Joel Neely joel.neely at gmail.com
Fri Jan 13 19:14:56 UTC 2017

```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],,[7,8],[10,11],]

So close but no cigar.

On Fri, Jan 13, 2017 at 10:05 AM, Saqib Shamsi <shamsi.saqib at gmail.com>
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
>                                       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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>
>

--
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- Plato
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