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In order to split an arbitrary set s into two subsets l and r, in which l has a fixed size n, I have written the following code:
parts :: Int -> [a] -> [([a], [a])]
parts n ls = (filter len . f) ls
where len (lhs,rhs) = length lhs==n -- the condition that the left subset is of fixed size n
f [] = [([],[])]
f (x:xs) = [ (x:l,r) | (l,r)<-f xs] ++ [ (l,x:r) | (l,r)<-f xs]
Here is a sample of its effect. Evaluating parts 2 "12345" yields:
[ ("12","345")
, ("13","245")
, ("14","235")
, ("15","234")
, ("23","145")
, ("24","135")
, ("25","134")
, ("34","125")
, ("35","124")
, ("45","123")
]
Please notice that my solution enumerates all subsets and then filters out the desired ones. I suppose the subsets function is familiar to you:
subsets :: [a] -> [[a]]
subsets [] = [[]]
subsets (x:xs) = map (x:) (subsets xs) ++ subsets xs
Personally, I find my solution disappointing. It filters the correct answers from a larger set. My question to the reader is:
Can you come up with a function that is equivalent to parts, but produces the answer directly without this a-posteriory filter?
Inspired by subsets, you might end up with
import Control.Arrow (first, second)
-- first f (x,y) = (f x, y) ; second f (x, y) = (x, f y)
parts n xs = parts' 0 xs where
parts' l (x:xs) | l < n = map (first (x:)) (parts' (l + 1) xs) ++ -- 1a
map (second (x:)) (parts' l xs) -- 1b
parts' l xs | l == n = [([],xs)] -- 2
parts' _ _ = [] -- 3
l contains the length of the first pair so far. As long as the pair isn't yet long enough, we take the first element of the list, and append it to all first elements in our pairs (1a). We're also going to map it onto the second elements (1b). Note that in this case the length of the first pairs didn't increase.
When the first pairs just happen to be long enough (2), we're going to put all other elements into the second half of the pair.
When the requirements for the guards do not hold (list exhausted), we return [] (3). This approach also retains the relative ordering of elements:
> parts 2 "12345"
[
("12","345"),
("13","245"),
("14","235"),
("15","234"),
("23","145"),
("24","135"),
("25","134"),
("34","125"),
("35","124"),
("45","123")
]
This approach will also work on infinite lists:
> map (second (const "...")) $ take 5 $ parts 3 [1..]
[([1,2,3],"..."),([1,2,4],"..."),([1,2,5],"..."),([1,2,6],"..."),([1,2,7],"...")]
(The second elements in the pairs will still be infinite lists)
Ok this is what I came up with:
parts :: Int -> [a] -> [([a],[a])]
parts n list = parts' 0 (length list) [] [] list where
parts' _ _ ls rs [] = [(ls,rs)]
parts' n' l ls rs as#(x:xs) | n' >= n = [(reverse ls, reverse rs ++ as)]
| n' + l <= n = [(reverse ls ++ as, reverse rs)]
| otherwise = parts' (n' + 1) (l - 1) (x : ls) rs xs
++ parts' n' (l - 1) ls (x : rs) xs
If it doesn't matter if the elements of the subsets are in the same order as they were in the original set, then you can remove the four uses of reverse.
Sliding split can be done using zipWith of inits and tails. For each split produce the solution for a smaller sublist, and append the element at the point of split to all such solutions.
parts 0 xs = [([],xs)]
parts n xs = concat $ zipWith f (inits xs) (init $ tails xs) where
f hs (h:ts) = [(h:t', hs++ts') | (t', ts') <- parts (n-1) ts]
-- f hs [] not possible - init $ tails xs does not produce empty lists
how can i get the most frequent value in a list example:
[1,3,4,5,6,6] -> output 6
[1,3,1,5] -> output 1
Im trying to get it by my own functions but i cant achieve it can you guys help me?
