算術表達式語法和解析器

Question

最近我正在尋找算術表達式的體面語法，但只發現了一些不重要的語法，例如忽略pow(..., ...)。然後我自己試了一下，但有時它並沒有像預期的那樣工作。例如，我錯過了允許表達式前面的一個單一的-並修復它。也許有人可以看看我目前的做法並改進它。此外，我認爲其他人可以利用，因爲解析算術表達式是一項常見任務。

import scala.math._ 
import scala.util.parsing.combinator._ 
import scala.util.Random 

class FormulaParser(val constants: Map[String,Double] = Map(), val userFcts: Map[String,String => Double] = Map(), random: Random = new Random) extends JavaTokenParsers { 
    require(constants.keySet.intersect(userFcts.keySet).isEmpty) 
    private val allConstants = constants ++ Map("E" -> E, "PI" -> Pi, "Pi" -> Pi) // shouldn´t be empty 
    private val unaryOps: Map[String,Double => Double] = Map(
    "sqrt" -> (sqrt(_)), "abs" -> (abs(_)), "floor" -> (floor(_)), "ceil" -> (ceil(_)), "ln" -> (math.log(_)), "round" -> (round(_)), "signum" -> (signum(_)) 
) 
    private val binaryOps1: Map[String,(Double,Double) => Double] = Map(
    "+" -> (_+_), "-" -> (_-_), "*" -> (_*_), "/" -> (_/_), "^" -> (pow(_,_)) 
) 
    private val binaryOps2: Map[String,(Double,Double) => Double] = Map(
    "max" -> (max(_,_)), "min" -> (min(_,_)) 
) 
    private def fold(d: Double, l: List[~[String,Double]]) = l.foldLeft(d){ case (d1,op~d2) => binaryOps1(op)(d1,d2) } 
    private implicit def map2Parser[V](m: Map[String,V]) = m.keys.map(_ ^^ (identity)).reduceLeft(_ | _) 
    private def expression: Parser[Double] = sign~term~rep(("+"|"-")~term) ^^ { case s~t~l => fold(s * t,l) } 
    private def sign:  Parser[Double] = opt("+" | "-") ^^ { case None => 1; case Some("+") => 1; case Some("-") => -1 } 
    private def term:  Parser[Double] = longFactor~rep(("*"|"/")~longFactor) ^^ { case d~l => fold(d,l) } 
    private def longFactor: Parser[Double] = shortFactor~rep("^"~shortFactor) ^^ { case d~l => fold(d,l) } 
    private def shortFactor: Parser[Double] = fpn | sign~(constant | rnd | unaryFct | binaryFct | userFct | "("~>expression<~")") ^^ { case s~x => s * x } 
    private def constant: Parser[Double] = allConstants ^^ (allConstants(_)) 
    private def rnd:   Parser[Double] = "rnd"~>"("~>fpn~","~fpn<~")" ^^ { case x~_~y => require(y > x); x + (y-x) * random.nextDouble } | "rnd" ^^ { _ => random.nextDouble } 
    private def fpn:   Parser[Double] = floatingPointNumber ^^ (_.toDouble) 
    private def unaryFct: Parser[Double] = unaryOps~"("~expression~")" ^^ { case op~_~d~_ => unaryOps(op)(d) } 
    private def binaryFct: Parser[Double] = binaryOps2~"("~expression~","~expression~")" ^^ { case op~_~d1~_~d2~_ => binaryOps2(op)(d1,d2) } 
    private def userFct:  Parser[Double] = userFcts~"("~(expression ^^ (_.toString) | ident)<~")" ^^ { case fct~_~x => userFcts(fct)(x) } 
    def evaluate(formula: String) = parseAll(expression,formula).get 
} 

所以可以評價如下：

val formulaParser = new FormulaParser(
    constants = Map("radius" -> 8D, 
        "height" -> 10D, 
        "c" -> 299792458, // m/s 
        "v" -> 130 * 1000/60/60, // 130 km/h in m/s 
        "m" -> 80), 
    userFcts = Map("perimeter" -> { _.toDouble * 2 * Pi })) 

println(formulaParser.evaluate("2+3*5")) // 17.0 
println(formulaParser.evaluate("height*perimeter(radius)")) // 502.6548245743669 
println(formulaParser.evaluate("m/sqrt(1-v^2/c^2)")) // 80.00000000003415 

