Haskell remove duplicates without recursion. My question about binding in pattern-matched, .
Haskell remove duplicates without recursion TrebledJ. – John La Rooy. This is a tricky question and we need to use recursion to find all the permutations of //WITHOUT DUPLICATE PERMUTATION OF STRING System. Most of the answers given below assume you need to preserve pairs without inverted duplicates, but the solution might be simpler if you don't. If we look at the type signature for nub, we see:. I originally read foldr as filter, but there are some useful details in that section that will help so I left it. If you pass your list of denominations in descending order (so that the last denomination is 1), that is a particularly fruitful optimisation (try it with a total of 400 or 800 to see what I'm talking about). filter is built into the standard Prelude, so it's a standard library function akin to elem. List module. If you can allow an Ord constraint you can use the sortBy function which implements a stable sorting algorithm, and then use groupBy to remove the contiguous duplicates:. Define first a function which given x and ys removes all the occurrences of x in the list ys. Set (Set) import qualified Data. Instead of accumulating the values on the way recursing to the end, you can collect the values on the way back up: let rem_from_right lst = let rec is_member n mlst = match mlst with | [] -> false | h::tl -> begin if h=n then true else is_member n tl end in let rec loop lbuf = match lbuf with | [] -> [] | h::tl -> begin let rbuf = loop tl in if is_member h rbuf then rbuf else h::rbuf end This is not how we do it in Haskell, there is no need to "declare" or "define" names before we use them. Here is one function that calls a another, recursive function and produces the result you want. Do while loop in Haskell. Start from the leftmost character and remove duplicates at the left corner if there are any. How to use recursive in Haskell. removeDuplicates method not remove all of the duplicated integers in a list. Every "snoc" and last will take longer as the answer of remdps grows, because Prelude lists are much better at "cons" and head. The result is the classic depth-first search for general graphs, which you can find in any algorithms textbook or in Russell & Norvig chapter 3. It's a rather ridiculous piece of code but it doesn't use any standard library functions and it modifies x in place. Then it will be evaluated like: Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about your product, service or employer brand; OverflowAI GenAI features for Teams; OverflowAPI Train & fine-tune LLMs; Labs The future of collective knowledge sharing; About the company The program works without the punctuation function, but not when I include it to (toWords fileContents) Please can someone look at what I have done and see what I am doing wrong. If I have got it I would end up with Removing duplicates from a list in Haskell without elem. You can work with foldr :: (a -> b -> b) -> b -> [a] -> b here. I want to implement a method that removes the max element from a list of Ints, or if there are duplicates of the max element, just remove one of them using recursion. Remove duplication elements in the list Haskell. Haskell:Check if first element in a list is a duplicate. 3. As we only have to remove the first space, we can return the remainder of the list (i. But I'm running into a bit of a problem, my method works and removes some elements How can I delete consecutive duplicates recursively without an int I am new to functional programming. We can use Map a Int to represent a bag (AKA multiset) of values of type a. The following approach can be followed to remove duplicates in O(N) time: Start from the leftmost character and remove duplicates at left corner if there are any. We bind results of effectful computations using. Can anyone help with this issues? -- delete the last character from a string in Haskell deleteLast [] = "Empty list!" -- if the string is enpty I am trying to write a function that removes all characters in the first string from the second string. nubBy is just nub using a custom function instead of (==). And indeed, in Haskell, there are often several other abstractions we can build upon to make our code simpler, such as applicatives and monads in the above example. f :: a -> v Before you can solve this problem, you need to have a good understanding of the functions elem, map and foldr. Unless someone else knows an easier way. However, recursing down two lists would be very awkward and difficult to do. elem. Without the return you keep increasing n (below at Explanation. For example: [1,1] Remove all duplicates from the given list(you can use I'm learning Haskell and am stuck this problem. Function(on) where s :: String is a string such as "Hello". I need to use recursion and another helper function that removes that removes all occurrences of a character from a string. except that you also wanted to do some sort of recursion to make sure that none of the elements in xs or ys is repeated. Ask Question Asked 3 years ago. out. Another very nice approach would be to push your recursion into existing Prelude functions; the more you can spot iteration patterns in practice the more bugs you can avoid by not re-implementing the same loops over and over. 11. List as List import Data. list comprehension unique values. Map. They both preserve the original order of the elements, but unique retains the last occurrence of each element, while nub retains the first occurrence. Find occurrences in a List using recursion in Haskell. The Overflow Blog Ryan Dahl explains why Deno had to evolve with version 2. ) I'm looking for a Python function similar to nubBy in Haskell, which removes duplicate but with a different equality test. – Dan Burton. answered Feb 11, 2013 at 22:48. 