my code:
del x [] = []
del x (y:ys) = if x /= y
then y:del x y
else del x ys
obj x []= []
obj x (y:ys) = if x== y then y:obj x y else(obj x ys)
tam [] = 0
tam (x:y) = 1+tam y
fun (n1:[]) (n:[]) [] =n1
fun (n1:[]) (n:[]) (x:s) =if (tam(obj x (x:s)))>n then fun (x:[]) ((tam(obj x (x:s))):[]) (del x (x:s)) else(fun (n1:[]) (n:[]) (del x (x:s)))
rep (x:s) = fun (x:[]) ((tam(obj x (x:s))):[]) (del x (x:s))
Expanding on Satvik's last suggestion, you can use (&&&) :: (b -> c) -> (b -> c') -> (b -> (c, c')) from Control.Arrow (Note that I substituted a = (->) in that type signature for simplicity) to cleanly perform a decorate-sort-undecorate transform.
mostCommon list = fst . maximumBy (compare `on` snd) $ elemCount
where elemCount = map (head &&& length) . group . sort $ list
The head &&& length function has type [b] -> (b, Int). It converts a list into a tuple of its first element and its length, so when it is combined with group . sort you get a list of each distinct value in the list along with the number of times it occurred.
Also, you should think about what happens when you call mostCommon []. Clearly there is no sensible value, since there is no element at all. As it stands, all the solutions proposed (including mine) just fail on an empty list, which is not good Haskell. The normal thing to do would be to return a Maybe a, where Nothing indicates an error (in this case, an empty list) and Just a represents a "real" return value. e.g.
mostCommon :: Ord a => [a] -> Maybe a
mostCommon [] = Nothing
mostCommon list = Just ... -- your implementation here
This is much nicer, as partial functions (functions that are undefined for some input values) are horrible from a code-safety point of view. You can manipulate Maybe values using pattern matching (matching on Nothing and Just x) and the functions in Data.Maybe (preferable fromMaybe and maybe rather than fromJust).
In case you would like to get some ideas from code that does what you wish to achieve, here is an example:
import Data.List (nub, maximumBy)
import Data.Function (on)
mostCommonElem list = fst $ maximumBy (compare `on` snd) elemCounts where
elemCounts = nub [(element, count) | element <- list, let count = length (filter (==element) list)]
Here are few suggestions
del can be implemented using filter rather than writing your own recursion. In your definition there was a mistake, you needed to give ys and not y while deleting.
del x = filter (/=x)
obj is similar to del with different filter function. Similarly here in your definition you need to give ys and not y in obj.
obj x = filter (==x)
tam is just length function
-- tam = length
You don't need to keep a list for n1 and n. I have also made your code more readable, although I have not made any changes to your algorithm.
fun n1 n [] =n1
fun n1 n xs#(x:s) | length (obj x xs) > n = fun x (length $ obj x xs) (del x xs)
| otherwise = fun n1 n $ del x xs
rep xs#(x:s) = fun x (length $ obj x xs) (del x xs)
Another way, not very optimal but much more readable is
import Data.List
import Data.Ord
rep :: Ord a => [a] -> a
rep = head . head . sortBy (flip $ comparing length) . group . sort
I will try to explain in short what this code is doing. You need to find the most frequent element of the list so the first idea that should come to mind is to find frequency of all the elements. Now group is a function which combines adjacent similar elements.
> group [1,2,2,3,3,3,1,2,4]
[[1],[2,2],[3,3,3],[1],[2],[4]]
So I have used sort to bring elements which are same adjacent to each other
> sort [1,2,2,3,3,3,1,2,4]
[1,1,2,2,2,3,3,3,4]
> group . sort $ [1,2,2,3,3,3,1,2,4]
[[1,1],[2,2,2],[3,3,3],[4]]
Finding element with the maximum frequency just reduces to finding the sublist with largest number of elements. Here comes the function sortBy with which you can sort based on given comparing function. So basically I have sorted on length of the sublists (The flip is just to make the sorting descending rather than ascending).
> sortBy (flip $ comparing length) . group . sort $ [1,2,2,3,3,3,1,2,4]
[[2,2,2],[3,3,3],[1,1],[4]]
Now you can just take head two times to get the element with the largest frequency.
Let's assume you already have argmax function. You can write
your own or even better, you can reuse list-extras package. I strongly suggest you
to take a look at the package anyway.
Then, it's quite easy:
import Data.List.Extras.Argmax ( argmax )
-- >> mostFrequent [3,1,2,3,2,3]
-- 3
mostFrequent xs = argmax f xs
where f x = length $ filter (==x) xs
I am trying a problem recently. And in this case I am having few problems.
