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 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]] _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

Hey Jacek, Thanks very much for this. I took a look around, and ended up using the *chop* function from that package. So there looks to be a lot of goodies in that package. Cheers Tim On Sun, Jul 26, 2015 at 2:01 PM, Jacek Dudek wrote:

Hi Timothy,

You might want to check out the split package. Here's the link: http://hackage.haskell.org/package/split

On 7/25/15, Timothy Washington wrote:

...

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

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]]