2. (:) has type a -> [a] and works better here. For information, the nub function takes a list and returns a list formed by removing all duplicates element from it; it is exported by the Data. One way to do this is to start a completely new list. Function seqToNubMap :: Ord a => Seq a -> Map a Int seqToNubMap = foldlWithIndex (\ m k v -> insertWith min v k m) I do not understand a sample solution for the following problem: given a list of elements, remove the duplicates. How can I convert a Cons list to a Haskell list without (explicit) Remove spaces from the 3rd line onwards in a file on linux (Removing the reverse if we don't care about the order) Recursion is a very handy skill to have, but it's not always the solution. Random main :: IO () main = do rs <- forM [1. List at the beginning of the file I am new to both Haskell and programming. I copied that part from the duplicate removal function. Recur for Well you can only filter out duplicates, if there is a way to check whether two values are duplicates. I can't seem to wrap my head around how to do recursion over a list of lists in Haskell. May I ask why you want to write it without explicit recursion? Explicit recursion makes it far easier to add the optimisation to treat the case of a one-element list specially. I'm trying to use a high order function to do this problem, so no recursion. Initially in the recursion, you haven't seen any elements so seen is empty. List (no, it's actually not in the Prelude) definitely does something like what you want, but it is not quite the same as your unique function. makeB @normanius, to remove all duplicates, the input would need to be sorted. And you'll get a list without that number as output. I need to write a predicate remove_duplicates/2 that removes duplicate elements form a given list. Commented Jan 3, Here is a solution without dependence on outside packages: list = [1,1,1,1,1,1,2,3,4,4,5,1,2] L = list + So I hope that was sufficiently helpful for you to see what's going on. Locating TIFF layers without displaying them I'd hesitate to give a type like XYZ an Ord instance, just to speed up this algorithm – any such instance should also make sense conceptually! You can achieve the same complexity by staying with lists and using map head . I think it might be more efficient to use Data. Ask Question Asked 11 years, 1 month ago. According to Thomas Guest at wordaligned. haskell; recursion; implementation; Share. group . permutations (delete i xs) inside the list comprehension should bring the flow closer to base case. Strict as M -- Add a value to a bag. So i changed the base case from Is there any function in Haskell that already do it? Not as far as I know no, and a search on Hoogle does not immediately lists such function. This would be quite helpful as I've seen plenty of times where the function, defined internally with a recursion scheme was first applying reverse to its input, clearly signalling the need for a foldl-like "front to back" execution. One common approach is to define a recursive function that checks each element of the list and adds it to a new list if it is not already I have a list of Ord a, and would like to "efficiently" determine whether or not it contains any duplicated elements. [I assume it from the example, you mention you want to print the specified output]. Use list comprehension to define a Haskell function Ted that takes a list of Ord items and removes all occurrences of the largest Can use the Prelude function without any restrictions as I am trying to remove the largest occurring items from 4 different lists and one of the lists How to remove a duplicate number in a list @PeCosta A duplicate for example: 1 penny 1 nickel vs 1 nickel 1 penny. I suppose you could call clean2 an auxiliary function, although, like you, I don't know exactly what that's supposed to mean. That said, you want a function rem which accepts a list xs and a function f :: a -> a -> Bool (on which elements are compared for removal from xs). Improve this question. List. Recursion in Haskell. The Python file below includes a function named remove_duplicates_recursion which takes two arguments: dupList and temp. Share. I am trying to learn recursion. In Haskell we separate effectful computations (IO is one type of effectful computations) from pure computations. Viewed 165 times Implementing recursion in Haskell without input variable. From what I can comprehend from your explanation is you want to filter based on some condition on the product of two list. Strings in Haskell are lists of characters; the generator c <- s feeds each character of s in turn to the left-hand expression toUpper c, building a new list. You can group the letters which are equal (via Data. Assuming the elements are in class Ord, I came up with the following: import Data. If I have the following: import System. println(set); } public static void NONDuplicatePermutation(String prefix ,String str I am assuming here you don't actually want to remove the element [as @aioobe mentioned, it cannot be done], but you actually want to: "print all elements in the