在哈斯克尔，有类似于分护人员的东西吗？
我正在写一个关于音乐音程分类的程序。概念结构相当复杂，我会尽可能清楚地说明这一点。前几行代码是一个工作正常的小摘录。第二种是伪代码，它能满足我对简洁性的需求。
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是否有一个可以像下面的伪代码那样简化代码的编程结构？
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提前感谢您的建议。
回答 2
Stack Overflow用户
发布于 2015-02-15 18:25:01
让我用一个比张贴的例子更短的例子：
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]我们的目标是将需要使用n < 12的两个分支“分组”，并将此条件考虑在内。(在original玩具示例中，这不是一个巨大的收获，但在更复杂的情况下可能是如此。)
我们可以天真地考虑将代码分成两种嵌套的情况：
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或者，等效地，使用MultiWayIf扩展：
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然而，这会带来令人惊讶的结果：
> map wrong1 [1..12]
*** Exception: Non-exhaustive patterns in case
> map wrong2 [1..12]
*** Exception: Non-exhaustive guards in multi-way if问题是，当n是1时，取n < 12分支，计算内部情况，然后那里的分支不考虑1。original代码只是尝试下一个分支，该分支处理它。然而，wrong1,wrong2并没有回溯到外部情况。
请注意，当您知道外部情况有不重叠的条件时，这不是一个问题。在OP发布的代码中，情况似乎是这样的，因此wrong1,wrong2方法将在那里工作(如Jefffrey所示)。
然而，在一般情况下，可能会有重叠，这又如何呢？幸运的是，Haskell很懒，所以很容易使用我们自己的控制结构。为此，我们可以如下所示利用Maybe 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 ]它有点冗长，但并没有太多。用这种风格编写代码要比看起来容易得多:一个简单的多向条件写成
foo n = fromJust $ msum
[ guard boolean1 >> return value1
, guard boolean2 >> return value2
, ...
]而且，如果您想要一个“嵌套”情况，只需将任何return value替换为msum [ ... ]即可。
这样做可以确保我们得到通缉的回溯。的确：
> map correct [1..12]
[4,4,4,3,3,2,2,1,1,2,2,4]这里的诀窍是，当一个guard失败时，它会生成一个Nothing值。库函数msum只是选择列表中的第一个非Nothing值。因此，即使内部列表中的每个元素都是Nothing，外部msum也会根据需要考虑外部列表中的下一个项目--回溯。
Stack Overflow用户
发布于 2015-02-15 14:00:43
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
...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 或者，您可以使用MultiWayIf语言扩展：
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")
...https://stackoverflow.com/questions/28526768
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