26 Mar
2018
26 Mar
'18
2:11 p.m.
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!