在Haskell中,是否有类似于副守卫的东西?

Alb*_*ani 12 haskell pattern-guards

我正在编写一个关于音程分类的程序.概念结构非常复杂,我会尽可能清楚地表达它.前几行代码是一个正常工作的小提取.第二个是伪代码,可以满足我的简洁需求.

interval pt1 pt2
  | gd == 0 && sd <  (-2) = ("unison",show (abs sd) ++ "d") 
  | gd == 0 && sd == (-2) = ("unison","dd")
  | gd == 0 && sd == (-1) = ("unison","d")
  | gd == 0 && sd == 0    = ("unison","P")
  | gd == 0 && sd == 1    = ("unison","A")
  | gd == 0 && sd == 2    = ("unison","AA")
  | gd == 0 && sd >  2    = ("unison",show sd ++ "A")

  | gd == 1 && sd <  (-1) = ("second",show (abs sd) ++ "d")
  | gd == 1 && sd == (-1) = ("second","dd")
  | gd == 1 && sd == 0    = ("second","d")
  | gd == 1 && sd == 1    = ("second","m")
  | gd == 1 && sd == 2    = ("second","M")
  | gd == 1 && sd == 3    = ("second","A")
  | gd == 1 && sd == 4    = ("second","AA")
  | gd == 1 && sd >  4    = ("second",show (abs sd) ++ "A")

  where
  (bn1,acc1,oct1) = parsePitch pt1
  (bn2,acc2,oct2) = parsePitch pt2
  direction = signum sd
  sd = displacementInSemitonesOfPitches pt1 pt2
  gd = abs $ displacementBetweenTwoBaseNotes direction bn1 bn2
Run Code Online (Sandbox Code Playgroud)

是否有一个编程结构可以像下面的伪代码那样简化代码?

interval pt1 pt2 
  | gd == 0  | sd <  (-2) = ("unison",show (abs sd) ++ "d") 
             | sd == (-2) = ("unison","dd")
             | sd == (-1) = ("unison","d")
             | sd == 0    = ("unison","P")
             | sd == 1    = ("unison","A")
             | sd == 2    = ("unison","AA")
             | sd >  2    = ("unison",show sd ++ "A")  
  | gd == 1  | sd <  (-1) = ("second",show (abs sd) ++ "d")
             | sd == (-1) = ("second","dd")
             | sd == 0    = ("second","d")
             | sd == 1    = ("second","m")
             | sd == 2    = ("second","M")
             | sd == 3    = ("second","A")
             | sd == 4    = ("second","AA")
             | sd >  4    = ("second",show (abs sd) ++ "A")
  | gd == 2  | sd ...     = ...
             | sd ...     = ...
  ...
  | mod gd 7 == 1 | mod sd 12 == ...
                  | mod sd 12 == ...
  ...
  | otherwise = ...

  where
  (bn1,acc1,oct1) = parsePitch pt1
  (bn2,acc2,oct2) = parsePitch pt2
  direction = signum sd
  sd = displacementInSemitonesOfPitches pt1 pt2
  gd = abs $ displacementBetweenTwoBaseNotes direction bn1 bn2
Run Code Online (Sandbox Code Playgroud)

提前感谢您的建议.

chi*_*chi 8

让我使用一个比发布的更短的例子:

original :: Int -> Int
original n
  | n < 10 && n > 7 = 1   -- matches 8,9
  | n < 12 && n > 5 = 2   -- matches 6,7,10,11
  | n < 12 && n > 3 = 3   -- matches 4,5
  | n < 13 && n > 0 = 4   -- matches 1,2,3,12
Run Code Online (Sandbox Code Playgroud)

代码在GHCi中运行如下:

> map original [1..12]
[4,4,4,3,3,2,2,1,1,2,2,4]
Run Code Online (Sandbox Code Playgroud)

我们的目标是将两个分支机构"组合"在一起,将n < 12这种情况考虑在内.(这在original玩具示例中并不是一个巨大的收获,但它可能在更复杂的情况下.)

