Question

我正在解析一个看起来像这样的文件：

Good
id123
^
Bad
id456
^
Middle
id789

记录由^\n分隔，并且该记录中的字段只需用newlines分隔即可。

读取此文件并拆分后，我得到一个看起来像这样的列表列表：

[["Good","id123",""],["Bad","id456",""],["Middle","id789",""]]

但是，我无法将其转换为Rec类型列表。

这是我的代码：

{-# LANGUAGE DeriveGeneric,  OverloadedStrings #-}
import Data.Text as T
import Data.Text.IO as T

data Rec = Rec Text Text Text deriving Show  -- simple

main :: IO ()
main = do
    contents <- T.readFile "./dat.txf" 
    let seps = splitOn "^\n" contents
    let recs = fmap (splitOn "\n") seps
    print recs

main

生产

[["Good","id123",""],["Bad","id456",""],["Middle","id789",""]]

符合预期。但是，尝试将其下一步，并将其转换为Recs，使用：

main_ :: IO ()
main_ = do
    contents <- T.readFile "./dat.txf" 
    let seps = splitOn "^\n" contents
    let recs = fmap (splitOn "\n") seps
    print recs
    print $ fmap (\(x, y, z) -> Rec x y z) recs
    -- print $ fmap (\r -> Rec r) recs
    -- let m = fmap (\x -> Rec [(x,x,x)]) recs 
    -- print m
    -- print $ fmap (\x -> Rec t3 x) recs
    -- where t3 [x,y,z] = (x,y,z)

main_

我得到：

<interactive>:7:44: error:
    • Couldn't match type ‘[Text]’ with ‘(Text, Text, Text)’
      Expected type: [(Text, Text, Text)]
        Actual type: [[Text]]
    • In the second argument of ‘fmap’, namely ‘recs’
      In the second argument of ‘($)’, namely ‘fmap (\ (x, y, z) -> Rec x y z) recs’
      In a stmt of a 'do' block: print $ fmap (\ (x, y, z) -> Rec x y z) recs

或

main__ :: IO ()
main__ = do
    contents <- T.readFile "./dat.txf" 
    let seps = splitOn "^\n" contents
    let recs = fmap (splitOn "\n") seps
    print recs
    print $ fmap (\x -> Rec (f x)) recs
                            where f [a,b,c] = (a,b,c)    
main__

<interactive>:7:30: error:
    • Couldn't match expected type ‘Text’ with actual type ‘(Text, Text, Text)’
    • In the first argument of ‘Rec’, namely ‘(f x)’
      In the expression: Rec (f x)
      In the first argument of ‘fmap’, namely ‘(\ x -> Rec (f x))’

将[[Text]]变成[Rec]时我缺少什么？

Answer 1

这有效：

main :: IO ()
main = do
    contents <- T.readFile "./dat.txf" 
    let seps = splitOn "^\n" contents
    let recs = fmap (splitOn "\n") seps
    -- print $ fmap (\[x, y, z] -> Rec x y z) recs
    let tada = fmap (\[x, y, z] -> Rec x y z) recs
    mapM_ print tada

main

产生：

Rec "Good" "id123" ""
Rec "Bad" "id456" ""
Rec "Middle" "id789" ""

\anonymous函数可以使用\[l,i,s,t]。

从列表列表到数据记录列表

1 个答案: