[Haskell-beginners] Filtering lists
Stephen Tetley
stephen.tetley at gmail.com
Tue Feb 9 07:27:50 EST 2010
Hello
If you have a list of String for each record, use partition from
Data.List, e.g.:
import Data.List ( partition )
partitionBySize :: [[String]] -> ([[String]],[[String]])
partitionBySize = partition (\xs -> length xs < 10)
-- for demo-ing:
printShortsLongs :: ([[String]],[[String]]) -> IO ()
printShortsLongs (xs,ys) = do
putStrLn "Shorts:"
mapM_ print xs
putStrLn "Longs:"
mapM_ print ys
demo1 = printShortsLongs $ partitionBySize [
["1", "2", "3", "4", "5", "6", "7", "8", "9", "10" ]
, ["1", "2", "3", "4"]
, ["1", "2", "3", "4", "5", "6", "7", "8", "9", "10" ]
, ["1", "2", "3", "4", "5", "6", "7", "8", "9", "10" ]
, ["1", "2", "3", "4"]
]
On 9 February 2010 10:53, <legajid at free.fr> wrote:
> Hi,
> i get a .csv file with two different record formats (too bad).
> The first one, identified by a "A" in the second field, contains 4 fields.
> The second one, identified by a "B", contains 10 fields.
>
> In order to put them in lists, i use lines then break to get a list of lists.
> Then i want to filter this list, one for the A list, the other one for the B list.
> Unfortunately, the pattern for the A type contains 4 fields; when a type B list occurs (10 fields), i get a message concerning the pattern : Non-exhaustive patterns in lambda
>
> How can i separate the two record formats?
>
> Thanks,
> Didier
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