array, without duplicates". In particular, it keeps only the first occurrence of each element. Here's another solution: Pair a normal list with a normal Set and get basically the same effect. , the tail) without further processing: removeSpace (' ' : cs) = cs What remains is to deal with the case in which the head character is not a space. However I think it can also be enlightening to see how to do it without an accumulator; the trick is just to check if the return value from the recursive call is 0 and maintain that value instead of incrementing if it is. Your first snippet produces the first x out of many but the third snippet produces the last x, that's the reason for the discrepancy. We can thus simply forward the type constraint further in the signatures of the functions: You attempt is on the right track for a bubble sort, which is a good starting place for sorting. It would look something like: I need to create a powerset function in haskell which takes a set and outputs the power set without duplicate entries, regardless of what is put in the input list. Anon is correct: nubBy is the function you are looking for, and can be found in Data. 1 Indexing lists. Test in GHCi: λ> reaction [Green, Purple, Green, Green] [Purple, Green, Blue, Yellow] 0 2 Haskell recursion loop counter. As far as useless goes; even for the bottom case of an infinite list without duplicates, this implementation provides the most useful possible behavior; it won't tell you that there are no duplicates (because, again, Halting Problem), but you can decide that if it doesn't give you an answer within a set time, then you assume that the list has no duplicates, and you can achieve Python code works, but Haskell code enters an infinite recursion. List(sortBy, groupBy) import Data. Buhr Commented May 19, 2017 at 14:13 You could look into the recursion-schemes package. The result of this list comprehension is "HELLO". Elements of the multiset are keys in the map, and they're paired with their counts. Basically you define your data type as if it were a functor. Runner will only see one dup per node, because if there were multiple duplicates they would have been removed already. alter $ \mc -> case mc of -- The item is not Note that the above implementation takes O(n^2) time, which is optimal for Eq instances. Basically by the time Haskell gets to evaluating the recursive call to getCommandsFromUser, it's already done with all of the data generated within the previous iteration, and so it gets garbage collected. For this particular problem, it might be more instructive to look at the C++ STL algorithm std::next_permutation. So far I have a working implementation: | otherwise = x : rmdups xs. g. If you don't mind changing the order of elements in the list, you can make all duplicates adjacent by sorting the list first. Follow edited Sep 18 , 2013 at 12: Remove pair duplicates Haskell. Haskell - Removing adjacent duplicates from a list. It returns triple: the first - all elements with removed duplicates (like sortUniq but the result is not sorted); the second - the elements that are repeated at least once in the list (result is the same as repeated but not sorted); the third - the unique elements that do not have duplicates (result is the same as unique but not sorted) I want to write a function which takes a list and constructs a subset of that list of a certain length based on the output of a function. Recur for a string of length n-1 (string without last character). Test Cases : "allcbcd" -> "alcbd" My Code (the one in which string has to be sorted) : I was asked to write my own implementation to remove duplicated values in an array. [1,2,1,2,1,2] is not changed): Removing duplicates from a list in Haskell without elem. Here is the code that I have so far: remove numbers from a string haskell. This amount to rewriting nub on your own. Ord(comparing) import Data. Then, take an item at a time from the original list and add it to the new list. asked Apr 25, 2011 at 7:43. Please keep in mind I'm only learning SWI-Prolog for two days and only understand the basics of Prolog. Commented Apr 21, Explicit Recursion for Determining If Duplicates in Haskell. 0. Here the first function takes an element of type a and the "result" of the function of the rest of the list (so here the unique list of elements of the remaining elements). Why does infinite recursion happen? Edit: @augustss says: Always beware when you have multiple base cases for a function over lists. I looked up and found a close example, but the answer found in this link: Remove adjacent duplicate elements from a list won't run the test cases for this problem. We can do that using recursion. Not a homework question. The good news is: you don't use list comprehensions. To faithfully render the first snippet as a right fold, we fold into functions so we can pass a state argument along, occasionally updated to the new unique element of the list: I'm trying to delete all instances of an item in a list using haskell. Then in the surrounding recursive calls this return value is used here: (modArrayAll xs ys) \\ modArray n m This tries to remove further elements (those returned by modArray n m) from the already empty list returned by modArrayAll xs ys. 0. A. elem x xs it can be more readable to write I can't use recursion either. 36. It is easy to take product of lists using list comprehension and then the filter function will reduce the Recursion to remove duplicate characters. Check the leftmost character in the starting substring. 