Re: [Haskell-cafe] function unique

Thans Gregory, But I sticked to the original solution because the exercise was about primitive recursion and pattern matching over lists. Gregory Propf wrote:

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) [1] [4,5,3,3,4] [3,5,4,1]

----- Original Message ---- From: Alexteslin To: haskell-cafe@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 [] = [] unique2 (x:xs) |elemNum2 x xs == 1 = x:unique2 xs |otherwise = unique2 xs

Any help to a right direction would be very appreciated, thanks. -- View this message in context: http://www.nabble.com/function-unique-tf4058328.html#a11528933 Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.

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