I want the intersection between two lists but it is only prints when they are at the same position.
inter [] _ = []
inter _ [] = []
inter (x:xs) (y:ys) = if (x == y) then x:inter xs ys else inter xs ys
As already mentioned in the comments, your function iterates over both lists at the same time and appends elements to the result list, if the same element appears in both lists at the same location.
From what I've understood, what you really want to do is to add all elements from the first list that appear anywhere in the second list. So you may want to define (or use) another function that checks whether an element appears somewhere in a list, e.g. if you are having a list of Ints:
elem :: Int -> [Int] -> Bool
or if you are already aware of what type classes are:
elem :: Eq a => a -> [a] -> Bool
Using that function, you can process the first list element-wise and filter out all elements from that list that appear somewhere in the second list.
As you are obviously trying to learn Haskell, I do not want to spoil the experience for you and provide a more complete solution.
Related
My task is to implement a function allCombinations, in Haskell, which returns every possible combination of a given list as an infinite list.
Since i couldn't find a solution, which does not include imports, i will ask it again.
I'm new to Haskell and my task is to create a function allCombinations :: [a] -> [[a]], which creates an infinite list of possible combinations. For Example: take 10 (allCombinations [True,False]),
in return I should get [[],[True],[False],[True,True],[False,True],[True,False],[False,False],[True,True,True],[False,True,True],[True,False,True]].
I have found a few solutions, but all of them included imports and i need to solve this without any imports.
You can work with recursion here. This starts with the empty list, and then you recurse where you prepend with the given list, and the items in the list, so:
allCombinations :: [a] -> [[a]]
allCombinations xs = zs
where zs = [] : [ (y:ys) | ys <- zs, y <- xs ]
I want to create a function that returns every third int from a list of ints without using any predefined functions. For example, everyThird [1,2,3,4,5] --> [1,4]
everyThird:: [a] -> [a]
Could I just continue to iterate over the list using tail and appending to a new list every third call? I am new to Haskell and very confused with all of this
One other way of doing this is to handle three different base cases, in all of which we're at the end of the list and the list is less than three elements long, and one recursive case, where the list is at least three elements long:
everyThird :: [a] -> [a]
everyThird [] = []
everyThird [x] = [x]
everyThird [x, _] = [x]
everyThird (x:_:_:xs) = x:everyThird xs
You want to do exactly what you said: iterate over the list and include the element only on each third call. However, there's a problem. Haskell is a funny language where the idea of "changing" a variable doesn't make sense, so the usual approach of "have a counter variable i which tells us whether we're on the third element or not" won't work in the usual way. Instead, we'll create a recursive helper function to maintain the count for us.
everyThird :: [Int] -> [Int]
everyThird xs = helper 0 xs
where helper _ [] = []
helper 0 (x : xs) = x : helper 2 xs
helper n (_ : xs) = helper (n - 1) xs
We have three cases in the helper.
If the list is empty, stop and return the empty list.
If the counter is at 0 (that is, if we're on the third element), make a list starting with the current element and ending with the rest of the computation.
If the counter is not at zero, count down and continue iteration.
Because of the way pattern matching works, it will try these three statements in order.
Notice how we use an additional argument to be the counter variable since we can't mutate the variable like we would in an imperative language. Also, notice how we construct the list recursively; we never "append" to an existing list because that would imply that we're mutating the list. We simply build the list up from scratch and end up with the correct result on the first go round.
Haskell doesn't have classical iteration (i.e. no loops), at least not without monads, but you can use similar logic as you would in a for loop by zipping your list with indexes [0..] and applying appropriate functions from Data.List.
E.g. What you need to do is filter every third element:
everyThirdWithIndexes list = filter (\x -> snd x `mod` 3 == 0) $ zip list [0..]
Of course you have to get rid of the indexes, there are two elegant ways you can do this:
everyThird list = map (fst) . everyThirdWithIndexes list
-- or:
everyThird list = fst . unzip . everyThirdWithIndexes list
If you're not familiar with filter and map, you can define a simple recursion that builds a list from every first element of a list, drops the next two and then adds another from a new function call:
everyThird [] = [] -- both in case if the list is empty and the end case
everyThird (x:xs) = x : everyThird (drop 2 xs)
EDIT: If you have any questions about these solutions (e.g. some syntax that you are not familiar with), feel free to ask in the comments. :)
One classic approach:
everyThird xs = [x | (1,x) <- zip (cycle [1..3]) xs]
You can also use chunksOf from Data.List.Split to seperate the lists into chunks of 3, then just map the first element of each:
import Data.List.Split
everyThird :: [a] -> [a]
everyThird xs = map head $ chunksOf 3 xs
Which works as follows:
*Main> everyThird [1,2,3,4,5]
[1,4]
Note: You may need to run cabal install split to use chunksOf.
let's say i have a list like this:
["Questions", "that", "may", "already", "have", "your", "correct", "answer"]
and want to have this:
[("Questions","that"),("may","already"),("have","your"),("correct","answer")]
can this be done ? or is it a bad Haskell practice ?
