On 10.02.2013 02:30, Nicolas Bock wrote:
Hi Aleksey,
could you show me how I would use ByteString? I can't get the script to
compile. It's complaining that:
No instance for (RegexContext
Regex Data.ByteString.ByteString
(AllTextSubmatches [] a0))
which is too cryptic for me. Is it not able to form a regular expression
with a ByteString argument? From the documentation of Text.Regex.Posix
it seems that it should be. Maybe it's because I am trying to "read
(r!!1) :: Double" which I am having issues with also. Is (r!!1) a
ByteString? And if so, how would I convert that to a Double?
It's error message from regex library you use. I can't say what exactly
it means, I never used it. But most likely it cannot work with bytestrings.
Most other languages rely on regexp as go to tool for parsing. In
haskell main parsing tools are parser combinators such as parsec[1] or
attoparsec[2]. Parsec is more generic and attoparsec is much faster.
In attachment program which uses attoparsec for parsing it's about
2times slower than C++ example posted in the thread.
[1] http://hackage.haskell.org/package/parsec
[2] http://hackage.haskell.org/package/attoparsec
{-# LANGUAGE OverloadedStrings #-}
import Control.Applicative
import Data.Histogram.Fill
import Data.Histogram (Histogram)
import Data.Attoparsec.Text.Lazy (parse,Result(..))
import Data.Attoparsec.Text hiding (parse,Done,Fail)
import qualified Data.Text.Lazy as T
import qualified Data.Text.Lazy.IO as T
import Prelude hiding (takeWhile)
hb :: HBuilder Double (Histogram LogBinD Int)
hb = forceInt -<< mkSimple (logBinDN 1e-8 10 10)
streamBS :: T.Text -> [Double]
streamBS bs
| T.null bs = []
| otherwise = case parse go bs of
Done rest x -> x : streamBS rest
Fail _ cxt e -> error $ e ++ " " ++ show cxt
where
num = decimal :: Parser Int
go = string "matrix("
*> num *> char ',' *> num *> char ')'
*> takeWhile (==' ') *> char '=' *> takeWhile (== ' ') *> double <* endOfLine
main :: IO ()
main = do
print . fillBuilder hb . streamBS =<< T.getContents
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