CODEWITHSUNDEEP
listhaskelltupleslist-comprehension
Chalupa
Chalupa

Reputation: 367

removing tuples from list (Haskell)

I want to write a function that takes as argument a list of tuples like this:

remove'  [ ("a", True), ("b", False), ("c", False), ("d", True) ]

I would like to return a list of tuples that have False as their second value, so I'd like my function to return

[ ("b", False), ("c", False) ]

Here's what I have so far but it won't load in GHCi. can anyone help me out? Thanks 

remove' :: [(a,b)] -> [(a,b)]
remove' [(a,b)] = [ c | c <- [(a,b)], c `notElem` True ]

Upvotes: 1

Views: 3708

Answers (3)

Emil
Emil

Reputation: 2167

I'm surprised no one has yet said

remove' = filter (not . snd)

Upvotes: 2

thefourtheye
thefourtheye

Reputation: 239473

Since you want to match the second element of the tuples, you need to pick the second element from the tuple and compare it with False like this

remove' :: [(a, Bool)] -> [(a, Bool)]
remove' xs = [c | c <- xs, snd c == False]

The snd function will get second element from each of the tuples and compare them with False. Only if they match, they will be gathered in the resulting list.

Since the second elements are booleans, you can make use of the not function like this

[c | c <- xs, (not . snd) c]

We can express the same wihtout the dot notation like this

[c | c <- xs, not(snd(c))]

Note: In your program, you have

remove' [(a,b)] = ...

it means that, it will be executed only when remove is called with a list of tuples of size 1. You can check that like this

remove' :: [(a, Bool)] -> [(a, Bool)]
remove' [(a, b)] = [(a, b)]

remove' [ ("a", True), ("b", False), ("c", False), ("d", True) ]

will print

Non-exhaustive patterns in function remove'

It means the patterns you specified in the function definition is not covering all the possible inputs. Since we need process a list of tuples of any length, we accept it in xs and use it in the List Comprehension.

Upvotes: 4

Mikhail
Mikhail

Reputation: 445

You just need filter and select second item in lambda:

filter (\x -> (snd x) == False) [("a", False), ("b", True)]

Response:

[("a",False)]

Upvotes: 0

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