All permutations with repetition using scala

问题

I am looking for the scala way to give all permutations without repetitions. I know there are some postings on this site already but they seem to have a slightly different problem.

I am searching for all permutations with repetitions. For example:

combine(List('A','C','G'))

Should yield:

List(List('A'.'A','A'),List('A'.'A','C'),List('A'.'A','G'),List('A'.'C','A'),
List('A'.'C',''C), ... List('G'.'G','G')

I am sorry if my problem is already solved but I was not able to find it.

Thanks in advance.

EDIT:

My own approach (doesn't compile):

def combine(size: Int = sym.length) : List[List[T]] = {
  size match {
    case 0 => List()
    case 1 => sym.toList.map(List(_))
    case _ => for (el <- sym) yield el :: combine(size-1)
  }
}

sym is an array member of a class which contains all the symbols to be combined.

回答1:

def combinations(size: Int = sym.length) : List[List[T]] = {
    if (size == 0)
        List(List())
    else {
        for {
            x  <- sym.toList
            xs <- combinations(size-1)
        } yield x :: xs
    }
}

回答2:

With Scalaz:

scala> import scalaz._
import scalaz._

scala> import Scalaz._
import Scalaz._

scala> def combine[A](xs: List[A]): List[List[A]] = {
     |   xs.replicate[List](xs.size).sequence
     | }
combine: [A](xs: List[A])List[List[A]]

scala> combine(List('A', 'C', 'G'))
res47: List[List[Char]] = List(List(A, A, A), List(A, A, C), List(A, A, G), List
(A, C, A), List(A, C, C), List(A, C, G), List(A, G, A), List(A, G, C), List(A, G
, G), List(C, A, A), List(C, A, C), List(C, A, G), List(C, C, A), List(C, C, C),
 List(C, C, G), List(C, G, A), List(C, G, C), List(C, G, G), List(G, A, A), List
(G, A, C), List(G, A, G), List(G, C, A), List(G, C, C), List(G, C, G), List(G, G
, A), List(G, G, C), List(G, G, G))

回答3:

This should work:

val input = List('A','C','G')

(input ++ input ++ input) combinations(3) toList

回答4:

scala> def comb(s:String)=(s * s.length).combinations(s.length)
comb: (s: String)Iterator[String]

scala> comb("ACG").toList
res16: List[String] = List(AAA, AAC, AAG, ACC, ACG, AGG, CCC, CCG, CGG, GGG)

And if you wanted the resulting permutations:

scala> comb("ACG").flatMap(_.toSeq.permutations.toList).toList
res11: List[Seq[Char]] = List(AAA, AAC, ACA, CAA, AAG, AGA, GAA, ACC, CAC, CCA, ACG, AGC, CAG, CGA, GAC, GCA, AGG, GAG, GGA, CCC, CCG, CGC, GCC, CGG, GCG, GGC, GGG)

You can leave out the toList but it's there so you can see the results.

回答5:

It seems no one has suggested the easiest---or, at least, easiest to read---solution. It is

myList = List("A", "C", "G")
for {
  i <- myList
  j <- myList
  k <- myList
} yield List(i,j,k)

(This is syntactic sugar for the following composition of maps:

myList.flatMap(i => myList.flatMap(j => myList.map(k => List(i,j,k))))

to which the Scala compiler translates the above for expression.)

回答6:

In ScalaZ 7

import scalaz._
import Scalaz._
def combinine[T](l: List[T]) = l.replicateM(l.size)

回答7:

Just making a more generic answers, from @opyate and @monnef:

// considering that we want a permutation_size
List.fill(permutation_size)(input).flatten.combinations(permutation_size).toList

This will generate the permutation with repetition with size permutation_size:

val a = List.fill(2)(List("A","B","C")).flatten.combinations(2).toList
a: List[List[String]] = List(List(A, A), List(A, B), List(A, C), List(B, B), List(B, C), List(C, C))

and

val a = List.fill(3)(List("A","B","C")).flatten.combinations(3).toList
a: List[List[String]] = List(List(A, A, A), List(A, A, B), List(A, A, C), List(A, B, B), List(A, B, C), List(A, C, C), List(B, B, B), List(B, B, C), List(B, C, C), List(C, C, C))

来源：https://stackoverflow.com/questions/7474709/all-permutations-with-repetition-using-scala

标签

scala

combinations

combinatorics

repeat