哈斯克爾平均誤差

我只需要一些幫助我的下面的代碼。我試圖給出一個給定導演的電影的網站評級的平均值。哈斯克爾平均誤差

但是，我運行下面的代碼似乎得到一個錯誤。錯誤是：無效的類型簽名：DirectorAverage :: String - > Float 應該是窗體::

我似乎無法發現代碼有什麼問題，因爲我相信我已經正確編寫了代碼，感謝它，如果有人可以幫助解決這個錯誤！

謝謝！

import Data.List 
import Text.Printf 

type Rating = (String, Int) 
type Title = String 
type Director = String 
type Year = Int 
type Film = (Title, Director, Year,[Rating]) 


testDatabase :: [Film] 
testDatabase = [("Blade Runner","Ridley Scott",1982,[("Amy",6), ("Bill",9), ("Ian",7), ("Kevin",9), ("Emma",4), ("Sam",5), ("Megan",4)]), 
       ("The Fly","David Cronenberg",1986,[("Megan",4), ("Fred",7), ("Chris",5), ("Ian",0), ("Amy",5)]), 
       ("Psycho","Alfred Hitchcock",1960,[("Bill",4), ("Jo",4), ("Garry",8), ("Kevin",7), ("Olga",8), ("Liz",10), ("Ian",9)]), 
       ("Body Of Lies","Ridley Scott",2008,[("Sam",3), ("Neal",7), ("Kevin",2), ("Chris",5), ("Olga",6)]), 
       ("Avatar","James Cameron",2009,[("Olga",2), ("Wally",8), ("Megan",9), ("Tim",5), ("Zoe",8), ("Emma",3)]), 
       ("Titanic","James Cameron",1997,[("Zoe",7), ("Amy",2), ("Emma",5), ("Heidi",3), ("Jo",8), ("Megan",5), ("Olga",7), ("Tim",10)]), 
       ("The Departed","Martin Scorsese",2006,[("Heidi",2), ("Jo",8), ("Megan",5), ("Tim",2), ("Fred",5)]), 
       ("Aliens","Ridley Scott",1986,[("Fred",8), ("Dave",6), ("Amy",10), ("Bill",7), ("Wally",2), ("Zoe",5)]), 
       ("Prometheus","Ridley Scott",2012,[("Garry",3), ("Chris",4), ("Emma",5), ("Bill",1), ("Dave",3)]), 
       ("E.T. The Extra-Terrestrial","Steven Spielberg",1982,[("Ian",7), ("Amy",2), ("Emma",7), ("Sam",8), ("Wally",5), ("Zoe",6)]), 
       ("The Birds","Alfred Hitchcock",1963,[("Garry",7), ("Kevin",9), ("Olga",4), ("Tim",7), ("Wally",3)]), 
       ("Goodfellas","Martin Scorsese",1990,[("Emma",7), ("Sam",9), ("Wally",5), ("Dave",3)]), 
       ("The Shawshank Redemption","Frank Darabont",1994,[("Jo",8), ("Sam",10), ("Zoe",3), ("Dave",7), ("Emma",3), ("Garry",10), ("Kevin",7)]), 
       ("Gladiator","Ridley Scott",2000,[("Garry",7), ("Ian",4), ("Neal",6), ("Wally",3), ("Emma",4)]), 
       ("The Green Mile","Frank Darabont",1999,[("Sam",3), ("Zoe",4), ("Dave",8), ("Wally",5), ("Jo",5)]), 
       ("True Lies","James Cameron",1994,[("Dave",3), ("Kevin",4), ("Jo",0)]), 
       ("Minority Report","Steven Spielberg",2002,[("Dave",5), ("Garry",6), ("Megan",2), ("Sam",7), ("Wally",8)]), 
       ("The Wolf of Wall Street","Martin Scorsese",2013,[("Dave",6), ("Garry",6), ("Megan",0), ("Sam",4)]), 
       ("War Horse","Steven Spielberg",2011,[("Dave",6), ("Garry",6), ("Megan",3), ("Sam",7), ("Wally",8), ("Zoe",8)]), 
       ("Lincoln","Steven Spielberg",2012,[("Ian",3), ("Sam",7), ("Wally",3), ("Zoe",4), ("Liz",7), ("Megan",4)]), 
       ("Vertigo","Alfred Hitchcock",1958,[("Bill",7), ("Emma",5), ("Zoe",9), ("Olga",6), ("Tim",10)]), 
       ("The Terminal","Steven Spielberg",2004,[("Olga",3), ("Heidi",8), ("Bill",2), ("Sam",6), ("Garry",8)]), 
       ("Jaws","Steven Spielberg",1975,[("Fred",3), ("Garry",0), ("Jo",3), ("Neal",9), ("Emma",7)]), 
       ("Hugo","Martin Scorsese",2011,[("Sam",4), ("Wally",3), ("Zoe",4), ("Liz",7)])]   


--Some functions needed later on: 
averageFilmRating :: [(String,Int)] -> Float 
averageFilmRating ratings 
    = (fromIntegral(sum $ map snd ratings))/(fromIntegral(length ratings)) 

--Formats the films for decimal, gives average rating of films instead of all users ratings. 
formatFilm :: Film -> String 
formatFilm (title, director, year, rating) 
    = printf "%s by %s. Year: %d, Average Rating: %.1f" (title) (director) (year) (averageFilmRating rating)  


displayByDirector :: String -> [Film] 
displayByDirector name 
    = filter(\(_,director,_,_) -> director == name) testDatabase 

DirectorAverage :: String -> Float 
DirectorAverage dir 
    = averageFilmRating [ (user,rating) | (title, director, year, [(user, rating)]) <- displayByDirector dir]

來源

2014-03-12 user3411002

的情況是在Haskell用來區分類型的名字，以大寫字母開始，從價值觀的名字，以小寫字母開頭。（請注意，函數是Haskell中的一個值。）

因此DirectorAverage是函數的無效名稱。按照慣例，你的函數應該命名爲directorAverage：

directorAverage :: String -> Float 
directorAverage dir 
    = averageFilmRating [ (user,rating) | (title, director, year, [(user, rating)]) <- displayByDirector dir]

來源

2014-03-12 14:36:25 svk

即便如此，修復錯誤，我不知道爲什麼我犯了一個愚蠢的錯誤一樣，我的成績都上來爲NaN的，我從來沒有之前沒有見過這個太確定爲什麼了。幫幫我？ – user3411002

@ user3411002您在'directorAverage'中的[[（用戶，評級）]'的模式匹配不正確，這意味着它只適用於擁有1等級電影的導演。你真的想要做一些像[評分| （title，director，year，ratings）< - displayByDirector dir，rating < - ratings]'，或者更簡單的'concat [ratings | （_，_，_，ratings）< - displayByDirecor dir]'。 – bheklilr

謝謝，我沒有閱讀過多的concat，但現在將閱讀並應用於此。感謝您的解決方案！現在運作良好！ – user3411002

哈斯克爾平均誤差

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