Question

我写了一个程序，结果发现使用列表太慢了，所以我试图切换到序列。但是，在查看文档之后，我似乎无法弄清楚正确的语法。

到目前为止，我尝试使用这个简单的代码学习：

import Control.Monad
import qualified Data.Sequence as S

main :: IO ()
main = do 
       let testSeq = S.empty
       testSeq S.|> 5
       testSeq S.|> 20
       testSeq S.|> 3
       let newSeq = S.update 2 3 testSeq
       let x = lookup 2 testSeq
       print x

我已经玩了一段时间没有运气的语法，但它仍然有很多错误：

test.hs:9:8:
    Couldn't match expected type ‘IO a0’
                with actual type ‘S.Seq Integer’
    In a stmt of a 'do' block: testSeq S.|> 5
    In the expression:
      do { let testSeq = S.empty;
           testSeq S.|> 5;
           testSeq S.|> 20;
           testSeq S.|> 3;
           .... }
    In an equation for ‘main’:
        main
          = do { let testSeq = ...;
                 testSeq S.|> 5;
                 testSeq S.|> 20;
                 .... }

test.hs:10:8:
    Couldn't match expected type ‘IO a1’
                with actual type ‘S.Seq Integer’
    In a stmt of a 'do' block: testSeq S.|> 20
    In the expression:
      do { let testSeq = S.empty;
           testSeq S.|> 5;
           testSeq S.|> 20;
           testSeq S.|> 3;
           .... }
    In an equation for ‘main’:
        main
          = do { let testSeq = ...;
                 testSeq S.|> 5;
                 testSeq S.|> 20;
                 .... }

test.hs:11:8:
    Couldn't match expected type ‘IO a2’
                with actual type ‘S.Seq Integer’
    In a stmt of a 'do' block: testSeq S.|> 3
    In the expression:
      do { let testSeq = S.empty;
           testSeq S.|> 5;
           testSeq S.|> 20;
           testSeq S.|> 3;
           .... }
    In an equation for ‘main’:
        main
          = do { let testSeq = ...;
                 testSeq S.|> 5;
                 testSeq S.|> 20;
                 .... }

test.hs:13:25:
    Couldn't match expected type ‘[(Integer, b)]’
                with actual type ‘S.Seq a3’
    Relevant bindings include x :: Maybe b (bound at test.hs:13:12)
    In the second argument of ‘lookup’, namely ‘testSeq’
    In the expression: lookup 2 testSeq

我是Haskell的新手，所以非常感谢任何帮助！

Answer 1

与Haskell中的几乎所有内容一样，Seq是一个纯粹的功能数据结构。它不像是一个必要的堆栈，你可以推送东西，改变原始结构。相反，与普通列表一样，您只需生成具有额外元素的新容器值，但这不会以任何方式影响旧的较短seq。所以，你问过的程序应该只是

testSeq :: S.Seq Int
testSeq = S.update 2 3 $ S.empty S.|> 5 S.|> 20 S.|> 30

main :: IO ()
main = print $ S.lookup 2 testSeq

（必须是S.lookup，或等效S.!?。lookup本身就是一个适用于普通列表的函数！）

请注意，这并没有真正优于

testSeq = S.update 2 3 $ S.fromList [5,20,30]

事实上，列表通常比<{em}更快，它们只是不允许有效的随机访问。

在Haskell

1 个答案: