[英]How to parse a Tuple (String,Int) in Haskell using parsec
我想我已經設法將字符串解析為字符串並將字符串解析為 Int,但我還需要解析 (String,Int) 類型,因為 userRatings 是為了正確讀取 textFile 並且我正在使用 Parsec
這是與導入一起的解析
import Text.Parsec
( Parsec, ParseError, parse -- Types and parser
, between, noneOf, sepBy, many1 -- Combinators
, char, spaces, digit, newline -- Simple parsers
)
-- Parse a string to a string
stringLit :: Parsec String u String
stringLit = between (char '"') (char '"') $ many1 $ noneOf "\"\n"
-- Parse a string to a list of strings
listOfStrings :: Parsec String u [String]
listOfStrings = stringLit `sepBy` (char ',' >> spaces)
-- Parse a string to an int
intLit :: Parsec String u Int
intLit = fmap read $ many1 digit
-- Or `read <$> many1 digit` with Control.Applicative
film :: Parsec String u Film
film = do
-- alternatively `title <- stringLit <* newline` with Control.Applicative
title <- stringLit
newline
director <- stringLit
newline
year <- intLit
newline
userRatings <- listOfStrings
newline
return (title, director, year, userRatings)
您可以通過使用Applicative
或Monad
接口從現有解析器進行組合來完成此操作。 Parser
具有兩者的實例。
使用Applicative
:
stringIntTuple :: Parser (String, Int)
stringIntTuple = (,) <$> yourStringParser <*> yourIntParser
這與以下內容相同:
stringIntTuple :: Parser (String, Int)
stringIntTuple = liftA2 (,) yourStringParser yourIntParser
使用Monad
:
stringIntTuple :: Parser (String, Int)
stringIntTuple =
do
theString <- yourStringParser
theInt <- yourIntParser
return (theString, theInt)
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