[Haskell-beginners] Haskell equivalent to Clojure's partition fn

[Timothy Washington]

While I can say A), what I really need is B)

*A)* > *take 5 $ chunksOf 10 [0..]*
[[0,1,2,3,4,5,6,7,8,9],[10,11,12,13,14,15,16,17,18,19],[20,21,22,23,24,25,26,27,28,29],[30,31,32,33,34,35,36,37,38,39],[40,41,42,43,44,45,46,47,48,49]]

*B)* > take 5 $ someFn 10 1 [0..]
[[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,10],[2,3,4,5,6,7,8,9,10,11],[3,4,5,6,7,8,9,10,11,12],[4,5,6,7,8,9,10,11,12,13]]


The music theory package indeed has a working partition function (source
here <http://rd.slavepianos.org/sw/hmt/Music/Theory/List.hs>). The
implementation simply *i)* takes `n` from the source list, *ii)* drops by
`m` then recurses.

segments :: Int -> Int -> [a] -> [[a]]
segments n m p =
    let q = take n p
        p' = drop m p
    in if length q /= n then [] else q : segments n m p'



But that's rather manual. So I played around with this using *chop*, and
came up with the *divvy* function. It does exactly what I need.

divvy :: Int -> Int -> [a] -> [[a]]
divvy n m lst =
  chop (\xs -> (take n xs , drop m xs)) lst


> *take 5 $ partitionClojure 10 1 [0..]*
[[0,1,2,3,4,5,6,7,8,9],[1,2,3,4,5,6,7,8,9,10],[2,3,4,5,6,7,8,9,10,11],[3,4,5,6,7,8,9,10,11,12],[4,5,6,7,8,9,10,11,12,13]]


Thanks guys. This helped a lot :)


Tim


On Sat, Jul 25, 2015 at 10:56 AM, Dan Serban <dserban01 at gmail.com> wrote:

> It looks like chunksOf will take you most of the way there. Here's my
> quick and dirty GHCi session output:
>
> λ> import Data.List.Split
> λ>
> λ> let clojurePartition n m = map (take n) $ chunksOf (n+m) [0..]
> λ>
> λ> take 3 $ clojurePartition 4 6
> [[0,1,2,3],[10,11,12,13],[20,21,22,23]]
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