24. dupodd on the other hand only makes one copy of the head, and then perform recursion on dupeven. Hint: you'll need reverse at some point. +1 our Haskell newbies should be introduced to things like Writer much sooner than we typically teach them. However the program I made is not working for inputs containing duplicates. Like: dupeven :: [a] -> [a] dupeven [] = [] dupeven (x:xs) = x : x : dupodd xs Take also a look learn recursion and at this Haskell Tutorial 3 - recursion. nub :: Eq a => [a] -> [a] So that means that in order to filter out duplicates in a list of as, we need a to be an instance of the Eq class. For example when reimplementing the "all" function I can choose between: Anyway, a classic problem in imperative programming is stack reversal "without using extra space". Is there Solution 1: To delete duplicate elements in a list in Haskell, we can use the nub function from the Data. I'm working on a program that is supposed to sort a vector of ints and then get passed to a recursive function to remove any duplicates by having an element check it's neighbor, and Removing duplicates from a vector, recursion c++. Strict (Map) import qualified Data. addElem :: Ord a => a -> Map a Int -> Map a Int addElem = M. The solution is typically using the call stack in double recursion, to get to the bottom of the stack, take the bottom element, and bring it back up to the outer recursion which re-builds the stack in reverse order. Without sorting the list. removing duplicates from a list in haskell. Haskell - Create Set (unique sorted list) - no recursion, no nub. The function would take the equality test and the list as parameters, and would return the list of elements with no duplicates. nubSeq combines these to generate a sequence without duplicates The whole thing should be O(n*log(n)), I believe: module NubSeq where import Data. i have to make Haskell function called markDups that processes a string, replacing all repeated occurrences of a character with the underscore, "_", character. When learning new functions in Haskell, The next function removes duplicates from a sorted list. E. ") I have this so far. Follow edited Nov 30, 2020 at 20:50. I have based my code off of this code (Remove all consecutive duplicates from the string; language used: C++). Again, you can assume that you get good (whatever that means) answers recursively for the components when you're building a good answer for the whole thing. I couldn't add the recursive part. My question about binding in pattern-matched, then it removes them and proceeds (sublist xs ys). Modified 10 years, 10 months ago. That's why the variable is called seen. Explicit Recursion for Determining If Duplicates in Haskell. Modified 1 year, 8 months ago. In general, any recursive algorithm can always be reduced to an iterative one through the use of stack or queue data structures. This makes sense - the second query after the UNION ALL will return the previous levels as well as the new level of recursion, and UNION ALL doesn't duplicate. Prelude> :t (++) (++) :: [a] -> [a] -> [a] Meaning you can only append lists together (Remember that String is an alias for [Char], so it's a list). Eliminating the duplicates completely in Haskell. elemIndices val list returns an in-order list of indices, giving the occurrences of val in list. Set module for removing duplicates. For example:?- remove_duplicates([a,a,b,c,c], List). – Recursion is important to Haskell because unlike imperative languages, you do computations in Haskell by declaring what something is instead of declaring how you get it. If you want a recursive solution without any string position variables, you can try this (though it's pretty ineffective): Recursion - Removing Duplicates. The question is asking to remove duplicates from the string, I came up with a solution to remove duplicates but for that, I sorted the string. The nub function removes duplicate elements from a list, keeping only the first occurrence of each element. empty nub2::Ord a => [a] -> [a] nub2 thelist = map fst $ You cannot sort a list without its elements having some ordering -- meaning that they must be instances of Ord. e. Given 2 files, an input file and output file, I want to read from the input file, remove duplicate lines then write (unique lines) into the output file. Basically I want to traverse through the list, iL, and get the numbers in range from v1 and v2. The repetition is a job for. However, when you recurse into duplicateHelper the sublist that is passed no longer contains t at index 0 but instead contains the element that was previously compared. How to recursively remove all adjacent duplicates. org, the basic implementation looks like this:. The nub function removes duplicate elements from a list. 8,977 7 7 gold badges 27 27 silver badges 49 49 bronze badges. This is a small part of a bigger function. 0 I'm working on a method which removes duplicates of an element in an ArrayList, recursively. So you can keep running this program indefinitely without needing more than a fixed amount of memory. I am self learning. @jozefg I think it's a matter of personal preference - you could even write x !! 0 < x !! 