我们可以天真地想到在两个嵌套的情况下拆分代码:

wrong1 :: Int -> Int
wrong1 n = case () of 
  _ | n < 10 && n > 7 -> 1
    | n < 12 -> case () of
                _ | n > 5 -> 2
                  | n > 3 -> 3
    | n < 13 && n > 0 -> 4
Run Code Online (Sandbox Code Playgroud)

或者,等效地,使用MultiWayIf扩展名:

wrong2 :: Int -> Int
wrong2 n = if 
  | n < 10 && n > 7 -> 1
  | n < 12 -> if | n > 5 -> 2
                 | n > 3 -> 3
  | n < 13 && n > 0 -> 4
Run Code Online (Sandbox Code Playgroud)

然而,这会带来惊喜:

> map wrong1 [1..12]
*** Exception: Non-exhaustive patterns in case

> map wrong2 [1..12]
*** Exception: Non-exhaustive guards in multi-way if
Run Code Online (Sandbox Code Playgroud)

问题是,当n1时,n < 12分支,内部情况评价,再没有分支有考虑1.该original代码只是尝试下一个分支,它处理它.但是,wrong1,wrong2不会回溯到外壳.

请注意,当您知道外壳具有非重叠条件时,这不是问题.在OP发布的代码中,情况似乎如此,因此这些wrong1,wrong2方法可以在那里工作(如Jefffrey所示).

但是,可能存在重叠的一般情况呢?幸运的是,Haskell是懒惰的,因此很容易推出自己的控制结构.为此,我们可以Maybe如下利用monad:

correct :: Int -> Int
correct n = fromJust $ msum 
   [ guard (n < 10 && n > 7) >> return 1
   , guard (n < 12)          >> msum
      [ guard (n > 5) >> return 2
      , guard (n > 3) >> return 3 ]
   , guard (n < 13 && n > 0) >> return 4 ]
Run Code Online (Sandbox Code Playgroud)

更详细一点,但不是很大.用这种风格编写代码比看起来容易:简单的多路条件写成

foo n = fromJust $ msum 
   [ guard boolean1 >> return value1
   , guard boolean2 >> return value2
   , ...
   ]
Run Code Online (Sandbox Code Playgroud)

并且,如果你想要一个"嵌套"的情况,只需return value用a 替换任何一个msum [ ... ].

这样做可以确保我们获得所需的回溯.确实:

> map correct [1..12]
[4,4,4,3,3,2,2,1,1,2,2,4]
Run Code Online (Sandbox Code Playgroud)

这里的诀窍是当guard失败时,它会生成一个Nothing值.库函数msum只是选择Nothing列表中的第一个非值.因此,即使内部列表中的每个元素都是Nothing,外部msum也将考虑外部列表中的下一个项目 - 回溯,如所需.


Sho*_*hoe 7

我建议将每个嵌套条件分组到一个函数中:

interval :: _ -> _ -> (String, String)
interval pt1 pt2
    | gd == 0 = doSomethingA pt1 pt2
    | gd == 1 = doSomethingB pt1 pt2
    | gd == 2 = doSomethingC pt1 pt2
    ...
Run Code Online (Sandbox Code Playgroud)

然后,例如:

doSomethingA :: _ -> _ -> (String, String)
doSomethingA pt1 pt2
    | sd <  (-2) = ("unison",show (abs sd) ++ "d") 
    | sd == (-2) = ("unison","dd")
    | sd == (-1) = ("unison","d")
    | sd == 0    = ("unison","P")
    | sd == 1    = ("unison","A")
    | sd == 2    = ("unison","AA")
    | sd >  2    = ("unison",show sd ++ "A")
    where sd = displacementInSemitonesOfPitches pt1 pt2  
Run Code Online (Sandbox Code Playgroud)

或者,您可以使用MultiWayIf语言扩展名:

interval pt1 pt2 =
    if | gd == 0 -> if | sd <  (-2) -> ("unison",show (abs sd) ++ "d") 
                       | sd == (-2) -> ("unison","dd")
                       | sd == (-1) -> ("unison","d")
                       ...
       | gd == 1 -> if | sd <  (-1) -> ("second",show (abs sd) ++ "d")
                       | sd == (-1) -> ("second","dd")
                       | sd == 0    -> ("second","d")
                       ...
Run Code Online (Sandbox Code Playgroud)

  • 请记住,此解决方案_not_始终等效于原始代码:如果内部情况/ if不匹配,则不会尝试外部情况下的下一个分支.当然,在发布的代码中,外壳具有不重叠的防护装置,所以这很好. (4认同)
  • @AlbertoCapitani,`case`*将允许它!`_的案例()c1 - > r1`等 (3认同)