
On 10/11/2010 07:20 PM, Thomas Miedema wrote:
Hi,
replace let fl = getAllLengths nums
by fl <- getAllLengths nums
, since getAllLengths returns a monadic action.
The author of the following book is much better at explaining why this is so than I am: http://learnonlineyouahaskell.com/ http://learnyouahaskell.com/. May I suggest you read it cover to cover, it's really really good. It is probably the best way to learn Haskell at the moment, together with trying out code snippets like you're doing right now.
Regards, Thomas
On Mon, Oct 11, 2010 at 6:06 PM, Lorenzo Isella
mailto:lorenzo.isella@gmail.com> wrote: Thanks a lot Daniel, but I am a bit lost (up to not long ago I did not even know the existence of a control monad...and some unstructured reading did not help). Some online research about mapM and fmap led me here http://en.wikibooks.org/wiki/Haskell/Category_theory and I think I am a bit astray at this point ;-)
Why does my "simple" snippet below raise a number of errors?
Cheers
Lorenzo
import Data.Ord
import Data.List
main :: IO ()
main = do
let nums=[1,2]
let fl = getAllLengths nums
putStrLn "fl is, " print fl
filename :: Int -> FilePath filename i = "file" ++ show i ++ ".dat"
fileLength :: FilePath -> IO Int fileLength file = fmap length (readFile file)
getAllLengths :: [Int] -> IO [Int] getAllLengths nums = mapM (fileLength . filename) nums
On 10/11/2010 05:21 PM, Daniel Fischer wrote:
On Monday 11 October 2010 16:56:58, Lorenzo Isella wrote:
Dear All, Another I/O question. Let us say that you are given a list of files file1.dat, file2.dat...file10.dat and so on (i.e. every file is indexed by a number and every file is a single column where every entry is a string without spaces). In the snippet below I read file1.dat, convert it to a list and then print out its length. Now, how can I iterate the process on file1.dat, file2.dat and file3.dat and store the lengths in a list?
fileLength :: FilePath -> IO Int fileLength file = fmap length (readFile file)
filename :: Int -> FilePath filename i = "file" ++ show i ++ ".dat"
getAllLengths :: [Int] -> IO [Int] getAllLengths nums = mapM (fileLength . filename) nums
If you want something other than the character count, instead of fileLength use e.g.
countLines :: FilePath -> IO Int countLines file = fmap (length . lines) (readFile file)
or whatever you're interested in.
Another nice thing is often forM (from Control.Monad)
forM nums $ \i -> do let filename = "file" ++ show i ++ ".dat" contents<- readFile filename let result = function contents doSomethingOrNot return result
I would like to map the file reading and following operations on the list [1,2.3], but that is giving me a headache. It is relatively easy to create the file name
filename="file"++(show i)++".dat" , for i=1,2,3
but it the the iteration part that is giving me troubles. Any suggestion is appreciated. Cheers
Lorenzo
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Thanks Thomas. Yep, I do need some extra reading unfortunately. One question: if I was to apply a function on many files file1, file2...using e.g. Python, this would be my pipeline read file1 do stuff on file 1 read file2 do stuff on file 2 ...... Now, due to the laziness of haskell, can I here resort to this approach read file1, file2... into a single list map (do-my-stuff) on list As far as I understand, this should not result e.g. into a huge RAM consumptions since files are read and processed only when needed (hence one at the time). Am I on the right track? Cheers Lorenzo