Можно ли "вывести" на карту?
Я пробовал
val rndTrans = for (s1 <- 0 to nStates;
s2 <- 0 to nStates
if rnd.nextDouble() < trans_probability)
yield (s1 -> s2);
(и , вместо ->), но я получаю ошибку
TestCaseGenerator.scala:42: error: type mismatch;
found : Seq.Projection[(Int, Int)]
required: Map[State,State]
new LTS(rndTrans, rndLabeling)
Я понимаю, почему, но я не вижу, как это решить: -/
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))
Альтернативное решение в 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>
Альтернатива (работает на 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))
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]])