以语义方式填充元组列表

本文介绍了以语义方式填充元组列表的处理方法，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 我正在处理一段代码，我必须处理元组列表，其中元组的键（ fst s）的顺序和名称）匹配某个模板。我通过验证和（如果需要）基于输入生成有效列表来实现容错。 以下是我的意思的示例： 给定键的模板， [hello，world，this，is，a，test] 和一个列表 [（hello，Just 1），（world，Just 2），（test，Just 3）] ，传递给我的函数 validate 会导致验证失败 - 因为键的顺序和值与模板不匹配。 验证失败后，我想生成一个新列表，它看起来像 [（hello，Just 1），（world，Just 2），（这个，Nothing），（is，Nothing），（a，Nothing），（test，Just 3）] 。 我尝试使用（不完整的）列表理解来执行最后一步： [（x，y）| x （显然，这是缺少步骤的地方空条目将被替换为 Nothing s，并且假设输入的类型为 [（String，Maybe Int）] 这样做最简单的语义方式是什么？ 你基本上想要将一个函数映射到你的字符串列表（你称之为模板），也就是 取一个字符串 xs ， 返回 n 与 （xs，Just n） （xs，Nothing）否则，c $ c> xs 。 以下是一种可能的方法： import Data.List（lookup） import Control.Monad（join） 合并:: [String] - > [（String，Maybe Int）] - > [（String，Maybe Int）] 合并temp l = map（\ xs - >（xs，join $ lookup xs l））temp 然而，如果你建立一个 Map ： / p> 将限定的Data.Map导入为M 导入Data.Maybe（也许） 合并:: [String] - > [（String，Maybe Int）] - > [（String，Maybe Int）] 合并temp l = map（\cs - >（cs，M.lookup cs $ fromList'l））temp fromList':: Ord a => [（a，Maybe b）] - > M.Map a b fromList'xs = foldr insertJust M.empty xs insertJust :: Ord a => （a，可能b） - > M.Map a b - > M.Map ab insertJust（xs，maybeVal）mp = maybe mp（\\\ - > M.insert xs n mp）maybeVal 在GHCi中： λ>让myTemplate = [hello，world，this，is，a，test] λ>让myList = [（hello，Just 1），（world，Just 2），（test，Just 3）] λ>合并myTemplate myList [（hello，Just 1），（world，Just 2），（this，Nothing），（is，Nothing），（a，Nothing）， （test，Just 3）] I'm working on a piece of code where I have to process lists of tuples where both the order and names of the "keys" (fsts of the tuples) match a certain template. I'm implementing fault tolerance by validating and (if needed) generating a valid list based on the input.Here's an example of what I mean:Given the template of keys, ["hello", "world", "this", "is", "a", "test"], and a list [("hello", Just 1), ("world", Just 2), ("test", Just 3)], passing it to my function validate would cause it to fail validation - as the order and values of the keys do not match up with the template.Upon failing validation, I want to generate a new list, which would look like [("hello", Just 1), ("world", Just 2), ("this", Nothing), ("is", Nothing), ("a", Nothing), ("test", Just 3)].I tried performing this last step using an (incomplete) list comprehension:[(x, y) | x <- template, y <- l](Obviously, this is missing the step where empty entries would be replaced with Nothings, and works under the assumption that the input is of type [(String, Maybe Int)]).What would be the easiest semantic way of doing this? 解决方案 You essentially want to map a function to your list of strings (which you call "template"), i.e. the function thattakes a string xs,returns(xs, Just n) if an integer n is associated to xs in your "list to validate",(xs, Nothing) otherwise.Here is one possible approach:import Data.List ( lookup )import Control.Monad ( join )consolidate :: [String] -> [(String, Maybe Int)] -> [(String, Maybe Int)]consolidate temp l = map (\xs -> (xs, join $ lookup xs l)) tempHowever, you will get faster lookup if you build a Map holding the key-value pairs of your association list (the "list to validate"):import qualified Data.Map as Mimport Data.Maybe (maybe)consolidate :: [String] -> [(String, Maybe Int)] -> [(String, Maybe Int)]consolidate temp l = map (\cs -> (cs, M.lookup cs $ fromList' l)) tempfromList' :: Ord a => [(a, Maybe b)] -> M.Map a bfromList' xs = foldr insertJust M.empty xsinsertJust :: Ord a => (a, Maybe b) -> M.Map a b -> M.Map a binsertJust (xs, maybeVal) mp = maybe mp (\n -> M.insert xs n mp) maybeValIn GHCi:λ> let myTemplate = ["hello", "world", "this", "is", "a", "test"]λ> let myList = [("hello", Just 1), ("world", Just 2), ("test", Just 3)]λ> consolidate myTemplate myList [("hello",Just 1),("world",Just 2),("this",Nothing),("is",Nothing),("a",Nothing),("test",Just 3)] 这篇关于以语义方式填充元组列表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！