On Thursday 23 December 2010 18:27:43, C K Kashyap wrote: > Hi all, > > Here's my attempt to convert a list of integers to a list of range > tuples - > > Given [1,2,3,6,8,9,10], I need [(1,3),(6,6),8,10)] > > My attempt using foldl yields me the output in reverse. I can ofcourse > reverse the result, but what would be a better way? > > f xs = foldl ff [] xs > where > [] `ff` i = [(i,i)] > ((s,e):ns) `ff` i = if i == e+1 then > (s,i):ns > else > (i,i):(s,e):ns >
A right fold? It's easier, at least: Prelude> let foo k [] = [(k,k)]; foo k xs@((l,h):t) = if l == k+1 then (k,h):t else (k,k):xs Prelude> foldr foo [] [1,2,3,6,8,9,10] [(1,3),(6,6),(8,10)] > Regards, > Kashyap _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
