haskell - Removing duplicate elements in a Seq -


wondering how implement nub on seq a

i 1 do:

nubseq :: seq -> seq nubseq = fromlist . nub . tolist

just wondering there standard not convert lists in order call nub :: [a]->[a]?

an implementation occurred me, based on nub, is:

nubseq :: (eq a) => seq -> seq nubseq = data.sequence.foldrwithindex             (\_ x -> case x `data.sequence.elemindexr` of                         _ ->                         nothing -> |> x) data.sequence.empty

but there must more elegant?

thanks.

not sure whether qualifies more elegant splits concerns in independent functions (caveat: need ord constraint on a):

  • seqtonubmap takes seq , outputs map associating each a smallest index @ appeared in sequence

  • maptolist takes map of values , positions , produces list of values in increasing order according specified positions

  • nubseq combines these generate sequence without duplicates

the whole thing should o(n*log(n)), believe:

module nubseq  import data.map      map import data.list     list import data.sequence seq import data.function  seqtonubmap :: ord => seq -> map int seqtonubmap = foldlwithindex (\ m k v -> insertwith min v k m) map.empty  maptolist :: ord => map int -> [a] maptolist = fmap fst . list.sortby (compare `on` snd) . map.tolist  nubseq :: ord => seq -> seq nubseq = seq.fromlist . maptolist . seqtonubmap

or simpler alternative following @davidfletcher's comment:

nubseq' :: forall a. ord => seq -> seq nubseq' xs = fold.foldr cons nil xs set.empty    cons :: -> (set -> seq a) -> (set -> seq a)   cons x xs seen     | x `elem` seen = xs seen     | otherwise     = x <| xs (set.insert x seen)    nil :: set -> seq   nil _ = seq.empty

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