任何改進的建議？我是否使用正確的語法，還是隻有用戶輸入一個有效的（關於我提供的函數）算術表達式才能解析的時間問題？
（運算符優先級What's？）

Answer 1

爲了更好的表現，我建議定義分析器時使用private lazy val，而不是private def。否則，只要解析器是引用，它就會再次創建。

不錯的代碼BTW。

Answer 2

嗯，也許在循環中添加變量：

import scala.math._ 
import scala.util.parsing.combinator._ 
import scala.util.Random 

class FormulaParser(val variables: Set[String] = Set(), 
        val constants: Map[String, Double] = Map(), 
        val unary: Map[String, Double => Double] = Map(), 
        val binary: Map[String, (Double, Double) => Double] = Map(), 
        val userFcts: Map[String, String => Double] = Map(), 
        random: Random = new Random) extends JavaTokenParsers { 
    require(constants.keySet.intersect(userFcts.keySet).isEmpty) 
    private val allConstants = constants ++ Map("E" -> E, "PI" -> Pi, "Pi" -> Pi) 
    // shouldn´t be empty 
    private val unaryOps = Map[String, Double => Double](
    "sqrt" -> (sqrt(_)), "abs" -> (abs(_)), "floor" -> (floor(_)), "ceil" -> (ceil(_)), "ln" -> (math.log(_)), "round" -> (round(_).toDouble), "signum" -> (signum(_)) 
    ) ++ unary 
    private val binaryOps1 = Map[String, (Double, Double) => Double](
     "+" -> (_ + _), "-" -> (_ - _), "*" -> (_ * _), "/" -> (_/_), "^" -> (pow(_, _)) 
    ) 
    private val binaryOps2 = Map[String, (Double, Double) => Double](
     "max" -> (max(_, _)), "min" -> (min(_, _)) 
    ) ++ binary 

    type Argument = Map[String, Double] 
    type Formula = Argument => Double 

    private def fold(d: Formula, l: List[~[String, Formula]]) = l.foldLeft(d) { case (d1, op ~ d2) => arg => binaryOps1(op)(d1(arg), d2(arg))} 
    private implicit def set2Parser[V](s: Set[String]) = s.map(_ ^^ identity).reduceLeft(_ | _) 
    private implicit def map2Parser[V](m: Map[String, V]) = m.keys.map(_ ^^ identity).reduceLeft(_ | _) 
    private def expression: Parser[Formula] = sign ~ term ~ rep(("+" | "-") ~ term) ^^ { case s ~ t ~ l => fold(arg => s * t(arg), l)} 
    private def sign: Parser[Double] = opt("+" | "-") ^^ { case None => 1; case Some("+") => 1; case Some("-") => -1} 
    private def term: Parser[Formula] = longFactor ~ rep(("*" | "/") ~ longFactor) ^^ { case d ~ l => fold(d, l)} 
    private def longFactor: Parser[Formula] = shortFactor ~ rep("^" ~ shortFactor) ^^ { case d ~ l => fold(d, l)} 
    private def shortFactor: Parser[Formula] = fpn | sign ~ (constant | variable | rnd | unaryFct | binaryFct | userFct | "(" ~> expression <~ ")") ^^ { case s ~ x => arg => s * x(arg)} 
    private def constant: Parser[Formula] = allConstants ^^ (name => arg => allConstants(name)) 
    private def variable: Parser[Formula] = variables ^^ (name => arg => arg(name)) 
    private def rnd: Parser[Formula] = "rnd" ~> "(" ~> fpn ~ "," ~ fpn <~ ")" ^^ { case x ~ _ ~ y => (arg: Argument) => require(y(arg) > x(arg)); x(arg) + (y(arg) - x(arg)) * random.nextDouble} | "rnd" ^^ { _ => arg => random.nextDouble} 
    private def fpn: Parser[Formula] = floatingPointNumber ^^ (value => arg => value.toDouble) 
    private def unaryFct: Parser[Formula] = unaryOps ~ "(" ~ expression ~ ")" ^^ { case op ~ _ ~ d ~ _ => arg => unaryOps(op)(d(arg))} 
    private def binaryFct: Parser[Formula] = binaryOps2 ~ "(" ~ expression ~ "," ~ expression ~ ")" ^^ { case op ~ _ ~ d1 ~ _ ~ d2 ~ _ => arg => binaryOps2(op)(d1(arg), d2(arg))} 
    private def userFct: Parser[Formula] = userFcts ~ "(" ~ (expression ^^ (_.toString) | ident) <~ ")" ^^ { case fct ~ _ ~ x => arg => userFcts(fct)(x)} 
    def evaluate(formula: String) = parseAll(expression, formula).get 
} 

所以，現在你必須通過地圖來評價，你可以這樣做：

val formulaParser = new FormulaParser(Set("x"), unary = Map(
    "sin" -> (math.sin(_)), "cos" -> (math.cos(_)), "tan" -> (math.tan(_)) 
)) 
val formula = formulaParser.evaluate("sin(x)^x") 
val function: Double => Double = x => formula(Map("x" -> x)) 
println(function(5.5)) 

正如你所看到的，我還添加參數添加一元和二元函數。

感謝您的代碼！