Re: [Haskell-beginners] Beginners Digest, Vol 103, Issue 8

Hello Francesco, Sorry for the late reply. Yes, your solution is elegant and does what I wanted. This is what I did -- zipped is the list of tuples formed as you mentioned λ> groupBy (\x y -> (snd x) == (snd y)) zipped [[(1,0),(2,0),(3,0)],[(7,3),(8,3)],[(10,4),(11,4),(12,4)]] λ> [[fst tpl | tpl <- x] | x <- it] [[1,2,3],[7,8],[10,11,12]] This works perfectly fine. Thank you. Regards, Saqib Shamsi On 14 January 2017 at 00:27, wrote:

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Today's Topics:

1. Better Code (Saqib Shamsi) 2. Re: Better Code (Francesco Ariis) 3. Re: Better Code (Joel Neely) 4. Re: Better Code (Joel Neely)

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Message: 1 Date: Fri, 13 Jan 2017 21:35:54 +0530 From: Saqib Shamsi To: beginners@haskell.org Subject: [Haskell-beginners] Better Code Message-ID: Content-Type: text/plain; charset="utf-8"

Hi,

The problem that I wish to solve is to divide a (sored) list of integers into sublists such that each sublist contains numbers in consecutive sequence.

For example, *Input:* [1,2,3,7,8,10,11,12] *Output:* [[1,2,3],[7,8],[10,11,12]]

I have written the following code and it does the trick.

-- Take a list and divide it at first point of non-consecutive number encounter divide :: [Int] -> [Int] -> ([Int], [Int]) divide first [] = (first, []) divide first second = if (last first) /= firstSecond - 1 then (first, second) else divide (first ++ [firstSecond]) (tail second) where firstSecond = head second

-- Helper for breaking a list of numbers into consecutive sublists breakIntoConsecsHelper :: [Int] -> [[Int]] -> [[Int]] breakIntoConsecsHelper [] [[]] = [[]] breakIntoConsecsHelper lst ans = if two == [] then ans ++ [one] else ans ++ [one] ++ breakIntoConsecsHelper two [] where firstElem = head lst remaining = tail lst (one, two) = divide [firstElem] remaining

-- Break the list into sublists of consective numbers breakIntoConsecs :: [Int] -> [[Int]] breakIntoConsecs lst = breakIntoConsecsHelper lst [[]]

-- Take the tail of the result given by the function above to get the required list of lists.

However, I was wondering if there was a better way of doing this. Any help would be highly appreciated.

Thank you. Best Regards, Saqib Shamsi