Create function to remove duplicates from Haskell list

Question

I am trying to build a function that takes the first element of a string, and removes all other elements equal to it from the string. Then does the same for the second character.

Ie - "Heello" would become "Helo" and "Chocolate" "Chlate"

My original attempt

removeSuccessor :: String -> String
removeSuccessor x = [c | c <- x, x ! `elem` c]

But that doesn't seem to work.. suggestions?

By x ! `elem` c, I think you probably mean not (elem c x). Haskell doesn't use ! for logical negation, and elem takes its arguments in the other order (its type is a -> [a] -> Bool, not [a] -> a -> Bool. — Seth Tisue
– Seth Tisue, Commented Oct 15, 2013 at 19:10
Rewrote the function as: removeDifferent :: String -> String removeDifferent x = [c | c <- x, not (elem c x)] Still not working :\ — MrD
– MrD, Commented Oct 15, 2013 at 19:17

Claudiu · Accepted Answer · 2013-10-15 19:17:28Z

5

You could keep a set of all elements seen and only keep the current one if it hasn't been seen yet:

import Data.Set
removeDups :: Ord a => [a] -> Set a -> [a]
removeDups [] sofar = []
removeDups (x:rest) sofar
     | member x sofar = (removeDups rest sofar)
     | otherwise      = x:(removeDups rest (insert x sofar))

Usage:

removeDups "Heello" empty    -- "Helo"
removeDups "Chocolate" empty -- "Choclate"

Run time is O(n log n), I think.

Or you can use nub from Data.List:

Prelude Data.List> import Data.List
Prelude Data.List> nub "Heello"
"Helo"
Prelude Data.List> nub "Chocolate"
"Choclate"

Run-time is O(n^2).

answered Oct 15, 2013 at 19:17

Claudiu

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