I *do* know what Packed Decimal is; at my previous job, I actually had a
whole Haskell library for parsing them. The only immediate suggestion that
pops to mind is to use Int instead of Integer (Int uses regular 32- or
64-bit integers, Integer uses arbitrary precision integers). If you send me
a sample Packed Decimal file, I can test out your code and get a better feel
for it that way.
Good luck with those mainframes, they can be beasts sometimes. Have you had
to parse EBCDIC yet? *That* was fun, manually copying all those character
codes from some IBM web page... ;)
Michael
On Fri, Jun 5, 2009 at 2:31 AM, Bartosz Wójcik
Hi Folks,
I had to transform Packed Decimal file into csv format (does anybody here know what this Mainframe format is?). Because of the file size I could not do this on mainframe directly. So I've created simply program using ByteString. Generally I'm happy with my solution: pgm processes arroud 2MB/s on my pc, so my 3GB file was transformed in reasonable 30 min time.
My question is: how to do this faster?
{code} module Main where
import qualified Data.ByteString.Lazy as B
main = B.getContents >>= myPrint . myConcat . B.unpack
myConcat = myConcat' 0
myConcat' :: (Integral a) => Integer -> [a] -> [Integer] myConcat' _ [] = [] myConcat' acc (x:xs) = case x `mod` 16 of 12 -> (10*acc + (restOf . fromIntegral) x) : myConcat' 0 xs 13 -> ((-10)*acc + (restOf . fromIntegral) x) : myConcat' 0 xs 15 -> (10*acc + (restOf . fromIntegral) x) : myConcat' 0 xs 10 -> fail $ show acc 11 -> fail $ show acc 14 -> fail $ show acc v -> myConcat' (100*acc + (numberOf . fromIntegral) x) xs where restOf n = (n - 12) `div` 16 numberOf n = n - 6 * (n `div` 16)
myPrint [] = return () myPrint xs = mapM_ myShow (take 14 xs) >> putStrLn "" >> myPrint (drop 14 xs)
myShow x = (putStr . show) x >> putStr ";" {code}
I knew that csv output had to be 14 fields per line.
Best, Bartek
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