2015-11-02 77 views
2

我一直在撞墙,试图找出如何让这个工作。最初的解析器工作得很好,但是当我试图从中得到一个Map时,它只给出了初始数值,但不是他重视块。Scala Combinator解析器映射问题

格式为 “浮动:[标签浮动:[optinallabel浮动:] ...]”,如:

 412285.556: [Label 0.0:[Label1 1.0:][Label2 2.0:] 

的解析器列表:

object ParseList extends JavaTokenParsers { 
    def sep : Parser[String] = ":" 
    def string : Parser[Any] = """[\.a-zA-Z0-9]+""".r 
    def num: Parser[Any] = floatingPointNumber <~ sep 
    def valueBlock: Parser[Any] = "["~>rep(valueBlock)<~"]" | string ~ floatingPointNumber <~ sep 
    def expr: Parser[Any] = num ~ rep(valueBlock) 
    def apply(in: String) = parseAll(expr,in) 
} 

测试给:

scala> ParseList("""412285.556: """) 
res150: ParseList.ParseResult[Any] = [1.13] parsed: (412285.556~List()) 

scala> ParseList("""412285.556: [Label 1.0:]""") 
res151: ParseList.ParseResult[Any] = [1.25] parsed: (412285.556~List(List((Label~1.0)))) 

scala> ParseList("""412285.556: [Label 0.0:[Label1 1.0:][Label2 2.0:]]""") 
res152: ParseList.ParseResult[Any] = [1.51] parsed: (412285.556~List(List((Label~0.0), List((Label1~1.0)), List((Label2~2.0))))) 

当我尝试使它成为一个Map,如果只返回数字,但调用成员例程。看到调试输出。

地图分析器:

object ParseMap extends JavaTokenParsers { 
    // Seperator 
    def sep : Parser[String] = ":" 
    // string 
    def string : Parser[String] = """[a-zA-Z][a-zA-Z0-9]+""".r 
    // Block within [] with label value: and option additional blocks 
    def valueBlock: Parser[(String,Any)] = 
     member <~ rep(obj) 
    // Member - value pair within a block 
    def member: Parser[(String, Any)] = 
    string ~ floatingPointNumber <~ sep ^^ 
     { case s ~ n => (s, n); println("In Member s=" +s+" n="+n); (s, n)} 
    def obj: Parser[Map[String,Any]] = 
    "["~> rep(valueBlock) <~"]" ^^ {Map() ++ _} 
    // Initial number value of the data 
    def num: Parser[(String, Any)] = 
      floatingPointNumber <~ sep ~ rep(obj) ^^ 
      { case floatingPointNumber => ("Num", floatingPointNumber) } 
    // order of operations 
    def value: Parser[Any] = (
     num 
     | obj 
     | member 
     | floatingPointNumber 
     | string 
    ) 
    def apply(in: String) = parseAll(value,in) 
} 

测试得出:

scala> ParseMap("""412285.556: """) 
res154: ParseMap.ParseResult[Any] = [1.13] parsed: (Num,412285.556) 

scala> ParseMap("""412285.556: [Label 1.0:]""") 
In Member s=Label n=1.0 
res155: ParseMap.ParseResult[Any] = [1.25] parsed: (Num,412285.556) 

scala> ParseMap("""412285.556: [Label 0.0:[Label1 1.0:][Label2 2.0:]]""") 
In Member s=Label n=0.0 
In Member s=Label1 n=1.0 
In Member s=Label2 n=2.0 
res156: ParseMap.ParseResult[Any] = [1.51] parsed: (Num,412285.556) 

我所有的尝试得到一个单一的地图出来,已经失败。任何帮助将不胜感激。

回答

2

的主要问题是您在num方法<~组合子的使用 - 这个扔掉所有解析数据跟随它的(包括你想为你的地图rep(obj)解析结果)。行修改为类似:

def num: Parser[(String, Any)] = 
     floatingPointNumber ~ sep ~ rep(obj) ^^ 
     { case floatingPointNumber ~ sep ~ objs => ("Num", (floatingPointNumber, objs)) } 

和你开始像结果:

scala> ParseMap("""412285.556: [Label 0.0:[Label1 1.0:][Label2 2.0:]]""") 
In Member s=Label n=0.0 
In Member s=Label1 n=1.0 
In Member s=Label2 n=2.0 
res3: ParseMap2.ParseResult[Any] = [1.51] parsed: (Num,(412285.556,List(Map(Label -> 0.0)))) 

,虽然它可能看起来并不很喜欢你可能会想什么,应该给你一个起点指向进一步发展。

+0

我认为这将在本猛冲得到我。 – user190941

2

Scala的解析器组合器不是最容易使用的东西,而且它们非常慢。改为尝试fastparse

我不能完全弄清楚你想从这些数据中得到什么(这是一个非常奇怪的格式!),但有一个很好的guide to getting started

核心标签解释可能会看起来像

val Pair = P(Label ~ " " ~ Num ~ ":") 
val MapLine = P("[" Pair ~ ("[" ~ Pair ~ "]").rep ~ "]"). 
    map{ case (pair, pairs) => pair :: pairs.toList } 
+0

谢谢,一旦我明白了这一点,我就打算看看快速分类。 – user190941