how do you change the following haskell code into a high order function?

Question

Define a function

rmdups :: Eq a => [a] -> [a]

that removes duplicates from a list. For example, rmdups "ababca" should return "abc". The order of the elements in the output list is not important

Here is what I have done so far: This works:

 rmdups :: Eq a => [a] -> [a]
 rmdups [] = []
 rmdups (x:xs) = x : rmdups (filter(/= x) xs)

This does not work, what am I doing wrong:

 rmdups = map head . group . sort

It works for the example you have given. What is an example of it not working? — David Young, Apr 21 '16 at 21:16
No the last line of code does not work. The compiler says group is not in scope. — B.John, Apr 21 '16 at 21:18
Ahh, as baxbaxwalanuksiwe says that is because you are missing an `import`. In future questions, you should include any error messages you are getting. — David Young, Apr 21 '16 at 21:38
Possible duplicate of http://stackoverflow.com/q/16108714/1523776 — Benjamin Hodgson, Apr 22 '16 at 08:13

score 2 · Answer 1 · edited May 23 '17 at 11:59

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You could do this:

import Data.List (nub)
rmdups = nub

But I assume that you're not allowed.

Otherwise, your solution (rmdups = map head . group . sort) works for me.

If the compiler says group or sort not in scope, just import it from Data.List.

PS: I believe you meant "point free style", not "higher order function"

Edit: Thanks user3217013 for pointing that out.

edited May 23 '17 at 11:59

Community

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answered Apr 21 '16 at 21:28

baxbaxwalanuksiwe

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You also need `Data.List` for `sort`, for some baffling reason. – Emil Apr 21 '16 at 21:30
Thank you for your answer, but like you said I am not allowed to do that. Is there any other way of solving the question, with the use of high order functions. – B.John Apr 21 '16 at 21:42
Why are you not allowed to do that ? My solution (which is actually yours) uses the higher order function `map`. – baxbaxwalanuksiwe Apr 21 '16 at 21:47
The tasks says that I only have to declare import Data.List (sort) and nothing else. – B.John Apr 21 '16 at 21:50
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Try writing your own version of `group` then. Or try writing the function another way. – baxbaxwalanuksiwe Apr 21 '16 at 21:53

how do you change the following haskell code into a high order function?

1 Answers1