1. The way to do it is maintain a Set containing all elements you have already encountered, and Write code to remove duplicates from an unsorted list WITHOUT a buffer. The first character must be different from its adjacent now. Currently my solution is basically as follows: hasDuplicates ts = (length I used Alexei Polkhanov's answer and came to the following, so you can remove duplicates from lists with a type that extends Eq class. Each time you emit a value however, you add it to the accumulator seen. Code is below: intersectB :: Eq a => [a] -> [a] -> [a] intersectB (x:xs) list | x `elem` list = x : intersectB xs list | otherwise = intersectB xs list Is there any way to use the structure of complex function is a complex investigation of the list. I'm really new to Haskell programming. So, in the type signature, we'll say a for the type of the accumulated value, v for the type of the value, and r for the type of the result. Splitting list into a list of possible tuples. Say for instance you call removeDuplicates [1,2,3,1,4,2]. How to implement if 'head' == " " haskell; recursion; or ask your own question. Time Complexity: O(N 2), In the worst case, the function may need to iterate almost through the entire string for each recursion, which makes the time complexity as O(N 2). Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about your product, service or employer brand; OverflowAI GenAI features for Teams; OverflowAPI Train & fine-tune LLMs; Labs The future of collective knowledge sharing; About the company Can we define a recursion scheme (without losing any of their generality) that constructs the value top-down, instead of bottom-up?. Try rewriting the function so that it uses "cons" instead. That's why there are no while loops or for loops in Haskell and instead we many times have to use recursion to declare what something is. elemIndex val list returns the index of the first occurrence, if any, of val in list as Just index. Related. A few notes: You handle the cases when the list is empty or has at least two elements (x and y), but you have forgotten the case when your list has exactly one element. – Removing duplicates from a list in Haskell without elem. I would solve it as such:-- Does not work for empty lists (so maybe needs to be wrapped in some logic) foldr1 min [-3,1,2,3] -- Works for empty but needs a "default value" (in this case 0) foldr min 0 [-3,1,2,3] If you want to learn by implementing it yourself, then this works for me complex function is a complex investigation of the list. I've tried . I'm working on a simple problem on Programming Praxis: remove all duplicates from a list without changing the order. You can do other things to deduplicate a list, Create function to remove duplicates from Haskell list. Haskell Remove duplicates from list. Or, if you really care about performance, define a However, my first attempt to solve the problem was the following, which only removes duplicates if they are consecutive (e. remove duplicate characters from a string in java without using string function. Ask Question Asked 11 years, 10 List2). How to place a heavy bike on a workstand without lifting more hot questions Question feed Subscribe to RSS Question feed To subscribe to How can I write a function in Haskell which takes a list and a number and it removes all the elements greater than that number and returns the list. An example would be: "|Hello|| My|| Name|| Is|| XYZ without using a let expression. dupeven thus will duplicate the first element, and the pass recursively to dupodd. If you want to solve it the most Haskell way. Auxiliary Space:O(N), As we are making a new string at every iteration. Then use recursion to scan the list and remove duplicates. , [2, 1, 4, 4, 3] should become [2, 1, 4, 3]. – hyleaus. Wwe can iterate with two pointers: “current” does a normal iteration, while “runner” iterates through all prior nodes to check for dups. If there is no value, or the list is empty, return -1. , findPrev 5 [1,2,3,4,5,6] The only issue is that I am doing this with explicit recursion. The function first checks if the length of dupList is 1 or 0. insert) Set. If it is present then. Hot Network Questions Am I somehow exempt from ETA and EES? Galfenol (FeGa) structure The code you posted is entirely relevant, it is exactly what you want Lets step through it: symEq takes two tuples and checks whether the first and last elements are equal OR if the elements across from each other are equal. Is it common practice to remove trusted certificate authorities (CA) Also note that this is a very slow approach. If you have some prior programming experience, elem should not be hard. So this is all I have so far: def remove_dups(thelist): When using recursion with list it is most of the time a problem of returning a sub-list or part of that list. I managed to read the file and remove duplicates now I'm having trouble writing to file. And there should only be 6 ways to make 17. (Of course, in this simple example you would just write map toUpper s. ] gives you 0 very quickly. In Haskell how do you remove a duplicate list with different order than the original? 