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问题描述 投票：0回答：5

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haskell pattern-matching
5个回答
4
投票

主要问题是您希望使用 

view
中的排列而不是单个值。我们只有一种允许排列的类型 - 记录。

所以，我们可以添加新数据、记录类型

data B = F|T -- just shorter name for Bool in patterns
data Palette = P {isW, isU, isB, isR, isG :: B}

bool2b :: Bool -> B
bool2b True  = T
bool2b False = F

viewColors :: Card -> Palette
viewColors (Card colors) = let m = bool2b . (`member` colors)
    in P {isW = m W, isU = m U, isB = m B, isR = m R, isG = m G}

foo :: Card -> String
foo (viewColors -> P {isW=T, isB=T}) = "card is white and black"
foo _ = "whatever"

已更新

我们也可以否认错误的模式。但这个解决方案更丑陋，但它允许使用“经典”模式

{-# LANGUAGE GADTs #-}
{-# LANGUAGE EmptyDataDecls #-}
{-# LANGUAGE RankNTypes #-}
data Color = W | U | B | R | G  deriving (Eq)

data W' 
data U' 
data B'
data R'
data G'

data Color' a where
      W' :: Color' W'
      U' :: Color' U'
      B' :: Color' B'
      R' :: Color' R'
      G' :: Color' G'

data M a = N | J a -- just shorter name for Maybe a in patterns

data Palette = Palette 
      (M (Color' W')) 
      (M (Color' U')) 
      (M (Color' B')) 
      (M (Color' R')) 
      (M (Color' G'))

并定义 

viewColor
:

viewColors :: Card -> Palette
viewColors (Card colors) = 
  let 
    m :: Color -> Color' a -> M (Color' a)
    m c e = if c `member` colors then J e else N
  in P (m W W') (m U U') (m B B') (m R R') (m G G')

foo :: Card -> String
foo (viewColors -> Palette (J W') N (J B') N N) = 
      "card is white and black"
foo _ = "whatever"
3
投票

我喜欢记录解决方案，但使用类型类很容易实现

{-# LANGUAGE ViewPatterns, ScopedTypeVariables #-}

import qualified Data.Set as Set

data Color = W' | U' | B' | R' | G' deriving (Show, Eq, Ord)
data Card = Card (Set.Set Color) 

newtype W a = W a
newtype U a = U a
newtype B a = B a
newtype R a = R a
newtype G a = G a

class ToColors x where
  toColors :: x -> [Color]
  reify :: x

instance ToColors () where
  toColors _ = []
  reify = ()

instance ToColors a => ToColors (W a) where
  toColors (W a) = W':toColors a
  reify = W reify

--other instances

members :: Set.Set Color -> [Color] -> Bool
members s = foldl (\b e -> b && (Set.member e s)) True

viewColors :: forall a. ToColors a => Card -> Maybe a
viewColors (Card s) = let a = reify :: a in 
  if members s (toColors a) then (Just a) else Nothing

foo :: Card -> String
foo (viewColors -> Just (W (B ()))) = "card is white and black"
foo _ = "whatever"

这可以很容易地被修改以获得其他语法。例如，您可以将颜色定义为不带参数的类型，然后使用中缀异构列表构造函数。无论哪种方式，它都不关心顺序。

编辑：如果你想匹配精确的集合，这也很容易——只需像这样替换 

members
函数

viewColors :: forall a. ToColors a => Card -> Maybe a
viewColors (Card s) = let a = reify :: a in 
  if s == (Set.fromList . toColors $ a) then (Just a) else Nothing
2
投票

编辑：进一步测试表明该解决方案实际上不起作用。

您实际上不需要更多的扩展，我想出了一个可以满足您想要的功能的解决方案，但您可能想要优化它，重命名一些东西，并使其不那么难看。你只需要创建一个新的数据类型并自己实现 

Eq
并让操作员使用 
infixr
:

{-# LANGUAGE ViewPatterns #-}
import Data.Set

data Color = W | U | B | R | G
    deriving (Show, Eq, Ord)

data Card = Card (Set Color) -- simplified Card type with only its colors

-- you may need to fiddle with the precedence here
infixr 0 :*
data MyList a = END | a :* (MyList a) deriving (Show)

myFromList :: [a] -> MyList a
myFromList [] = END
myFromList (x:xs) = x :* myFromList xs

instance Eq a => Eq (MyList a) where
    END == END = True
    END == _   = False
    _   == END = False
    l1  == l2  = allElem l1 l2 && allElem l2 l1
        where
            -- optimize this, otherwise it'll just be really slow
            -- I was just too lazy to write it correctly
            elemMyList :: Eq a => a -> MyList a -> Bool
            elemMyList a ml = case ml of
                END -> False
                (h :* rest) -> if a == h then True else elemMyList a rest
            allElem :: Eq a => MyList a -> MyList a -> Bool
            allElem END l = True
            allElem (h :* rest) l = h `elemMyList` l && allElem rest l

viewColors :: Card -> MyList Color
viewColors (Card colors) = myFromList $ toList colors

fuz :: Card -> String
fuz (viewColors -> (W :* END)) = "it's white"
fuz (viewColors -> (W :* U :* END)) = "it's white and blue"
fuz (viewColors -> (W :* B :* END)) = "it's white and black"
fuz (viewColors -> (W :* B :* R :* END)) = "it's white, black, and red"
fuz (viewColors -> (W :* U :* B :* R :* G :* END)) = "it's all colors"
fuz _ = "I don't know all my colors"

main = do
    putStrLn $ fuz $ Card $ fromList [W, B]
    putStrLn $ fuz $ Card $ fromList [B, W]

编辑：只是稍微修复了代码

0
投票

我认为你应该首先专注于准确表达卡片的颜色，然后再担心其他问题，例如稍后让事情变得简洁。在我看来，您的 

Bool
元组解决方案几乎是完美的，但我猜测卡片必须有一种颜色，对吗？

在这种情况下，类似这样的东西可能会起作用，并且很容易进行模式匹配：

data CardColors = W' BlackBool GreenBool ...
                | B' WhiteBool GreenBool ...
                | G' BlackBool WhiteBool ...
                ....

data BlackBool = B 
               | NotB
-- etc.

您可以相当轻松地创建具有定义顺序的异构列表，但我认为这种多态性在这里不会为您服务。

0
投票

（不是您问题的答案，但希望能解决您的问题！）

我会选择最愚蠢但可能有效的方法：

is :: Card -> Color -> Bool
is card col = col `elem` (viewColors card) -- can be optimized to use the proper elem!

然后

foo :: Card -> String
foo c
    | c `is` B && c `is` W = "card is black and white"
    | c `is` R || c `is` G = "card is red or green"
    | otherwise = "whatever"

如果拼写出整个列表来检查卡片是否具有所有 5 种颜色太长，那么您可以定义额外的组合器，例如

hasColors :: Card -> [Color] -> Bool
hasColors card = all (`elem` (viewColors card))

有什么理由不可接受吗？

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