Question

我使用Data.Sequence代替列表以获得更好的效果。使用列表我们可以执行以下操作

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

如何使用Data.Sequence完成此操作。我尝试过以下方法：

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.viewl和S.viewr，但似乎无法弄清楚如何。

Answer 1

从GHC 7.8开始，您可以将pattern synonyms与view patterns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，您还可以将其转换为双向模式同义词，以便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。您的代码无效，因为您需要先在viewl上致电viewr或Seq，以获取ViewL或ViewR类型的内容。 ViewPatterns可以很好地处理：

{-# 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) -> ...

模式匹配Data.Sequence之类的列表

2 个答案: