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
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
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