Haskell：引用更新函数中以前更新的列表元素

本文介绍了Haskell：引用更新函数中以前更新的列表元素的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 假设我有以下定义： pre $ data Book = Book {id :: Int，title :: String} type Shelf = [Book] 假设我有一个假设的函数（upd用于更新） updShelf :: Shelf - >书架 updShelf all @（book：books）= updBook book：updShelf books 所有罚款至今。现在我们假设updateBook函数需要在它之前引用 更新的 三本书，也就是说，书架中位置5的book的updateBook需要参考位置2的book（假设前三本书不需要这样的参考来更新）。没问题，我说，并修改我的代码： updShelf :: Shelf - >书架 updShelf所有@（书：书）prevBook = updBook书prevBook：updShelf书其中prevBook = ??? 我需要帮助的是prevBook函数。虽然我甚至不确定我是否以正确的方式处理这个问题。所以，如果你们有更好的建议来解决这个问题，我们将非常感激。 编辑： 解决方案不适用于我。这是为什么：假设初始货架（全部）状态为 Book {id = 1，title =a } Book {id = 2，title =b} Book {id = 3，title =c} Book {id = 4，title =d} Book {id = 5，title =e} Book {id = 6，title =f} Book {id = 7，title =g} $ b $ （book）{b = Book {title =h} 然后（drop 3 partialUpdate）是 updBook 4 updBook 5 updBook 6 updBook 7 updBook 8 zipWith'（$）（drop 3 partialUpdate）（所有）是： updBook 4 1 updBook 5 2 updBook 6 3 updBook 7 4 - >哎呀！旧版本的书4！ updBook 8 5 - >哎呀！旧版本的书5！ 在我的情况中，我需要书籍7和8根据已更新版本的书4进行更新， 5，而不是未更新的。我希望你明白我的意思。 .haskell.org/haskellwiki/Tying_the_Knot>绑结：我们将在计算答案时使用答案。出于说明的目的，我将使用 type Book = Int 来替代。 updateShelf :: Shelf - >货架 updateShelf shelf = answer其中 answer = zipWith updateBook移动货架移动=复制3 Nothing ++ map只需回答 - 一些愚蠢的实现只是为了说明 updateBook ::可能预订 - >书 - > Book updateBook Nothing current = current + 1 updateBook（Just threeBack）current = current + threeBack + 1 现在，在 ghci 中，我们可以验证 updateShelf 确实在使用更新的版本： *主要> updateShelf [1,10,100,1000,10000] [2,11,101,1003,10012] 1 + 1 ， 10 + 1 和 100 + 1 ，其余两个是 1000+（1 + 1）+1 和 10000+（ 10 + 1）+1 ，因此正如你所希望的那样使用更新的以前的值。 Say I have the following definitionsdata Book = Book {id :: Int, title :: String}type Shelf = [Book]Assuming I have a hypothetical function (upd is for update)updShelf :: Shelf -> ShelfupdShelf all@(book : books) = updBook book : updShelf booksAll fine so far. Now let's say the updateBook function needs to refer to the updated book three books before it i.e updateBook for book at position 5 in bookshelf need to refer to book at position 2 (assume first three books need no such reference to update). No problem, I say, and modify my code as such:updShelf :: Shelf -> ShelfupdShelf all@(book : books) prevBook = updBook book prevBook : updShelf books where prevBook = ???What I need help is with is the prevBook function. Although I am not even sure if I am approaching this problem the right way. So, if you guys have any better suggestion to approach this problem differently, it would be highly appreciatedEDIT:Thomas M. DuBuisson: Your solution won't work for me. Here's why:Assume initial shelf (all) state asBook {id=1, title="a"}Book {id=2, title="b"}Book {id=3, title="c"}Book {id=4, title="d"}Book {id=5, title="e"}Book {id=6, title="f"}Book {id=7, title="g"}Book {id=8, title="h"}then (drop 3 partialUpdate) is (using only ids rather than entire book statement):updBook 4updBook 5updBook 6updBook 7updBook 8zipWith' ($) (drop 3 partialUpdate) (all) is :updBook 4 1updBook 5 2updBook 6 3updBook 7 4 -> YIKES! Older version of book 4!updBook 8 5 -> YIKES! Older version of book 5!In my case, I need books 7 and 8 to be updated against already updated versions of book 4 and 5,not the un-updated ones. I hope you understand what I mean to convey. 解决方案 This trick is related to tying the knot: we'll use the answer while computing the answer. For the purposes of illustration, I'll use type Book = Int instead.updateShelf :: Shelf -> ShelfupdateShelf shelf = answer where answer = zipWith updateBook shifted shelf shifted = replicate 3 Nothing ++ map Just answer-- some stupid implementation just for illustrationupdateBook :: Maybe Book -> Book -> BookupdateBook Nothing current = current + 1updateBook (Just threeBack) current = current + threeBack + 1Now, in ghci, we can verify that updateShelf is really using the updated versions:*Main> updateShelf [1,10,100,1000,10000][2,11,101,1003,10012]As you can see, the first three are 1+1, 10+1, and 100+1, and the remaining two are 1000+(1+1)+1 and 10000+(10+1)+1, and are therefore using the updated previous values, just as you'd hope. 这篇关于Haskell：引用更新函数中以前更新的列表元素的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！