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 <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 <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
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