For a simple method (that fails for a odd number of elements) you can use
combine :: [a] -> [(a, a)]
combine (x1:x2:xs) = (x1,x2):combine xs
combine (_:_) = error "Odd number of elements"
combine [] = []
Live demo
Or you could use some complex method like in an other answer that I don't really want to understand.
More generic:
map2 :: (a -> a -> b) -> [a] -> [b]
map2 f (x1:x2:xs) = (f x1 x2) : map2 f xs
map2 _ (_:_) = error "Odd number of elements"
map2 _ [] = []
Here is one way to do it, with the help of a helper function that lets you drop every second element from your target list, and then just use zip. This may not have your desired behavior when the list is of odd length since that's not yet defined in the question.
-- This is just from ghci
let my_list = ["Questions", "that", "may", "already", "have", "your", "correct", "answer"]
let dropEvery [] _ = []
let dropEvery list count = (take (count-1) list) ++ dropEvery (drop count list) count
zip (dropEvery my_list 2) $ dropEvery (tail my_list) 2
[("Questions","that"),("may","already"),("have","your"),("correct","answer")
The helper function is taken from question #6 from 99 Questions., where there are many other implementations of the same idea, probably many with better recursion optimization properties.
To understand dropEvery, it's good to remember what take and drop each do. take k some_list takes the first k entries of some_list. Meanwhile drop k some_list drops the first k entries.
If we want to drop every Nth element, it means we want to keep each run of (N-1) elements, then drop one, then do the same thing again until we are done.
The first part of dropEvery does this: it takes the first count-1 entries, which it will then concatenate to whatever it gets from the rest of the list.
After that, it says drop count (forget about the N-1 you kept, and also the 1 (in the Nth spot) that you had wanted to drop all along) -- and after these are dropped, you can just recursively apply the same logic to whatever is leftover.
Using ++ in this manner can be quite expensive in Haskell, so from a performance point of view this is not so great, but it was one of the shorter implementations available at that 99 questions page.
Here's a function to do it all in one shot, which is maybe a bit more readable:
byTwos :: [a] -> [(a,a)]
byTwos [] = []
byTwos xs = zip firsts seconds
where enumerated = zip xs [1..]
firsts = [fst x | x <- enumerated, odd $ snd x]
seconds = [fst x | x <- enumerated, even $ snd x]
In this case, I started out by saying this problem will be easy to solve with zip if I just already had the list of odd-indexed elements and the list of even-indexed elements. So let me just write that down, and then worry about getting them in some where clause.
In the where clause, I say first zip xs [1..] which will make [("Questions", 1), ("that", 2), ...] and so on.
Side note: recall that fst takes the first element of a tuple, and snd takes the second element.
Then firsts says take the first element of all these values if the second element is odd -- these will serve as "firsts" in the final output tuples from zip.
seconds says do the same thing, but only if the second element is even -- these will serve as "seconds" in the final output tuples from zip.
In case the list has odd length, firsts will be one element longer than seconds and so the final zip means that the final element of the list will simply be dropped, and the result will be the same as though you called the function on the front of the list (all but final element).
A simple pattern matching could do the trick :
f [] = []
f (x:y:xs) = (x,y):f(xs)
It means that an empty list gives an empty list, and that a list of a least two elements returns you a list with a couple of these two elements and then application of the same reasoning with what follows...
Using chunk from Data.List.Split you can get the desired result of pairing every two consecutive items in a list, namely for the given list named by xs,
import Data.List.Split
map (\ys -> (ys!!0, ys!!1)) $ chunk 2 xs
This solution assumes the given list has an even number of items.
I have an exercise to do, however and since I'm new to the language, I can not find any way on how to do it.
I have this function "repeated" that's defined as such, given under this paragraph. It receives a list of Int and gives back a Bool value. It's supposed to check if a list has any repeated elements. If so, it's TRUE, if not, it's false. There is one extra: I have to define the function by recursion, so it has to be a recursive function. Would appreciate any help.
repeated :: [Int] -> Bool
EDIT1: So far, I've only managed to succeed with this amount of code
repeated :: [Int] -> Bool
repeated [] = False
repeated (h:t) =
Which gives me back the empty list, only. The rest, I've not been able to figure out so far...
