[Haskell-cafe] Copying Arrays

[Jefferson Heard]

Exactly.  Someone on the list gave me this example awhile back for
reading CSV files.  I can process a gigabyte (simple unpack and print
to dev/null for IO testing purposes) in about two and a half seconds
using this code.

import Data.ByteString.Lazy.Char8 as C

-- | Read a datafile and turn it into lists of columns
readDatafileAndTranspose name = do
    sheet <- (transpose . map (C.split '\t') . C.lines) `fmap` C.readFile name
    return $ foldl' go M.empty sheet
  where go m (x:xs) = M.insert (C.unpack x) xs m


2008/5/29 Adrian Neumann <aneumann at inf.fu-berlin.de>:
> Isn't fast IO what ByteStrings where invented for?
>
> Adrian
>
> Tom Harper schrieb:
>>
>> I'm trying to implement some file I/O where I can read in a file to an
>> array, but do so without having to know how much space I will need.
>> (Before you suggest it, lists are too slow/space consuming.)  I was
>> thinking that one possible solution is to use a strategy used in
>> dynamic arrays, where everytime you run out of space you double the
>> capacity of the array.  Trouble is, most arrays in Haskell would
>> require you to traverse the entire array and copy each individual
>> cell, which would make it worthless.
>>
>> Are there any low-level array types (MutableByteArray#, for example)
>> that support a memcpy-like function where I could copy the entire
>> array into the new one instead of copying it one value at a time?  Is
>> there another solution that I'm missing?
>>
>>
>
>
>
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