I have been trying to make a function that concatenates a list of lists, sorts it, and gives back the duplicated values.
The issue I'm facing is that it tells me to change (Eq a) to (Ord a) for the last function, but I cannot do this. How can I solve this without changing (Eq a) to (Ord a) ?
This is the code I have:
group :: Eq a => [a] -> [[a]]
group = groupBy (==)
groupBy :: (a -> a -> Bool) -> [a] -> [[a]]
groupBy _ [] = []
groupBy eq (x:xs) = (x:ys) : groupBy eq zs
where (ys,zs) = span (eq x) xs
uniq :: Eq b => [b] -> [b]
uniq = map head . group
insert :: Ord a => a -> [a] -> [a]
insert x [] = [x]
insert x (y:ys) | x < y = x:y:ys
| otherwise = y:(insert x ys)
isort :: (Eq a, Ord a) => [a] -> [a]
isort [] = []
isort (x:xs) = insert x (isort xs)
kms :: Ord a => [a]
kms xss = uniq (isort (concat xss))
pairwiseIntersections :: (Eq a) => [[a]] -> [a]
pairwiseIntersections xss = kms xss
You cannot sort a list without its elements having some ordering -- meaning that they must be instances of Ord.
You can do other things to deduplicate a list, like nub, but if you want it sorted you need Ord or an equivalent ordering.
You mention no complexity requirements, so the simplest approach could be
keepOnlyDups [] = []
keepOnlyDups (x:xs) = [x | elem x xs] ++ keepOnlyDups [ y | y <- xs, y /= x]
removeExtras [] = []
removeExtras (x:xs) = [x] ++ removeExtras [ y | y <- xs, y /= x]
answering both your question's text and the implied meaning of the code.
Implementing keepOnlyUniques is left as an exercise, if you're interested in that.
Related
Currently, I am making a function that can take the values in two lists, compare them, and print any similar values. If a value is duplicated in both lists, it is not printed a second time.
EXAMPLE
INPUT: commons [1,2,2,3,4] [2,3,4,5,2,6,8]
EXPECTED OUTPUT: [2,3,4]
What I currently have causes a repeated values in both lists to repeat in the output. So, in the above example, the 2 would print twice.
Here is the current code I am working on:
commons :: Eq a => [a] -> [a] -> [a]
commons [] [] = []
commons x y = commons_Helper x y
commons_Helper :: Eq a => [a] -> [a] -> [a]
commons_Helper [] [] = []
commons_Helper x [] = []
commons_Helper [] y = []
commons_Helper (x:xs) y =
if elem x y then x : commons_Helper xs y
else commons_Helper xs y
Any and all help would be greatly appreciated.
EDIT: This must remain as commons :: Eq a => [a] -> [a] -> [a], and I cannot import and libraries
You could make your traversal of the xs list a stateful one, the state keeping track of elements that have been already seen. The state begins life as an empty list.
It is possible to do that by adding a 3rd parameter to your helper function, which currently is not very useful.
commons_Helper :: Eq a => [a] -> [a] -> [a] -> [a]
commons_Helper [] ys st = []
commons_Helper (x:xs) ys st =
if (elem x ys && (not $ elem x st)) -- extra test here
then x : (commons_Helper xs ys (x:st))
else commons_Helper xs ys st
commons :: Eq a => [a] -> [a] -> [a]
commons xs ys = commons_Helper xs ys []
This state-based technique is very common in Haskell. There is even a library function: mapAccumL :: (s -> a -> (s, b)) -> s -> [a] -> (s, [b]) to support it.
I have created a program to remove first smallest element but I dont how to do for second largest:
withoutBiggest (x:xs) =
withoutBiggestImpl (biggest x xs) [] (x:xs)
where
biggest :: (Ord a) => a -> [a] -> a
biggest big [] = big
biggest big (x:xs) =
if x < big then
biggest x xs
else
biggest big xs
withoutBiggestImpl :: (Eq a) => a -> [a] -> [a] -> [a]
withoutBiggestImpl big before (x:xs) =
if big == x then
before ++ xs
else
withoutBiggestImpl big (before ++ [x]) xs
Here is a simple solution.
Prelude> let list = [10,20,100,50,40,80]
Prelude> let secondLargest = maximum $ filter (/= (maximum list)) list
Prelude> let result = filter (/= secondLargest) list
Prelude> result
[10,20,100,50,40]
Prelude>
A possibility, surely not the best one.
import Data.Permute (rank)
x = [4,2,3]
ranks = rank (length x) x -- this gives [2,0,1]; that means 3 (index 1) is the second smallest
Then:
[x !! i | i <- [0 .. length x -1], i /= 1]
Hmm.. not very cool, let me some time to think to something better please and I'll edit my post.
