Re: [Haskell-beginners] removing duplicate tuples (including symmetrical ones)

Hi Edgar, thanks for the help. I wanted to first test with nub, but I still cannot get it to work. I cannot see any missing parenthesis, but I retyped it this way to check: removeDuplTuples :: (Eq a) =>[(a,a)] -> [(a,a)] removeDuplTuples [] = [] removeDuplTuples [b] = [b] removeDuplTuples x:xs = nub(if ((snd x,fst x) `elem` xs) then (removeDuplTuples xs) else ([x] ++ removeDuplTuples xs)) still the same error What am I missing? Martin On 9/28/2010 11:45 AM, edgar klerks wrote:

...

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 mailto:edgar.klerks@gmail.com> wrote:

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 mailto:martin.tomko@geo.uzh.ch> 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 mailto:Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners