Here's one that does what you want, doesn't require the list to be sorted, and groups together consecutive and equal terms: groupConsecutive :: (Enum a,Eq a) => [a] -> [[a]] groupConsecutive = foldr go [] where go x ls@(hd@(y:_):yss) | x == y || x == pred y = (x:hd):yss | otherwise = [x]:ls go x [] = [[x]] go x ([]:yss) = [x]:yss Then
groupConsecutive [1,2,3,7,8,10,11,12] [[1,2,3],[7,8],[10,11,12]]
groupConsecutive [1,2,2,3,2,3] [[1,2,2,3],[2,3]]
and
groupConsecutive "bookkeeper understudy" ["b","oo","kk","ee","p","e","r"," ","u","n","de","rstu","d","y"]
The third case of go will never be reached. If you use a type that is also an instance of Bounded, and if your list contains the minimum element of the type, you'll get a runtime error on the use of pred. For example:
groupConsecutive [True,False,True] *** Exception: Prelude.Enum.Bool.pred: bad argument
Graham On 13-Jan-2017 11:05 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