3. Furthermore, this function will not work when used with infinite lists, because of the way foldl Thats very helpful, however is there anyway of doing this without using group, considering I need to use only prelude functions? If not, I suppose I can try to implement it. Then remove function takes out the list before the number and after the number using inbuilt 'take' and 'drop' function of the Remove explicit recursion from function with dependent values. If I do another round of recursion, the level 2 employees are also duplicated, and so on. Haskell function that tests if First let's consider that function f that you have. What I see is that when you first enter duplicateHelper, elements t is at position 0 of the passed list. List = [a,b,c] Yes. Checking for adjacent duplicates is much cheaper -- O(n). Ask Question Asked 11 years, 9 months ago. If the length of the string is zero or one then return the string. – Arthur. So essentially replace every recursive occurrence of Expr with a type parameter a, and then your Expr type becomes the least fixed point of this functor. Good luck and happy coding! I'm trying to do removing space (" ") from a string without using the strip function. Note also that instead of. Then count the unique digits of a number. I'm trying to find all ways to combine number in a certain group size without any repeats using Haskell. 32. I understand that I can change UNION ALL to UNION in order to remove duplicates, Beginner Haskell question: how do I find indices in Haskell (manually using recursion without using findIndices) Hot Network Questions A website asks you to enter a Microsoft/Google/Facebook password. 5. 1. but if you want to do the explicit recursion yourself, deleteAllInstances :: Eq a => a -> [a] -> [a] deleteAllInstances a (x:xs) Removing duplicates from a list in Haskell without elem. (define (remove-duplicates xs) (fold-right cons-if-not-duplicate '() (sort xs >))) Every time you hit an item, first check whether it's been seen before; if it is, stop the recursion; if it's not, add it to the set and proceed. 960 6 6 gold badges 17 17 silver badges 30 30 bronze badges. sortBy adHocOrd instead of nub, with an "inline definition" of your proposed Ord instance. If so, it would trivially imply that there is no possibility of any duplicates occurring. Follow edited Mar 27, 2019 at 12:24. If no, it removes head from the second list (sublist (x:xs I want to note that you can often avoid explicit recursion to remove unnecessary items from the front of the list with dropWhile. Haskell all possible ways to remove 1 element from the list. Filter Duplicate elements from a Here is some duplicate removing code, just for example (it doesn´t do the same thing as yours though) java how to avoid duplicate in recursion. Right now I'm working on a problem in Haskell in which I'm trying to check a list for a particular pair of values and return True/False depending on whether they are present in said list. Glove Glove. If the item we pick is already in the Yes, it is possible to remove duplicates from a list using pattern matching in Haskell. Skip to main content. nubBy is in the Data. – K. However it is working for inputs which doesn't have consecutive duplicates in them. Im trying to write a function to get the Intersect of two sets. Then on inserts, check membership first, using the measure of your full fingertree. A I'm trying to build a function which will remove all duplicates from a list. 4. However, the problem is that elements in my list rely on other elements in the same list. What would be the best method for this? I'd like to do this using my own function and not nub When it comes to removing duplicates from a list, multiple approaches are available. iterate :: (a -> a) -> a -> [a] Removing duplicates from a list in Haskell without elem. 17. Then i want to combine that with set B and get rid of all the duplicates in that set. text <- readFile I don't see how foldl could be used for this. Then sorted. Basically i combined the sets. You can do classic recursion (that reads like Haskell) without the global by recursing on the tail of the list: And recursion (this does not work Removing duplicates from a list in Haskell without elem. Commented Mar 24, 2012 at 20:59 Printing a result during a recursion in haskell. Create function to remove duplicates from Haskell list. I think it is rather strange, since usually one aims to either add, or delete, but not both depending on a condition. Here is what I have created. I have to remove consecutive characters in a string by recursion. To remove duplicates, we want something that takes a list [a] and removes things from it, returning [a] based on some predicate a -> a -> Bool. (I'm refraining from posting the corrected code since you would presumably like to fix it yourself, and thereby learn more - let me know if you need I am new to haskell and I'm trying to work out this problem. removing direct duplicates on a list. So I wanted to know is there a way of removing duplicates from a string without sorting the string. Recur for The terminating case of the recursion is modArrayAll [] [] = [], so there an empty list is returned. You currently use tail recursion and an accumulator which you grow on the right: it would be cleaner to grow directly on the left of the result of a (non-tail-) recursive call. What's the motivation for writing this function without explicit recursion? – Chris Taylor. To obtain only unique elements, first you need to sort each list of your list and then use remove duplicate elements : This library defines some lesser-used operations over lists. 