将字符串转换为自定义类型的Haskell

Question

你好我想这个字符串

Blade Runner,Ridley Scott,1982,Amy,5,Bill,8,Ian,7,Kevin,9,Emma,4,Sam,7,Megan,4​ 

转换为电影类型

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

从包含25部电影

这个文本文件是我试图做的

maybeReadTup :: String ->(String, Int) 
maybeReadTup s = do 
    [(n, [c])] <- return $ reads s 
    return [(n, [c])] 

    parseLines :: [String] -> Film 
    parseLines list 
      | isInt(list !! 3) = (list !! 0,(list !! 1), read (list !! 2), maybeReadTup [ (list!!1,read (list !! 2))]) 

isInt :: String ->Bool 
isInt[] = True 
isInt (x:xs) 
    | isNumber x = True && isInt xs 
    | otherwise = False 

parseChars :: String -> String -> [String] 
parseChars [] _ = [] 
parseChars (x:xs) stringCount 
    | x == ',' = [stringCount] ++ parseChars xs "" 
    | otherwise = (parseChars xs (stringCount ++ [x])) 

parseAll :: [String] -> [Film] 
parseAll [] = [] 
parseAll (x:xs) = parseLines (parseChars x "") : (parseAll xs) 

但我得到错误的类型可以有人请帮我解析这个UserRatings元组类型[(String,Int)]？你能帮我理解parseLines的工作原理吗？我在Haskell

Answer 1

这里新是一个使用Text.Parsec一个解决方案：

import   Text.Parsec 
import   Text.Parsec.String 

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

str :: Parser String 
str = many1 (noneOf ",") 

int :: Parser Int 
int = read <$> many1 digit 

tup :: Parser UserRatings 
tup = do user <- str 
     _ <- oneOf "," 
     rating <- int 
     return (user, rating) 

parser :: Parser Film 
parser = do title <- str 
      _ <- oneOf "," 
      director <- str 
      _ <- oneOf "," 
      year <- int 
      _ <- oneOf "," 
      ratings <- sepBy tup (oneOf ",") 
      eof 
      return (title, director, year, ratings) 

testString :: String 
testString = "Blade Runner,Ridley Scott,1982,Amy,5,Bill,8,Ian,7,Kevin,9,Emma,4,Sam,7,Megan,4" 

main :: IO() 
main = print $ runParser parser() "testString" testString