¿Puedo usar for-comprensión/rendimiento para crear un mapa en Scala?

for-comprehension scala-2.7

Question

¿Puedo "ceder" en un mapa?

He intentado

val rndTrans = for (s1 <- 0 to nStates;
s2 <- 0 to nStates
if rnd.nextDouble() < trans_probability)
yield (s1 -> s2);

(y con , lugar de -> ) pero me sale el error

TestCaseGenerator.scala:42: error: type mismatch;
found : Seq.Projection[(Int, Int)]
required: Map[State,State]
new LTS(rndTrans, rndLabeling)

Puedo ver por qué, pero no puedo ver cómo resolver esto: - /

Answer 1

Alternativa ( funciona en 2.7 ):

scala> Map((for(i <- 0 to 10; j <- 0 to 10) yield (i -> j)): _*)
res0: scala.collection.immutable.Map[Int,Int] = Map((0,10), (5,10), (10,10), (1,10), (6,10), (9,10), (2,10), (7,10), (3,10), (8,10), (4,10))

Answer 2

Una solución alternativa en Scala 2.8:

Welcome to Scala version 2.8.1.r23457-b20101106033551 (Java HotSpot(TM) Client VM, Java 1.6.0_22).
Type in expressions to have them evaluated.
Type :help for more information.
scala> import scala.collection.breakOut
import scala.collection.breakOut
scala> val list: List[(Int,Int)] = (for(i<-0 to 3;j<-0 to 2) yield(i->j))(breakOut)
list: List[(Int, Int)] = List((0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,1), (2,2), (3,0), (3,1), (3,2))
scala> val map: Map[Int,Int] = (for(i<-0 to 3;j<-0 to 2) yield(i->j))(breakOut)
map: Map[Int,Int] = Map((0,2), (1,2), (2,2), (3,2))
scala> val set: Set[(Int,Int)] = (for(i<-0 to 3;j<-0 to 2) yield(i->j))(breakOut)
set: Set[(Int, Int)] = Set((2,2), (3,2), (0,1), (1,2), (0,0), (2,0), (3,1), (0,2), (1,1), (2,1), (1,0), (3,0))
scala> 

Answer 3

val rndTrans = (
for {
s1 <- 0 to nStates
s2 <- 0 to nStates if rnd.nextDouble() < trans_probability
} yield s1 -> s2
) (collection.breakOut[Any, (Int, Int), Map[Int, Int]])

Answer 4

scala> (for(i <- 0 to 10; j <- 0 to 10) yield (i -> j)) toMap
res1: scala.collection.immutable.Map[Int,Int] = Map((0,10), (5,10), (10,10), (1,10), (6,10), (9,10), (2,10), (7,10), (3,10), (8,10), (4,10))