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
Regards, Kashyap
I suggested:
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)]
Lazier is better: bar k xs = (k,m):t where (m,t) = case xs of [] -> (k,[]) (l,h):u | l == k+1 -> (h,u) | otherwise -> (k,xs) Prelude> foldr foo [] [1 .. 1000000] [(1,1000000)] (15.46 secs, 226229904 bytes) Prelude> foldr bar [] [1 .. 1000000] [(1,1000000)] (3.47 secs, 242992364 bytes)