haskell - 模式匹配 Data.Sequence 类似列表

Question

我正在使用Data.Sequence列表来获得更好的性能。使用列表，我们可以执行以下操作

foo :: [Int] -> Int
foo [] m = m
foo (x:xs) m = ...

怎么能做到这一点Data.Sequence。我尝试了以下方法：

foo:: S.Seq Int -> Int
foo S.empty m = m
foo (x S.<: xs) m = ...

我认为解决方案涉及使用S.viewland S.viewr，但似乎无法弄清楚如何。

Answer 1

从 GHC 7.8 开始，您可以将模式同义词与视图模式一起用于此目的：

{-# LANGUAGE ViewPatterns, PatternSynonyms #-}

import qualified Data.Sequence as Seq

pattern Empty   <- (Seq.viewl -> Seq.EmptyL)
pattern x :< xs <- (Seq.viewl -> x Seq.:< xs)
pattern xs :> x <- (Seq.viewr -> xs Seq.:> x)

从 GHC 7.10 开始，您还可以将其变成双向模式的同义词，这样Empty,(:<)和(:>)也可以用作“构造函数”：

{-# LANGUAGE ViewPatterns, PatternSynonyms #-}

import qualified Data.Sequence as Seq

pattern Empty   <- (Seq.viewl -> Seq.EmptyL)  where Empty = Seq.empty
pattern x :< xs <- (Seq.viewl -> x Seq.:< xs) where (:<)  = (Seq.<|) 
pattern xs :> x <- (Seq.viewr -> xs Seq.:> x) where (:>)  = (Seq.|>) 

Answer 2

ViewPatterns可能是去这里的方式。您的代码不起作用，因为您需要先调用orviewl来获取类型or的东西。可以很好地处理：viewrSeqViewLViewRViewPatterns

{-# LANGUAGE ViewPatterns #-}

foo (S.viewl -> S.EmptyL)    = ... -- empty on left
foo (S.viewl -> (x S.:< xs)) = ... -- not empty on left

这相当于：

foo seq = case S.viewl seq of
    S.EmptyL    -> ...
    (x S.:< xs) -> ...