EDIT
Moreover my previous solution was wrong. This one should be correct, but again not the best one:
import Data.Permute (rank, elems, inverse)
ranks = elems $ rank (length x) x
iranks = elems $ inverse $ rank (length x) x
>>> [x !! (iranks !! i) | i <- filter (/=1) ranks]
[4,2]
An advantage is that this preserves the order of the list, I think.
Here is a solution that removes the n smallest elements from your list:
import Data.List
deleteN :: Int -> [a] -> [a]
deleteN _ [] = []
deleteN i (a:as)
| i == 0 = as
| otherwise = a : deleteN (i-1) as
ntails :: Int -> [a] -> [(a, Int)] -> [a]
ntails 0 l _ = l
ntails n l s = ntails (n-1) (deleteN (snd $ head s) l) (tail s)
removeNSmallest :: Ord a => Int -> [a] -> [a]
removeNSmallest n l = ntails n l $ sort $ zip l [0..]
EDIT:
If you just want to remove the 2nd smallest element:
deleteN :: Int -> [a] -> [a]
deleteN _ [] = []
deleteN i (a:as)
| i == 0 = as
| otherwise = a : deleteN (i-1) as
remove2 :: [a] -> [(a, Int)] -> [a]
remove2 [] _ = []
remove2 [a] _ = []
remove2 l s = deleteN (snd $ head $ tail s) l
remove2Smallest :: Ord a => [a] -> [a]
remove2Smallest l = remove2 l $ sort $ zip l [0..]
It was not clear if the OP is looking for the biggest (as the name withoutBiggest implies) or what. In this case, one solution is to combine the filter :: (a->Bool) -> [a] -> [a] and maximum :: Ord a => [a] -> a functions from the Prelude.
withoutBiggest l = filter (/= maximum l) l
You can remove the biggest elements by first finding it and then filtering it:
withoutBiggest :: Ord a => [a] -> [a]
withoutBiggest [] = []
withoutBiggest xs = filter (/= maximum xs) xs
You can then remove the second-biggest element in much the same way:
withoutSecondBiggest :: Ord a => [a] -> [a]
withoutSecondBiggest xs =
case withoutBiggest xs of
[] -> xs
rest -> filter (/= maximum rest) xs
Assumptions made:
You want each occurrence of the second-biggest element removed.
When there is zero/one element in the list, there isn't a second element, so there isn't a second-biggest element. Having the list without an element that isn't there is equivalent to having the list.
When the list contains only values equivalent to maximum xs, there also isn't a second-biggest element even though there may be two or more elements in total.
The Ord type-class instance implies a total ordering. Otherwise you may have multiple maxima that are not equivalent; otherwise which one is picked as the biggest and second-biggest is not well-defined.
I'd like to design an algorithm in Haskell using tail recursion and first order programming for insertion sort
I've came up with this solution
isort :: Ord a => [a] -> [a]
isort [] = []
isort [x] = [x]
isort ( x:xs ) = insert ( isort xs )
where insert [] = [x]
insert ( y:ys )
| x < y = x : y : ys
| otherwise = y : insert ys
But I'm not sure if it uses first order and tail recursion.
Can someone come up with an alternative solution using
tail recursion
first order programming
Thanks,
This is the simple version, not using tail recursion nor HOF
sort :: Ord a => [a] -> [a]
sort [] = []
sort [x] = [x]
sort (x:xs) = insert x (sort xs)
insert :: Ord a => a -> [a] -> [a]
insert x [] = [x]
insert x (y:ys) | x < y = x:y:ys
| otherwise = y:insert x ys
You can add an accumulator, that allows us to rewrite the sort using tail recursion:
sort' :: Ord a => [a] -> [a]
sort' xs = sortAcc xs []
sortAcc :: Ord a => [a] -> [a] -> [a]
sortAcc [] acc = acc
sortAcc (x:xs) acc = sortAcc xs (insert x acc)
insert is pretty nicely defined the way it is; but just for the purpose
of using higher order functions, we can define it like:
insert' :: Ord a => a -> [a] -> [a]
insert' x xs = menores ++ x:mayores
where (menores,mayores) = span (<x) xs
where the section (<x) is a function of type Ord a => a -> Bool passed to span.
