Re: [Haskell-beginners] Better Code

Hi Saqib, perhaps something like this? divide :: [Int] -> [[Int]] divide [] = [] divide (x:xs) = divideAux xs [] [x] x divideAux :: [Int] -> [[Int]] -> [Int] -> Int -> [[Int]] divideAux [] result current max = result ++ [current] divideAux (x:xs) result current max = if x - max > 1 then divideAux xs (result ++ [current]) [x] x else divideAux xs result (current ++ [x]) x This can of course be generalised to other types (as per Joel's requirement) just by replacing the check on line 7, i.e., 'x - max > 1'. Happy to explain/talk about the code if you'd like me to :-) Cheers, Carlo On Fri, Jan 13, 2017 at 7:14 PM, Joel Neely wrote:

Had a chance to chat with ghci, so earlier conjecture not confirmed:

Prelude Data.List> groupBy (\x y -> x == y-1) [1,2,3,7,8,10,11,12]

[[1,2],[3],[7,8],[10,11],[12]]

So close but no cigar.

On Fri, Jan 13, 2017 at 10:05 AM, Saqib Shamsi wrote:

...

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

_______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

-- Beauty of style and harmony and grace and good rhythm depend on simplicity. - Plato

_______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

-- Carlo Matteo Scalzo Mobile: +44 (0) 7934 583582 E-mail: cmscalzo@gmail.com