来自升序的连续子列表

Question

given

xs = [1,2,3,4,6,7,9,10,11]

I am aiming to return

[[1,2,3,4],[6,7],[9,10,11]]

I thought I could do:

groupBy (\\xy -> succ x == y) xs

but this returns:

[[1,2],[3,4],[6,7],[9,10],[11]]

a little bit of searching returned the following from the Haskell Data.List suggestion page .

groupBy                 :: (a -> a -> Bool) -> [a] -> [[a]]
 groupBy rel []          =  []
 groupBy rel (x:xs)      =  (x:ys) : groupBy rel zs
   where (ys,zs) = groupByAux x xs
         groupByAux x0 (x:xs) | rel x0 x = (x:ys, zs)
           where (ys,zs) = groupByAux x xs
         groupByAux y xs = ([], xs)

One of the examples they give is exacly what I am looking for:

groupBy (\a b -> a+1 == b) [1,2,3,4,6]
[[1,2,3,4],[6]]

So My question... Is there another approach to this, as opposed to re-defining groupBy as it seems a little dramatic?

EDIT...

I have decided to implement it as follows:

pattern :: (Enum a, Eq a) => (a -> a) -> [a] -> [[a]]
pattern f = foldr g []
  where g a [] = [[a]]
        g a xs | f a == head (head xs) = (a : head xs): tail xs
               | otherwise = [a]:xs

which allows for such things:

*Main Map> pattern succ "thisabcdeisxyz"
["t","hi","s","abcde","i","s","xyz"]
*Main Map> pattern (+ 3) [3,6,9,12,1,2,3,2,5,8,23,24,25]
[[3,6,9,12],[1],[2],[3],[2,5,8],[23],[24],[25]]

or to function exactly like group -- not that there is any reason:

*Main Map> let xs = [1,1,1,2,3,4,5,6,6,6,5]
*Main Map> group xs == pattern id xs
True

Answer 1

There are many ways to do that. One way can be using foldr

f = foldr g []
  where g a [] = [[a]]
        g a xs@(x:xs') | a+1 == head x = (a : x): xs'
                       | otherwise = [a]:xs

Now trying this in action

*Main> f [1,2,3,4,6,7,9,10,11]
[[1,2,3,4],[6,7],[9,10,11]]

Answer 2

If xs is strictly increasing then

 myGrouping = map (map snd) . groupBy (\(u, v) (x, y) -> u - v == x - y) . zip [0..]

solve your problem.

Prelude> myGrouping [1,2,3,4,6,7,9,10,11]
[[1,2,3,4],[6,7],[9,10,11]]