[Haskell-beginners] subset - a little add
luca_ciciriello at hotmail.com
Fri Jan 29 05:01:36 EST 2010
Yes my function operate only in a set-theory contest and your solution:
subset xs ys = all (`elem` ys) xs
is indeed more elegant than mine.
Thanks again for your help.
> From: daniel.is.fischer at web.de
> To: beginners at haskell.org
> Subject: Re: [Haskell-beginners] subset - a little add
> Date: Fri, 29 Jan 2010 10:06:29 +0100
> CC: luca_ciciriello at hotmail.com
> Am Freitag 29 Januar 2010 08:36:35 schrieb Luca Ciciriello:
> > Just a little add to may previous mail.
> > The solution I've found from myself is:
> > subset :: [String] -> [String] -> Bool
> > subset xs ys = and [elem x ys | x <- xs]
> subset xs ys = all (`elem` ys) xs
> but is that really what you want? That says subset [1,1,1,1]  ~> True.
> If you regard your lists as representatives of sets (as the name suggests),
> then that's correct, otherwise not.
> However, this is O(length xs * length ys). If you need it only for types
> belonging to Ord, a much better way is
> import qualified Data.Set as Set
> import Data.Set (fromList, isSubsetOf, ...)
> subset xs ys = fromList xs `isSubsetOf` fromList ys
> or, if you don't want to depend on Data.Set,
> subset xs ys = sort xs `isOrderedSublistOf` sort ys
> xxs@(x:xs) `isOrderedSublistOf` (y:ys)
> | x < y = False
> | x == y = xs `isOrderedSublistOf` ys
> | otherwise = xxs `isOrderedSublistOf` ys
>  `isOrderedSublistOf` _ = True
> _ `isOrderedSublistOf`  = False
> > My question is if exists a more elegant way to do that.
> > Luca.
Not got a Hotmail account? Sign-up now - Free
Send us your Hotmail stories and be featured in our newsletter
-------------- next part --------------
An HTML attachment was scrubbed...
More information about the Beginners