Re: [Haskell-beginners] Beginners Digest, Vol 45, Issue 35

Or, with a one-liner (inefficient, though): unique xs = [x | x <- xs, length (filter (== x) xs) == 1] L. On Wed, Mar 28, 2012 at 9:38 AM, wrote:

Indeed the second snipper contains quite an obvious mistake. Thanks for noticing!

It doesn't seem to me it utilises a lambda expression though? You mean the '.' operator for chaining function? If that's it, it could be rewritten

unique :: [Integer] -> [Integer] unique [] = [] unique (x:xs) | elem x xs = unique (filter (/= x) xs)

| otherwise = x : unique xs

----- Original Message -----

From: Ramesh Kumar

Sent: 03/28/12 10:14 AM

To: franco00@gmx.com, beginners@haskell.org

Subject: Re: [Haskell-beginners] Beginners Digest, Vol 45, Issue 35

Thanks Franco, Your (first) solution is the only one which has worked so far although it utilizes a lambda expression. The problem is indeed tricky.

------------------------------ *From:* "franco00@gmx.com" *To:* beginners@haskell.org *Sent:* Wednesday, March 28, 2012 3:39 PM *Subject:* Re: [Haskell-beginners] Beginners Digest, Vol 45, Issue 35

gah sorry I obviously meant to reply to the "Unique integers in a list" message

----- Original Message ----- From: franco00@gmx.com Sent: 03/28/12 09:36 AM To: beginners@haskell.org Subject: Re: Beginners Digest, Vol 45, Issue 35

unique :: [Integer] -> [Integer] unique [] = [] unique (x:xs) | elem x xs = (unique . filter (/= x)) xs | otherwise = x : unique xs

-- This is a simpler to read version (albeit inefficient?) unique :: [Integer] -> [Integer] unique [] = [] unique (x:xs) | elem x xs = unique xs | otherwise = x : unique xs

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