问题
I'm parsing a file that looks like this:
Good
id123
^
Bad
id456
^
Middle
id789
Records are separated by ^\n
, and fields within that record
are separated simply by newlines
.
Reading this file and splitting, I end up with a list of lists that looks like this:
[["Good","id123",""],["Bad","id456",""],["Middle","id789",""]]
However, I fail to turn this into a list of Rec types.
Here's my code:
{-# 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
Produces
[["Good","id123",""],["Bad","id456",""],["Middle","id789",""]]
As expected. But trying to take this to the next step and turn these into Recs, with:
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_
I get:
<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
or
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))’
What am I missing to turn the [[Text]]
into [Rec]
?
回答1:
This works:
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
Produces:
Rec "Good" "id123" ""
Rec "Bad" "id456" ""
Rec "Middle" "id789" ""
The \anonymous
function can take a \[l,i,s,t]
.
来源:https://stackoverflow.com/questions/55643003/from-a-list-of-lists-to-a-list-of-data-records