b :: rest => (a,b) :: zipPairs(rest)
case _ => Nil
}
scala> zipPairs(a)
res1: List[(java.lang.String, java.lang.String)] = List((A,B), (C,D), (E,F))
Edit: here's another briefer approach that works on 2.7 also:
scala> (a zip a.drop(1)).zipWithIndex.filter(_._2 % 2 == 0).map(_._1)
res2: List[(java.lang.String, java.lang.String)] = List((A,B), (C,D), (E,F))
(Note the use of drop(1)
instead of tail
so it works with empty lists.)
in Scala 2.8 you can do:
def pairs[T](xs: List[T]) =
xs.grouped(2)
.map{case List(a, b) => (a,b)}
.toList
Untested:
def ziptwo(l: List[String]): List[(String, String)] = l match {
case Nil => Nil
case a :: b :: rest =>
Pair(a,b) :: ziptwo(rest)
}
The only advantage of having everyone come up with the most obvious ways of doing it is that I have to think harder about alternate solutions. So here is one that works on Scala 2.8. On Scala 2.7, replace view
with projection
.
def everyNofM[T](l: List[T], n: Int, m: Int) =
l.view.zipWithIndex.filter(_._2 % m == n).map(_._1)
def evens[T](l: List[T]) = everyNofM(l, 0, 2)
def odds[T](l: List[T]) = everyNofM(l, 1, 2)
def zip[T](l: List[T]) = evens(l) zip odds(l) toList
Strictly speaking, view
/projection
is unnecessary, but it avoids unnecessary creation of intermediate results.
Other fun ways of doing it:
def zip[T](l: List[T]) = l.indices.partition(_ % 2 == 0).zipped.map(
(x: Int, y: Int) => (l(x), l(y))
).toList
def zip[T](l: List[T]) = l.zipWithIndex.partition(_._2 % 2 == 0).zipped.map(
(x, y) => (x._1, y._1)
)
PS: Bonus point to whoever gets the pun. ;-)
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