Question

我的 remStudentdup 功能有问题。我希望它从列表中提取并删除所有多余的学生，然后返回整个列表而不重复学生。我会修改它吗？

import Data.List
type CourseData = [(String,String,String,String,String)]
l :: CourseData

--list contains name of student, year, programme and personal tutor
l = [("fondi","201202378","2012","Bsc280","mr tautology"),
     ("fondi","201202378","2012","Bsc280","mr tautology"),
     ("Sylvee","200801245","2008","Bsc209","ms puma"),
     ("dijeje","201307845","2013","Bsc205","mr tautology"),
     ("heron","201002567","2010","Bsc280","mr setlhako"),
     ("slow","201198746","2011","Bsc205","mr mampu"),
     ("Sylvee","201198746","2008","bsc209","ms puma"),
    ("Sylvee","201198746","2008","bsc209","ms puma")]

rmdups :: Eq a => [a] -> [a]
rmdups [] = []
rmdups (x:xs)   | x `elem` xs   = rmdups xs
                | otherwise     = x : rmdups xs

remStudentdup :: CourseData -> [String]
remStudentdup list  =rmdups [name|(name,id,yr,prog,ptut)<-list]

remStudentdup 函数只返回名称，但我希望它返回所有字段

Answer 1

好吧，让它返回所有字段：

remStudentdup :: CourseData -> CourseData
remStudentdup = rmdups

你可以摆脱remStudentdup函数，因为它只是将调用的参数传递给rmdups函数，而是使用rmdups。

Answer 2

您的列表理解[name|(name,id,yr,prog,ptut)<-list]返回仅包含名称的列表，因此rmdups除了名称之外永远不会看到任何其他内容。

一种可能的解决方案是将完整列表传递给rmdups并稍后提取名称：

rmdups :: Eq a => [a] -> [a]
rmdups [] = []
rmdups (x:xs)   | x `elem` xs   = rmdups xs
                | otherwise     = x : rmdups xs

remStudentdup :: CourseData -> [String]
remStudentdup list  = rmdups list

但是，这会比较学生记录的记录，如果所有字段都相同，则两条记录相同。一个更好的解决方案是通过将比较函数传递给它来使rmdups更通用：

elemBy :: (a -> a -> Bool) -> a -> [a] -> Bool
elemBy isEqual x (y:ys) = isEqual x y || elemBy isEqual x ys
elemBy _ _ [] = False

rmdups' :: (a -> a -> Bool) -> [a] -> [a]
rmdups' _ [] = []
rmdups' isEqual (x:xs) | elemBy isEqual x xs   = rmdups xs
                       | otherwise     = x : rmdups xs

studentName (name,_id,_yr,_prog,_ptut) = name

remStudentdup :: CourseData -> [String]
remStudentdup list  = rmdups' (\s1 s2 -> studentName s1 == studentName s2) list

如何使用从列表中删除多余学生的函数返回整个列表

2 个答案: