
You forgot the parenthesis. Parse error in pattern usually means a type in the input of one of your functions. Nub needs >elements that can be equal.
type has to be typo.
On Tue, Sep 28, 2010 at 11:44 AM, edgar klerks
Hi Martin,
You have some typos:
import Data.List removeDuplTuples :: (Eq a) => [(a,a)] -> [(a,a)]
removeDuplTuples [] = [] removeDuplTuples [b] = [b] -- using the syntactic sugar for single element in list removeDuplTuples (x:xs) = nub (if elem (snd x,fst x) xs then removeDuplTuples xs else [x] ++ removeDuplTuples xs) -------- You forgot the parenthesis. Parse error in pattern usually means a type in the input of one of your functions. Nub needs elements that can be equal.
Nub is quitte inefficient, if your elements can be ordered, there is a more efficient version. It is something like:
fmap head.group.sort $ [1,1,1,1,4,4,5,6,6,7,8,9] [1,4,5,6,7,8,9]
But I haven't test it thoroughly.
Greets,
Edgar
On Tue, Sep 28, 2010 at 11:33 AM, Martin Tomko
wrote: Hi all, I apologize for spamming this forum so frequently, but there is noone I can turn to around here... I have a list of (a,a) tuples, and am trying something like nub, but also matching for symmetrical tuples. I implemented it using the template from delete from Prelude. Seems like my typse signature has some troubles (Paarse error in pattern) but I am not sure where the problem is.
removeDuplTuples :: [(a,a)] -> [(a,a)] removeDuplTuples [] = [] removeDuplTuples [b] = [b]
-- using the syntactic sugar for single element in list removeDuplTuples x:xs = nub (if elem (snd x,fst x) xs then removeDuplTuples xs else [x] ++ removeDuplTuples xs)
I assume the problem lies in elem (snd x,fst x) xs but I am not sure how to rewrite it.
Thanks for all help, Martin _______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners