Hello, I try to achieve this but I can not find a convenient (elegant solution) let l =[1,2, 3, 4, 6, 7, 9, 10] I want this [[1, 2, 3, 4][6, 7],[9, 10]] In fact I want to split a list with all consecutive series. I imagine that it existe a one-liner for this but I did not find it :)) Cheers and thanks for your help Frederic
On Mon, Mar 26, 2018 at 05:52:58PM +0000, PICCA Frederic-Emmanuel wrote:
Hello, I try to achieve this but I can not find a convenient (elegant solution)
let l =[1,2, 3, 4, 6, 7, 9, 10]
I want this
[[1, 2, 3, 4][6, 7],[9, 10]]
There's the usual zip trick λ> l = [1,2, 3, 4, 6, 7, 9, 10] λ> zipWith (-) l [1..] [0,0,0,0,1,1,2,2] λ> zip l it [(1,0),(2,0),(3,0),(4,0),(6,1),(7,1),(9,2),(10,2)] Now you can groupBy on the second element. I am not sure there is a shorter method!
On Mon, Mar 26, 2018 at 08:11:44PM +0200, Francesco Ariis wrote:
There's the usual zip trick
λ> l = [1,2, 3, 4, 6, 7, 9, 10] λ> zipWith (-) l [1..] [0,0,0,0,1,1,2,2] λ> zip l it [(1,0),(2,0),(3,0),(4,0),(6,1),(7,1),(9,2),(10,2)]
To be more explicit: λ> :m +Data.List (groupBy) λ> :m +Data.Function (on) λ> l = [1,2, 3, 4, 6, 7, 9, 10] λ> map (map snd) $ groupBy ((==) `on` snd) $ zip l (zipWith (-) l [1..]) [[0,0,0,0],[1,1],[2,2]] Maybe with some lens or with a foldr you can end up with something shorter than this (69 characters).
To be more explicit:
λ> :m +Data.List (groupBy) λ> :m +Data.Function (on) λ> l = [1,2, 3, 4, 6, 7, 9, 10] λ> map (map snd) $ groupBy ((==) `on` snd) $ zip l (zipWith (-) l [1..]) [[0,0,0,0],[1,1],[2,2]]
I imagine that you mean fst instead of snd in order to produce the right output :) How can I check if this code produce list copies and is efficient with big lists ? thanks Frederic
On Mon, Mar 26, 2018 at 06:26:19PM +0000, PICCA Frederic-Emmanuel wrote:
I imagine that you mean fst instead of snd in order to produce the right output :)
Indeed :P
How can I check if this code produce list copies and is efficient with big lists ?
`time` and profiling tools provided by GHC [1] If you feel lost, shout and I'll set up and example [1] http://book.realworldhaskell.org/read/profiling-and-optimization.html
Just one question about this. I do not want to put a print in order to generate the newList object. let l = [] let newList = superChnker l print newList How Can I tell to haskell, produce the result even if I do not use it.
Wow... after attempting a simpler solution, I now understand a discussion from a little while back on one of the other lists about how terribly unexpected the default groupBy behavior is. On Mon, Mar 26, 2018 at 12:21 PM, PICCA Frederic-Emmanuel < frederic-emmanuel.picca@synchrotron-soleil.fr> wrote:
Just one question about this.
I do not want to put a print in order to generate the newList object.
let l = [] let newList = superChnker l print newList
How Can I tell to haskell, produce the result even if I do not use it. _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
On Mon, Mar 26, 2018 at 07:21:22PM +0000, PICCA Frederic-Emmanuel wrote:
How Can I tell to haskell, produce the result even if I do not use it.
deepseq [1]! [1] https://hackage.haskell.org/package/deepseq-1.4.3.0/docs/Control-DeepSeq.htm...
participants (3)
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Francesco Ariis -
PICCA Frederic-Emmanuel -
Theodore Lief Gannon