模式匹配Data.Sequence之类的列表

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 = ...

如何实现这一目标Data.Sequence.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)

作为GHC 7.10的,你也可以把它做成双向模式代名词,这样Empty,(:<)并且(:>)可以作为"构造",以及:

{-# 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.您的代码不工作,因为你需要打电话viewl或viewr上你的Seq第一个拿到类型的东西ViewL或ViewR.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

这等同于:

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