I want to write a function that takes a list of sorted lists, then merges everything together and sorts them again.
I managed to write this so far:
merge_:: Ord a => [[a]] -> [a] --takes in the list and merges it
merge_ [] = []
merge_ (x:xs) = x ++ merge_ xs
isort:: Ord a => [a] -> [a] --Sorts a list
isort [] = []
isort (a:x) = ins a (isort x)
where
ins a [] = [a]
ins a (b:y) | a<= b = a:(b:y)
| otherwise = b: (ins a y)
I haven't been able to find a way to combine these two in one function in a way that makes sense. Note that I'm not allowed to use things such as ('.', '$'..etc) (homework)
We start simple. How do we merge two sorted lists?
mergeTwo :: Ord a => [a] -> [a] -> [a]
mergeTwo [] ys = ys
mergeTwo xs [] = xs
mergeTwo (x:xs) (y:ys)
| x <= y = x : mergeTwo xs (y:ys)
| otherwise = y : mergeTwo (x:xs) ys
How do we merge multiple? Well, we start with the first and the second and merge them together. Then we merge the new one and the third together:
mergeAll :: Ord a => [[a]] -> [a]
mergeAll (x:y:xs) = mergeAll ((mergeTwo x y) : xs)
mergeAll [x] = x
mergeAll _ = []
Allright. Now, to sort all elements, we need to create a list from every element, and then merge them back. Let's write a function that creates a list for a single item:
toList :: a -> [a]
toList x = -- exercise
And now a function to wrap all elements in lists:
allToList :: [a] -> [[a]]
allToList = -- exercise
And now we're done. We simply need to use allToList and then mergeAll:
isort :: Ord a => [a] -> [a]
isort xs = mergeAll (allToList xs)
Note that this exercise got a lot easier since we've split it into four functions.
Exercises (which might not be possible for you(r homework))
Write toList and allToList.
Try a list comprehension for allToList. Try a higher order function for allToList.
Write isort point-free (with (.)).
Check whether there is already a toList function with the same type. Use that one.
Rewrite mergeAll using foldr
Try this (not tested):
merge :: Ord a => [a] -> [a] -> [a]
merge [] l1 = l1
merge l1 [] = l1
merge (e1:l1) (e2:l2)
| e1<e2 = e1:merge l1 (e2:l2)
| otherwise = e2:merge (e1:l1) l2
I made this code where I need to find elements in a list that appears only once
for example: for input [1,2,2,3,4,4], the output will be: [1,3]
unique :: =[a]->[a]
unique xs =[x|x<-xs, elemNum x xs ==1]
elemNum :: Int -> [Int]->[Int]
elemNum x (y:ys)
| x==y =1+ elemNum x ys
| otherwise =elemNum x ys
However I am getting an error :
Not in scope: `unique'
Is this the correct way to use 2 function in Haskell? (define them at the same file), What'a wrong with the code?
There are a few problems in your code:
type signature of unique is wrong, it should be
unique :: (Eq a) => [a] -> [a]
that type constraint (Eq a) comes from elemNum
type signature of elemNum also wrong, it should be
elemNum :: (Eq a) => a -> [a] -> Int
that type constraint comes from ==, and the type of its first parameter no need to be Int, but its return type should be Int because you want to find out how many x in xs.
Also, you forgot to deal with empty list in that definition.
Here is a fixed version of your code:
unique :: (Eq a) => [a] -> [a]
unique xs =[x| x<-xs, elemNum x xs == 1]
elemNum :: (Eq a) => a -> [a] -> Int
elemNum x [] = 0
elemNum x (y:ys)
| x==y = 1 + elemNum x ys
| otherwise = elemNum x ys
Here is another implementation:
onlyOnce [] = []
onlyOnce (x:xs)
| x `elem` xs = onlyOnce $ filter (/=x) xs
| otherwise = x : onlyOnce xs
If x occurs in xs, then the result of onlyOnce (x:xs) should be the same as the result of applying onlyOnce to the result of removing all occurrences of x from xs; otherwise, x occurs in (x:xs) only once, so x should be part of the final result.
You have an equals sign in the Type declaration for unique:
unique :: =[a]->[a]
should be
unique :: [a] -> [a]
In my opinion it is much easier to implement this function using functions from Data.List module:
import Data.List
unique :: (Ord a) => [a] -> [a]
unique = map (\(y,_) -> y) . filter (\(x,l) -> l == 1) . map (\l#(x:xs) -> (x,length l)) . group . sort