[Haskell-cafe] function unique
gregorypropf at yahoo.com
Tue Jul 10 23:19:44 EDT 2007
Well there's this approach. Granted you need to split the input list into *both* args of the fold (i.e. the "input list" is really [1,4,5,3] but you can get this with head and tail). I'm just learning about the fold family myself. - Greg
Prelude> foldl (\a b -> if (any (\x -> x == b) a) then a else b:a)  [4,5,3,3,4]
----- Original Message ----
From: Alexteslin <alexteslin at yahoo.co.uk>
To: haskell-cafe at haskell.org
Sent: Tuesday, July 10, 2007 1:40:56 PM
Subject: [Haskell-cafe] function unique
Hi, i am a beginner to Haskell and i have a beginner's question to ask.
An exercise asks to define function unique :: [Int] -> [Int], which outputs
a list with only elements that are unique to the input list (that appears no
more than once). I defined a function with list comprehension which works
but trying to implement with pattern matching and primitive recursion with
lists and doesn't work.
unique :: [Int] -> [Int]
unique xs = [x | x <- xs, elemNum2 x xs == 1]
elemNum2 :: Int -> [Int] -> Int
elemNum2 el xs = length [x| x <- xs, x == el]
//This doesn't work, I know because the list shrinks and produces wrong
result but can not get a right //thinking
unique2 :: [Int] -> [Int]
unique2  = 
|elemNum2 x xs == 1 = x:unique2 xs
|otherwise = unique2 xs
Any help to a right direction would be very appreciated, thanks.
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