7. You get better constant factors, but O(log(n)) deletes. Ask Question Asked 8 years ago. The function isolate deletes the duplicates of a list Recursion is assisted by pre and post conditions. (Edit: maybe you won't need filter. If you want to avoid library functions, you have to reimplement them in some form. Nothing is returned if not (val `elem` list). – If you're supposed to be removing duplicates, union is indeed a useful tool: if you apply it to two lists without duplicates, you get a list without duplicates. It is goining in infinite recursion and hence gives segmentation fault. I have a question though mainly in regards to when it is better to choose recursion over iteration. template<typename Iter> bool How many elements are the same in two lists, which have duplicate elements Hot Network Questions Why do BJT datasheets give a DC version and a small signal version of hFE? One way of achieving this is that we can store it in an array or so and then remove the duplicates. But it, too, is not hard to write. List (group) -- so we write group instead of You're not passing new_list as a parameter to the recursive function; therefore, each level of the recursion is unaware of the results you have gathered so far, and the result of the recursion is only the result of the first level. About; Haskell Recursion Function. If you're at the highestnumber, just add that to your current total and stop, otherwise, add the number to your total total + n and move on to the next one above n: add n total |highestnumber n = total + n |otherwise = add (above n) (total + n) Then you can do I would like to know how to remove specific duplicates from a string. removeDuplicates :: Eq a => [[a]] -> [[a]] uniq which removes duplicates from a list uniq :: Eq a => [a] -> [a] -> [a] uniq x [] = x uniq [] (a:xs) = uniq [a] xs uniq x (a:xs) = if a `elem` x then uniq x xs else uniq (a:x) xs filt Assume that we need to write a function to remove duplicated items in a list as a code snippet bellow So far I have this, but I feel that there should be a better way to remove duplicates from a list and to filter by non-letters: -- load file to be indexed and stop words. import Data. 51. 6] $ \_x -> randomRIO (1, 40) :: (IO Int) print rs this is halfway. [Expected Approach] Using Recursion – O(N 2) Time and O(N) Space: . ) What you seem to intend is: repeating the removeAdjDups, until no duplicates are found anymore. However I'd like to rework this without using elem. For me, coming from an imperative background, using x !! 1 implies that it's a O(1) operation, whereas for Haskell lists, it's O(n). Perhaps you'd be better off with an example involving addition where you stop as soon as the accumulator's zero, so you can't tell from the data that it will be: you want to ask for mult _ 0 = 0 not the other way round if you Two hints. This gives you O(log(n)) cons and snoc, O(1) deletes. I'm creating a function called getInRange that takes two ints and a list as parameters, v1, v2, and iL . Things that are always true at start and finish. Since the definition is recursive, you need a base case and a recursive case. One option is to utilize the Data. find returns the first element of a list that satisfies a predicate but I can't use list comprehensions, only high-order functions and lambdas. This means we're looking for a function with the following form: If the head character is a space, it will be the first space that we encounter and we need to remove it. elements_from_nth_areUnique(N, List) :- (length(List, Len), N > Len) %stoping condition for recursion ; (nth_elem_isUnique(N, List), M is N + 1 Now the haskell wiki has two suggestion for solving mutual recursion issues, but both of them are really more about mutually recursive types and neither of them is going to work here. From PHP to JavaScript to Kubernetes: how one Im practicing recursion on Haskell and need to do concat method with own impelentation on a nested list. dreamcrash dreamcrash. Here's one solution: Use a fingertree with the Set monoid as your measure. Alternatively, you can create your own This means we remove duplicates by taking a list (a:as), keeping the a, and to the rest, remove duplicates after removing all copies of a. rmCharsRec is the function that I am trying to write and rmChar is the other function. You can use foldr for this example, because foldr is non-strict, as is your mult operator, so foldr1 mult $ [1,2] ++ [0. In order to write remdups, it would be useful to have a function that tells us if a list contains an element. Ask Question Asked 10 years, 10 months ago. The nub function from Data. I'm trying to define a function which will remove duplicates from a list. if it hits the end of the list without ever seeing a, it returns False. head (tail x), on the other hand, clearly tells me that I'm performing two operations on a linked list. In the base case xs = [] and rem f xs = [], since the I am writing a recursive function called threeN that takes a list with one element and adds a new number depending on if it's even or odd. That's why Chris's answer is better, and the more Haskell-y way to approach it. If you check the type of (++) in ghci you get:. Here's my issue: type Symbol = String type Sentence = [[Symbol]] getSymbols :: devise a method to remove duplicates that uses sort :: Ord a => [a] -> [a] from Data. i want remove elements from a list but if i do this remove element from list without removing all duplicates. This solution also removes adjacent This is a perfect time to use Hoogle!Let's think about what you're looking for. You can do this to make nub act exactly like I'm guessing that you're a student, and this is a homework problem, so I'll give you part of the answer and let you finish it. haskell; recursion; or ask your own In order to remove duplicates, you need to somehow store what values you have already seen. recursion-schemes provides a