
Prelude Data.List.Split> (splitOn [0] . map read . splitOn ",") "2,15,33,0,8,1,16,18" [[2,15,33],[8,1,16,18]] -Brent On Tue, Feb 22, 2011 at 12:03:45AM -0700, Luke Palmer wrote:
See the http://hackage.haskell.org/package/split package. You should be able to do this by splitting the string on comma, and then splitting the result on 0, plus some plumbing.
Luke
On Mon, Feb 21, 2011 at 11:46 PM, z_axis
wrote: I want to split "2,15,33,0,8,1,16,18" to ([2,15,33],[8,1,16,18]). the "0" will by discarded.
However, it seems that there isnot any standard function to do it.
Sincerely!
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