递归元组函数在Haskell中的困难

我有一个小程序，可以读取文件并将数据处理为自定义数据类型。正在读取的文件包含如下所示的数据行：

cat    10    20    dog
hamster    12    2    wombat
monkey    1    9    zebra
lion    30    60    rhino
...

main :: IO ()
    main = do
    contents <- readFile "myfile.xyz"
    let process = clean contents
    let f = processFile process
    print f

clean :: String -> [[String]]
clean x = Prelude.map words $ lines x

processFile :: [[String]] -> [(XyzData)]
processFile [[a,b,c,d]] = [(XyzData a (read b :: Int) (read c :: Int) d)]
processFile (x:xs) = processFile xs 

data XyzData = XyzData { animal1 :: String, 
                         number1 :: Int,
                         number2 :: Int,
                         animal2 :: String
                       } deriving (Show)

我处理文件的程序如下所示：

cat    10    20    dog
hamster    12    2    wombat
monkey    1    9    zebra
lion    30    60    rhino
...

main :: IO ()
    main = do
    contents <- readFile "myfile.xyz"
    let process = clean contents
    let f = processFile process
    print f

clean :: String -> [[String]]
clean x = Prelude.map words $ lines x

processFile :: [[String]] -> [(XyzData)]
processFile [[a,b,c,d]] = [(XyzData a (read b :: Int) (read c :: Int) d)]
processFile (x:xs) = processFile xs 

data XyzData = XyzData { animal1 :: String, 
                         number1 :: Int,
                         number2 :: Int,
                         animal2 :: String
                       } deriving (Show)

谢谢你抽出时间

你可能是有意的

processFile :: [[String]] -> [(XyzData)]
processFile ([a,b,c,d]:xs) = (XyzData a (read b :: Int) (read c :: Int) d) : processFile xs
processFile (x:xs) = processFile xs 
processFile [] = []

---

您的第一个模式

[[a，b，c，d]]

只匹配具有一个元素的列表，即具有四个元素的列表。

非常感谢您的帮助。我明白我做错了什么。您的解决方案工作正常。另外，我认为解决方案中的第三行（（

processFile（x:xs）=processFile xs

）是多余的？它不是。他的第三行确保所有（内部）列表与

[a，b，c，d]

不匹配（即没有四个元素）被忽略，处理继续到文件的下一行。谢谢你的解释Riccardo，我现在明白了那行的目的。汉普斯特是什么？某种朗登？现在更正，谢谢你的回答！