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Re: sorting and grouping by

by rg0now (Chaplain)
on Mar 13, 2008 at 21:25 UTC ( #674076=note: print w/ replies, xml ) Need Help??


in reply to sorting and grouping by

Here is some (admittedly, very inmature) Haskell to solve this problem.

type Data = (Int, Int, String) bottom (b, _, _) = b top (_, t, _) = t ide (_, _, i) = i contains :: Data -> Data -> Bool x `contains` y = bottom x <= bottom y && top x >= top y rangeSort1 :: [Data] -> [(Data, [Data])] rangeSort1 r = map (\x -> (x, (filter (contains x) (filter (/=x) r)))) + r ---- test :: [Data] test = [ (11, 22, "A") , (22, 45, "B") , (22, 33, "C") , (25, 28, "D") , (47, 49, "E") ]
And here is the corresponding Perl, or at least, this is the closest I could get to this 'declarational' style in Perl:
sub bottom { shift->[0] }; sub top { shift->[1] }; sub ide { shift->[2] }; sub contains { bottom($_[0]) <= bottom($_[1]) && top($_[0]) >= top($_[ +1]) }; sub rangeSort1 { map { my $x = $_; [$x, [grep {contains($x, $_)} (grep { ide($_) ne ide($x) } @_) ] ] } @_ }; my $test = [ [11, 22, "A"] , [22, 45, "B"] , [22, 33, "C"] , [25, 28, "D"] , [47, 49, "E"] ];
If there's a Haskell/Perl guru out there with some free time, could you please improve on this?

Update: slightly improved Haskell:

rangeSort1 :: [Data] -> [(Data, [Data])] rangeSort1 r = mapMaybe (\x -> let l = filter (contains x) . filter (/ +=x) in case l r of [] -> Nothing otherwise -> Just (x, l r) ) r


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