Question

我正在努力展示信息并将它们格式化为Haskell中的数据库。我正在使用的是...

type Title = String
type Actor = String
type Cast = [Actor]
type Year = Int
type Fan = String
type Fans = [Fan]
type Period = (Year, Year)
type Film = (Title, Cast, Year, Fans)
type Database = [Film]

testDatabase :: Database
testDatabase = [("Casino Royale", ["Daniel Craig", "Eva Green", "Judi Dench"], 2006,          ["Garry", "Dave", "Zoe", "Kevin", "Emma"]),
("Cowboys & Aliens", ["Harrison Ford", "Daniel Craig", "Olivia Wilde"], 2011, ["Bill", "Jo", "Garry", "Kevin", "Olga", "Liz"]),     
    ("Catch Me If You Can", ["Leonardo DiCaprio", "Tom Hanks"], 2002, ["Zoe", "Heidi", "Jo", "Emma", "Liz", "Sam", "Olga", "Kevin", "Tim"])

以上是示例数据库 - 最终目标是编写一个输出“displayAllFilms”的函数：

Casino Royale，Daniel Craig，Eva Green，Judi Dench，2006,5

牛仔＆amp; Aliens，Harrison Ford，Daniel Craig，Olivia Wilde，2011,6，

注意：制作方法很重要，每部影片都会显示在单独的线条和最后的粉丝数量上，而不是粉丝列表。

这是我尝试解决这个问题 -

displayAllFilms :: [Film] -> String -> String
displayAllFilms [] filmString = filmString
displayAllFilms ((Title cast year fans _):films) filmString = displayAllFilms films (filmString ++ title ++ "\n" ++ (show cast) ++ "\n" ++ (show year) ++ "\n")

但它似乎没有编译但我似乎无法找到我的代码有什么问题...是类型错误？

提前致谢！

Answer 1

有几个错误：

testDatabase :: Database
testDatabase = [("Casino Royale", ["Daniel Craig", "Eva Green", "Judi Dench"], 2006,          ["Garry", "Dave", "Zoe", "Kevin", "Emma"]),
    -- indentation should be consistent - this next line was too far left:
    ("Cowboys & Aliens", ["Harrison Ford", "Daniel Craig", "Olivia Wilde"], 2011, ["Bill", "Jo", "Garry", "Kevin", "Olga", "Liz"]),     
    ("Catch Me If You Can", ["Leonardo DiCaprio", "Tom Hanks"], 2002, ["Zoe", "Heidi", "Jo", "Emma", "Liz", "Sam", "Olga", "Kevin", "Tim"])]

并且您忘记了该列表末尾的小括号]。

displayAllFilms :: [Film] -> String -> String
displayAllFilms [] filmString = filmString
displayAllFilms ((title,cast,year,fans):films) filmString = displayAllFilms films (filmString ++ title ++ "\n" ++ (show cast) ++ "\n" ++ (show year) ++ "\n")

当您的意思是(Title cast year fans)时，您已(title,cast,year,fans)。

Haskell区分大小写，大写字母意味着Title是一个构造函数，但你的意思是它是一个变量。因为你的Film类型是一个元组，所以你需要这里的逗号作为元组。

正确输出

现在可以输出

putStrLn $ displayAllFilms testDatabase "=========="
==========Casino Royale
["Daniel Craig","Eva Green","Judi Dench"]
2006
Cowboys & Aliens
["Harrison Ford","Daniel Craig","Olivia Wilde"]
2011
Catch Me If You Can
["Leonardo DiCaprio","Tom Hanks"]
2002

现在这与您要求的输出不符，

Casino Royale, Daniel Craig, Eva Green, Judi Dench, 2006, 5

因为有新行而不是逗号和括号你不想要。

listStuff:: [String] -> String -> String
listStuff strings separator = concat [string++separator| string <-strings]

这里我使用list comprehension来制作一个带有分隔符的字符串列表，然后concat将它们连接在一起。最后你会得到一个备用的", "，但这没关系，因为我们需要一个演员和一年之间。

现在我们可以把它放在一起，使用length fans来显示粉丝的数量。

displayAllFilms' :: [Film] -> String -> String
displayAllFilms' [] filmString = filmString
displayAllFilms' ((title,cast,year,fans):films) filmString = 
       displayAllFilms' films (filmString ++ "\n" ++ title ++ ", " ++ listStuff cast ", " ++ (show year) ++ ", " ++ show (length fans))

测试为

>putStrLn $ displayAllFilms' testDatabase ""

Casino Royale, Daniel Craig, Eva Green, Judi Dench, 2006, 5
Cowboys & Aliens, Harrison Ford, Daniel Craig, Olivia Wilde, 2011, 6
Catch Me If You Can, Leonardo DiCaprio, Tom Hanks, 2002, 9

稍微清理代码

我认为displayAllFilms不需要递归 - 我们可以重用listStuff但是有一个库函数可以完成我们想要的更多工作。如果你在hoogle上搜索[String] -> String -> String，你会得到这些结果http://www.haskell.org/hoogle/?hoogle=[String]+-%3E+String+-%3E+String和第二个，{ {1}}很方便：

intercalate

你必须把

> intercalate "_" ["Hello","Mum,","how","are","you?"]
"Hello_Mum,_how_are_you?"

在您的代码顶部使用它。所以现在我们可以做到

import Data.List

因此，我们可以显示单个电影，然后使用它来显示电影列表。

commas :: [String] -> String
commas = intercalate ", "

我们可以写

showFilm :: Film -> String
showFilm (title,cast,year,fans) = commas [title, commas cast, show year, show (length fans)]

但是做得更好

showDatabase :: Database -> String
showDatabase fs = intercalate "\n" [showFilm f | f<-fs]

因为你只想在每部电影上使用showDatabase :: Database -> String showDatabase = unlines.map showFilm，然后showFilm，这只是unlines，但更容易看到。

intercalate "\n"

Haskell格式化和显示数据

1 个答案:

正确输出

稍微清理代码