过滤 Haskell 中的元组列表

Question

我是 Haskell 的新手，我正在尝试过滤包含一个字符串和一个字符串列表的元组列表。 我想根据字符串列表是否包含某个字符串进行过滤。 我想创建一个列表列表，其中每个列表都是每个元组的第一个元素，该元素对过滤器的计算结果为真。

例子：
给定这个元组列表
程序语言 =
[（“CptS121”，[“C”]），
("CptS122", ["C++"]),
("CptS223", ["C++"]),
("CptS233", ["Java"]),
("CptS321", ["C#","Java"]),
("CptS322", ["Python","JavaScript"]),
("CptS355", ["Haskell", "Python", "PostScript", "Java"]),
("CptS360", ["C"]),
("CptS370", ["Java"]),
("CptS315", ["Python"]),
("CptS411", ["C", "C++"]),
("CptS451", ["Python", "C#", "SQL","Java"]),
（“CptS475”，[“Python”，“R”]）
]

我想过滤包含 ["Python","C","C++"] 的元组并创建一个列表列表，其中每个子列表是使用每种语言的课程。

所以，["Python","C","C++"] 会 output [["CptS322","CptS355","CptS315","CptS451","CptS475"], ["CptS121","CptS360"," CptS411"], ["CptS122","CptS223","CptS411"] ]

到目前为止，我有以下内容：

    filter_courses [] courses = []
    filter_courses list [] = []
    filter_courses list courses = map snd (filter ((`elem` courses).fst) list)

这会过滤 progLanguages 并输出使用“课程”中任何列出的语言的所有元组的列表。

从上面的列表中取出 output，我正在尝试执行以下操作......

    filter_more [] langs = []
    filter_more list [] = []
    filter_more list langs = map fst (filter ((`elem` langs).snd) list)

这将输出使用任何语言的所有课程的列表，而不是每种语言的课程列表。

Answer 1

以下情况如何：

progLanguages = [ ("CptS121", ["C"])
                , ("CptS122", ["C++"])
                , ("CptS223", ["C++"])
                , ("CptS233", ["Java"])
                , ("CptS321", ["C#","Java"])
                , ("CptS322", ["Python","JavaScript"])
                , ("CptS355", ["Haskell", "Python", "PostScript", "Java"])
                , ("CptS360", ["C"])
                , ("CptS370", ["Java"])
                , ("CptS315", ["Python"])
                , ("CptS411", ["C", "C++"])
                , ("CptS451", ["Python", "C#", "SQL","Java"])
                , ("CptS475", ["Python", "R"])
                ]

filterCourses :: Eq a => [a] -> [(b, [a])] -> [[b]]
filterCourses langs courses = map coursesUsing langs
  where coursesUsing lang = map fst . filter (elem lang . snd) $ courses

main = do
  print $ filterCourses [] progLanguages
-- []
  print $ filterCourses ["Python","C","C++"] progLanguages
-- [["CptS322","CptS355","CptS315","CptS451","CptS475"],["CptS121","CptS360","CptS411"],["CptS122","CptS223","CptS411"]]

我将任务分为两个步骤：首先，我制作了一个 function coursesUsing ，它列出了给定语言的课程。 然后我使用map将 function 应用于每种语言。

或者，使用列表理解：

filterCourses langs courses = [[course | (course, langs) <- courses, lang `elem` langs] | lang <- langs]