EDIT2: Forgot about the singular lists... Also, possible answer?
repeated :: [Int] -> Bool
repeated [] = False
repeated [_] = False
repeated (h:t) = if elem h t then True
else repeated t
That's pretty much it. I've compiled the .hs and it worked perfectly. Thank you all for the suggestions and hints! :)
You want to find if a list has any duplicates. This means that you'll have to keep up with a list of elements that you've already visited so you can check this. So first, write a function that checks if a single element exists in a list of already visited values:
alreadyVisited :: Int -> [Int] -> Bool
alreadyVisited x [] = False
alreadyVisited x (v:visited) = ???
(Note: this is known as elem in Prelude, but you should be able to implement it yourself, and it's good practice)
Then you'll want to write the main function that loops over all elements in your target list, building a set of visited elements until it finds a duplicate. Once a duplicate is found, the function can exit without checking the rest of the list.
-- Using a helper hides the fact that the visited list is needed
repeated :: [Int] -> Bool
repeated xs = go xs []
-- ^-- initial visited list is empty
where
-- same base case that you came up with,
-- an empty list does not have duplicate elements
go [] _ = False
-- The recursive step, think about what you need this function to do
go (x:xs) visited =
if alreadyVisited x visited
then ??? -- If it's already visited, do what?
else ??? -- Otherwise?
Here I've just set up the structure for you, you'll have to fill in the details yourself. Keep in mind that this is not an efficient implementation, particularly because of how slow alreadyVisited will become as visited grows in size, but if you are interested in speed then you can swap out the visited list for a Data.Set.Set, which has much better lookup time.
This is my approach for this (using Sets and comparing lengths)
import qualified Data.Set as Set -- From the 'containers' library
hasDuplicates :: (Ord a) => [a] -> Bool
hasDuplicates list = length list /= length set
where set = Set.fromList list
I'm using the containers Haskell Package.
Try using nub.
import Data.List
hasDuplicates :: (Ord a) => [a] -> Bool
hasDuplicates xs = length (nub xs) /= length xs
Essentially, nub will return the unique elements of a list.
The language I'm using is a subset of Haskell called Core Haskell which does not allow the use of the built-in functions of Haskell. For example, if I were to create a function which counts the number of times that the item x appears in the list xs, then I would write:
count = \x ->
\xs -> if null xs
then 0
else if x == head xs
then 1 + count x(tail xs)
else count x(tail xs)
I'm trying to create a function which outputs a list xs with its duplicate values removed. E.g. remdups (7:7:7:4:5:7:4:4:[]) => (7:4:5:[])
can anyone offer any advice?
Thanks!
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. In order to write remdups, it would be useful to have a function that tells us if a list contains an element. We can do that using recursion. When using recursion, start by asking yourself what the "base case", or simplest possible case is. Well, when the list is empty, then obviously the answer is False (no matter what the character is). So now, what if the list isn't empty? We can check if the first character in the list is a match. If it is, then we know that the answer is True. Otherwise, we need to check the rest of the list -- which we do by calling the function again.
elem _ [] = False
elem x (y:ys) = if x==y
then True
else elem x ys
The underscore (_) simply means "I'm not going to use this variable, so I won't even bother to give it a name." That can be written more succinctly as:
elem _ [] = False
elem x (y:ys) = x==y || elem x ys
Writing remdups is a little tricky, but I suspect your teacher gave you some hints. One way to approach it is to imagine we're partway through processing the list. We have part of the list that hasn't been processed yet, and part of the list that has been processed (and doesn't contain any duplicates). Suppose we had a function called remdupHelper, which takes those two arguments, called remaining and finished. It would look at the first character in remaining, and return a different result depending on whether or not that character is in finished. (That result could call remdupHelper recursively). Can you write remdupHelper?
remdupHelper = ???
Once you have remdupHelper, you're ready to write remdups. It just invokes remdupHelper in the initial condition, where none of the list has been processed yet:
remdups l = remdupHelper l [] -- '
This works with Ints:
removeDuplicates :: [Int] -> [Int]
removeDuplicates = foldr insertIfNotMember []
where
insertIfNotMember item list = if (notMember item list)
then item : list
else list
notMember :: Int -> [Int] -> Bool
notMember item [] = True
notMember item (x:xs)
| item == x = False
| otherwise = notMember item xs
How it works should be obvious. The only "tricky" part is that the type of foldr is:
(a -> b -> b) -> b -> [a] -> b
but in this case b unifies with [a], so it becomes:
(a -> [a] -> [a]) -> [a] -> [a] -> [a]
and therefore, you can pass the function insertIfNotMember, which is of type:
Int -> [Int] -> [Int] -- a unifies with Int