[英]Struggling with IO monad in Haskell
所以我有兩個文件,其內容如下:
File 1:
Tom 965432145
Bill 932121234
File 2:
Steve 923432323
Tom 933232323
我想將它們合並,並將結果輸出到名為“ out.txt”的文件中。 我編寫了此函數來處理重復項(當同一個名字出現多次時,它選擇將哪個數字輸入最終文件)。
該函數稱為選擇:
choosing :: [String] −> Int −> Int −> Int
choosing ("Name_of_person":_) num1 _ = num1
choosing _ num1 num2
| num2 ‘div‘ 100000000 == 2 = num2
| otherwise = num1
這是我的嘗試:
import System.IO
import Data.Char
choosing :: [String] −> Int −> Int −> Int
choosing name num1 _ = num1
choosing _ num1 num2
| num2 `div` 100000000 == 2 = num2
| otherwise = num1
main :: IO ()
main = do
in1 <- openFile "in1.txt" ReadMode
in2 <- openFile "in2.txt" ReadMode
out <- openFile "out.txt" WriteMode
processData in1 in2 out
hClose in1
hClose in2
hClose out
processData :: Handle -> Handle -> Handle -> IO ()
processData in1 in2 out =
do ineof <- hIsEOF in1
ineof2 <- h2IsEOF in2
if ineof && ineof2
then return ()
else do inpStr <- hGetLine in1
inp2Str <- h2GetLine in2
num1Int <- num1GetNumber in1
num2Int <- num2GetNumber in2
if inpStr = inp2Str
then PutStrLn out (impStr choosing inpStr num1Int num2Int )
else PutStrLn out (inpStr num1Int)
PutStrLn out (inp2Str num2Int)
processData in1 in2 out
但是,這對我來說很有意義,它無法編譯,並且在嘗試調試此功能后一段時間,我現在開始認為這里存在一些嚴重錯誤,因此,非常感謝您的幫助。
這是我首先嘗試的一些簡單的嘗試:
import System.IO
import Data.Char
choosing name num1 _ = num1
choosing _ num1 num2
| num2 `div` 100000000 == 2 = num2
| otherwise = num1
main :: IO ()
main = do
in1 <- openFile "in1.rtf" ReadMode
in2 <- openFile "in2.rtf" ReadMode
out <- openFile "out.rtf" WriteMode
mainloop in1 out
mainloop in2 out
hClose in1
hClose in2
hClose out
mainloop :: Handle -> Handle -> IO ()
mainloop _ out =
do ineof <- hIsEOF in
if ineof
then return ()
else do inpStr <- hGetLine in
hPutStrLn out (inpStr)
mainloop in out
但這也不起作用...
更新:
所以基本上我一直在嘗試解決我的問題,並獲得了所有提示,我設法做到了:
import System.IO
import Data.Char
- Main function to run the program
main = do
entries1 <- fmap parseEntries $ readFile "in1.txt"
entries2 <- fmap parseEntries $ readFile "in2.txt"
writeFile "out.txt" $ serializeEntries $ mergeEntries entries1 entries2
- Function to deal with duplicates
choosing name num1 _ = num1
choosing _ num1 num2
| num2 `div` 100000000 == 2 = num2
| otherwise = num1
- Function to read a line from a file into a tuple
Now i need help making this function 'cover' the whole file, and not just one line of it.
parseLine :: String -> (String, Int)
parseLine xs = (\(n:i:_) -> (n, read i)) (words xs)
- A function that receives entries, merges them into a single string so that it can be writen to a file.
import Data.Char
tupleToString :: (Int, Char) -> [Char]
tupleToString x = (intToDigit.fst) x:(snd x):[]
tuplesToStrings [] = []
tuplesToStrings (x:xs) = tupleToString x : tuplesToStrings xs
tuplesToString xs = (concat . tuplesToStrings) xs
我認為問題在於您的思考勢在必行。 在Haskell中,您通常將解決方案分成幾個小塊,每個塊只能做一件事。 推理一個小塊要容易得多,也可以在其他部分重用該塊。 例如,這是我如何分解此問題的代碼:
parseEntries :: String -> [(String, Int)]
接收文件內容並解析條目的函數。 如果是in1.txt
的內容,它將返回[("Tom", 965432145), ("Bill", 932121234)]
mergeEntries :: [(String, Int)] -> [(String, Int)] -> [(String, Int)]
從兩個文件接收條目並將其合並的函數。
serializeEntries :: [(String, Int)] -> String
接收條目的函數,將它們合並為單個字符串,以便可以將其寫入文件。
定義了這些功能后, main
變得如此簡單:
main = do
entries1 <- fmap parseEntries $ readFile "in1.txt"
entries2 <- fmap parseEntries $ readFile "in2.txt"
writeFile "out.txt" $ serializeEntries $ mergeEntries entries1 entries2
回答您的更新代碼:
既然您具有解析行的功能,則parseEntries
變得很簡單。 使用lines
函數將內容按行分割,然后將 parseLine
映射到每一行。
tuplesToStrings
可以寫得更簡單,因為tuplesToStrings = map tupleToString
我看不到tuplesToString
將如何幫助您。 它的類型不匹配的返回類型parseLine
( parseLine
返回的列表(String, Int)
而tuplesToString
預計的列表(Int, Char)
)。 而且它甚至不會在單詞之間或行之間插入空格。 這是serializeEntries
的可能實現(使用Text.Printf模塊):
serializeEntries entries = concatMap (uncurry $ printf "%s %d\\n") entries
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