Gracjan Polak wrote:
Data.List.foldr (Data.Set.delete) s [1,3,5]
or
Data.List.foldl' (flip Data.Set.delete) s [1,3,5]
There will be a day when I finally grasp foldr/foldl :)
Maybe
http://en.wikibooks.org/wiki/Haskell/Performance_Introduction
http://www.haskell.org/haskellwiki/Stack_overflow
helps to make that day dawn earlier :)
Note that in your original approach of fold (.) id , it doesn't really
matter whether you use foldr or foldl' (except for argument order!)
because the latter would only evaluate to a function Data.Set ->
Data.Set which is not very useful. In contrast, Duncan's code evaluates
to Data.Set , that's what you want to be strict in.
Regards,
apfelmus
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