Zip elements with odd and even indices in a list

问题

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?

回答1:

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.)

回答2:

in Scala 2.8 you can do:

def pairs[T](xs: List[T]) =
  xs.grouped(2)
    .map{case List(a, b) => (a,b)}
    .toList

回答3:

Untested:

def ziptwo(l: List[String]): List[(String, String)] = l match {
    case Nil => Nil
    case a :: b :: rest =>
         Pair(a,b) :: ziptwo(rest)
}

回答4:

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. ;-)

回答5:

def pairify[T](list: List[T]): List[(T, T)] = list match {
  case Nil => Nil
  case x :: y :: xs => (x, y) :: pairify(xs)
  case _ => error("odd length list!")
}

回答6:

This allows incomplete pairs:

def pairwise [T] (xs: List[T]) : List [(Option[T], Option[T])] = xs match {
     case (x :: y :: xsr) => (Some (x), Some (y)) :: pairwise (xsr)             
     case (x :: Nil) => List ((Some (x), None))                                 
     case (_) => Nil }

回答7:

Using Scala version 2.11.8

val s = Seq("A", "B", "C", "D", "E", "F")
s.sliding(2,2).map( l => (l.head,l.tail.head) ).toArray

output:

Array[(String, String)] = Array((A,B), (C,D), (E,F))

if the input sequence can have an odd number of elements, a little more verbose is required:

val s = Seq("A", "B", "C", "D", "E", "F", "ODD")
val defaultValue = "_"
s.sliding(2,2).map( l => (l.head,l.tail.headOption.getOrElse(defaultStr)) ).toArray

output:

Array[(String, String)] = Array((A,B), (C,D), (E,F), (ODD,_))

来源：https://stackoverflow.com/questions/2601611/zip-elements-with-odd-and-even-indices-in-a-list

标签

scala

functional-programming