如何从scala中的字符串值生成规则？答案

【问题标题】：How to generate rules from a string values in scala?如何从scala中的字符串值生成规则？
【发布时间】：2014-12-02 21:36:22
【问题描述】：

我想用字符串制定规则。例如，我的字符串是：

Ball Cat Egg
Cat Egg
Apple Cat
Ball Egg
Ball Cat

我希望规则为，

Ball -> Cat Egg
Cat -> Ball Egg
Egg -> Ball Cat
Ball Cat -> Egg
Cat Egg -> Ball
Ball Egg -> Cat
Cat -> Egg
Egg -> Cat
Apple -> Cat
Cat -> Apple
Ball -> Egg
Egg -> Ball
Ball -> Cat
Cat -> Ball

【问题讨论】：

标签： string scala dictionary scala-collections

【解决方案1】：

scala> val strs = Seq("BCE", "CE", "AC", "BE", "BC")
strs: Seq[String] = List(BCE, CE, AC, BE, BC)

scala> val rules = for(s <- strs;i <- 1 to s.size-1;p <- s combinations i) yield p -> s.diff(p)
rules: Seq[(String, String)] = 
List((B,CE), 
     (C,BE), 
     (E,BC), 
     (BC,E), 
     (BE,C), 
     (CE,B), 
     (C,E), 
     (E,C), 
     (A,C), 
     (C,A), 
     (B,E), 
     (E,B), 
     (B,C), 
     (C,B))

编辑这是新问题的代码：

scala> case class Rule(src: Seq[String], dst: Seq[String]) {
     |   override def toString = s"${src.mkString(" ")} -> ${dst.mkString(" ")}"
     | }
defined class Rule

scala> val strs = Seq("Ball Cat Egg", "Cat Egg", "Apple Cat", "Ball Egg", "Ball Cat")
strs: Seq[String] = List(Ball Cat Egg, Cat Egg, Apple Cat, Ball Egg, Ball Cat)

scala> val rules = for(s <- strs;l = s.split(" ");i <- 1 to l.size-1;p <- l combinations i) 
                     yield Rule(p, l diff p)
rules: Seq[Rule] = List(Ball -> Cat Egg, Cat -> Ball Egg, Egg -> Ball Cat, 
                        Ball Cat -> Egg, Ball Egg -> Cat, Cat Egg -> Ball, 
                        Cat -> Egg, Egg -> Cat, Apple -> Cat, Cat -> Apple, 
                        Ball -> Egg, Egg -> Ball, Ball -> Cat, Cat -> Ball)

【讨论】：

如何在(B, CE)之间放“->”，即(B -> CE)
非常感谢先生......还有1个疑问......实际上它是一个规则类型，我们如何单独获取每个项目。即，c - > b，我想要单独的c和b。 .这是下一个目的所需要的吗？