template Haskell command that will do this for you - then you can use the functions defined If you want to do this without lists, keep a running total, and use recursion. Sequence Removing duplicates from a list in Haskell without elem. Duplicating elements in a list. Improve this answer. Haskell list remove duplicates doesn't work completely correctly. Recursion - Removing Duplicates. (Generally, foldl pretty much combines the disadvantages of foldr and foldl' those, or foldMap, are the folds you should normally be using, not foldl. group), and then take the first of each group (via map head, which applies head to each element of the list and gives you back the list of the results):. this is my thought process so far: I am studying a bit of functional programming using haskell and I am trying to go through some concepts by re-implementing some library functions. I need to remove consecutive duplicates from a string with a recursion method (for example, turning "aabbcddeghhi" into "abcdefghi. This question is only about removing consecutive duplicates, so you would not want to sort in this case. remove [5, 4, 3, 9, 1 ] 5 should return [5,4,3, Skip to main content. Haskell: replace characters in a string. . Here is an example code snippet demonstrating how to delete duplicate elements in a list: The other answers discuss how to achieve this by adding an accumulator. name <- I want to eliminate consecutive duplicates from a string like f "aaabbbcccdeefgggg" = "abcdefg". It returns triple: the first - all elements with removed duplicates (like sortUniq but the result is not sorted); the second - the elements that are repeated at least once in the list (result is the same as repeated but not sorted); the third - the unique elements that do not have duplicates (result is the same as unique but not sorted) Is there a convenient Haskell function for removing duplicate tuples from a list? Or perhaps it is something a bit more complicated such as iterating . How to delete an element from list and all his duplicates? 0. filter is built into the standard Prelude, so it's a The starting point is to work out a method to remove duplicates from a list. 2k 26 26 gold badges 109 109 silver badges 130 130 bronze badges. Removing duplicates from a list in Haskell without elem. or in my case I believe the function is counting the same combination more than once. Stack Overflow. If it's even divide it by 2, and if it's odd multiply it by three and subtract 1. You did his homework for him without really explain what he did wrong, or why it was wrong. You currently use map head and map tail to take whole lists of things apart: it might be better to use zipWith (:) to put whole lists of things together. I'm having trouble with my base case which will check the list to see if an element is already contained in the list. No explicit recursion may be used for eit Implement a search algorithm that searches through a list for Int n and returns the value in the list before n. This means we remove duplicates by taking a list (a:as), keeping the a, and to the rest, remove duplicates after removing all copies of a. The list has to be composed of elements that can be compared by some > operator. If I have the removeDuplicates() line that's commented, I get a strange I just started learning functional programming in SML and I want to know how I can combine the following two functions into a single function. Finding the first duplicate element in a list. here is my code i did so far. The type of hGetLine is Handle -> IO String, and the type of hGetLines should be IO [String] So you can not append these values. Is there anyway to get around this mutual recursion Well we can define two functions that perform mutual recursion: dupeven :: [a] -> [a] and dupodd :: [a] -> [a]. Map as Map import Data. But after tests with 1,000,000 elements it took very long time to finish. Suppose I start with the list [1, 2, 1, 3, 2, 4] and I only want to get [3,4] without repeated elements. dupList is the list possibly containing duplicate elements, and temp is initially an empty list. You will always reach this case because you are calling your function recursively on smaller lists. myConcat :: [[a]] Implement a function like concat in Haskell without using any prelude functions. It works with a global variable, however, I find that rather weak work around. Sequence as Seq import Data. counting recursive calls - Haskell. Replacing explicit recursion with higher-order functions. Call 'remove' function with a number and a list as parameters. Set as Set buildsets::Ord a => [a] -> [Set a] buildsets = scanl (flip Set. Deleting elements of a Simulating a lottery of 6 numbers chosen from 40, I want to create a list of numbers in Haskell using the system random generator but eliminate duplicates, which often arise. If I were simply interested in the first 50 elements of the sorted list xs, then I would use fst (splitAt 50 (sort xs)). e. In the above code, remove_temp function returns the index at which the number is present in the list. Commented Nov 22, Python Removing duplicates ( and not keeping them) in a list. List package, just do import Data. It takes some sort of accumulated value, a plain value, and combines them into a result. replacing an element in a list of lists in haskell. everyday Haskell is not about recursion, it's more about recursive patterns and precise formulation of a problem. hsbo sjqr pqpdlhl rhmgz dvfkcu thsf fpex ejwp oykcumz leipdz