Input: generatingListforRightShifting 2 [1,2,3,4,5,6,7,8]
Output: [[1,2,3,4,5,6,7,8],[8,1,2,3,4,5,6,7],[7,8,1,2,3,4,5,6]]
As you understand this program will shift an element in right direction. The 1st argument indicates how many times it will do shifting.
As a newbie I am trying solving it few well known list functions. and using recursion. But to me recursion idea is not clear. My code is:
generatingListforRightShifting' _ []=[]
generatingListforRightShifting' 0 x=x
generatingListforRightShifting' n xs= newList where
newList=takeWhile(\i->[1..n]
<=n)(reverse(take
i(reverse xs))++reverse(drop i(reverse xs)))
I understand that the main mistake I'm doing is in the part takeWhile. But how can I iterate through n times. I have already made a program which directly shows the shifted result such as
Input:generatingListforRightShifting 2 [1,2,3,4,5,6,7,8]
Output: [7,8,1,2,3,4,5,6]
But when I try to get all previous shifting I cannot.
Can anyone help me out here. I also welcome you if you give me the solving idea.
This is more commonly known as rotating instead of shifting. Rotating the list right once is simple, as there are methods to get the last element and the the sublist of all elements but the last.
rotateOnce lst = (last lst):(init lst)
Also note that the rotating twice is equivalent to calling rotateOnce twice. Therefore, the method could be implemented simply as a recursion from the previous result:
rotateN 0 lst = [lst]
rotateN n lst = lst : rotateN (n-1) ((last lst):(init lst))
(Note: that may not be the most optimal solution.)
You can define "shift" recursively: shift 0 is a no-op, shift 1+n (x:xs) is shift n xs.
Something like:
shift 0 = \x -> x
shift n = \lst#(x:xs) -> (shift (n-1) xs)
-- example:
sh3 = shift 3
Then the 'rotate' problem becomes easier:
rotate n = \lst -> (shift lst) ++ (take n lst)
You seem to prefer that we fix your code than start again, so
let's have a look at your code. Firstly, the main list chopping:
reverse (take i (reverse xs)) ++ reverse (drop i (reverse xs))
Now reverse (take i (reverse xs)) takes i elements from the end of the list,
but you reverse the list twice to achieve this, and it would be better to do
drop (length xs - i) xs. Similarly, you can implement reverse (drop i (reverse xs)))
as take (length xs - i) xs. That gives us
drop (length xs - i) xs ++ take (length xs - i) xs
Now your code \i->[1..n]<=n doesn't make sense because it compares the list [1..n]
with n, which can't work. I think you're trying to make a loop where i runs from
1 to n, which is a good plan. Let's use a list comprehension to get the ones we wanted:
[drop (length xs - i) xs ++ take (length xs - i) xs | i <- [1 .. length xs], i <= n]
but now we're running from 1 to the length of the list but throwing away numbers above n,
which would be better written
[drop (length xs - i) xs ++ take (length xs - i) xs | i <- [1..n]]
This does allow n to be more than length xs, but I don't see a big issue there, we could check that at first.
Notice now that we're only using i in the form (length xs - i), and really we're recalculating
length xs an awful lot more than we should, so instead of letting i run from 1 to n, and using
length xs - i, why don't we just have j=length xs -i so j runs from length xs to length xs - n:
[drop j xs ++ take j xs | j <- [length xs,length xs - 1 .. length xs - n]]
which works because for example [6,5..1] == [6,5,4,3,2,1]
It would be neater to do
let l = length xs in
[drop j xs ++ take j xs | j <- [l,l - 1 .. l - n]]
or maybe you like to take more than you like to do arithmetic, so we could use:
let l = length xs in
take n [drop j xs ++ take j xs | j <- [l,l - 1 .. 0]]
which has the added benefit of stopping you doing too many, stopping
you when you get back to the start.
I'd rename your function from generatingListforRightShifting to rotationsR, giving
rotationsR n xs = let l = length xs in
take n [drop j xs ++ take j xs | j <- [l,l - 1 ..]]
Which gives rotationsR 6 [1..4] == [[1,2,3,4],[4,1,2,3],[3,4,1,2],[2,3,4,1],[1,2,3,4]].
Left rotation would look simpler:
rotationsL n xs = take n [drop j xs ++ take j xs | j <- [0..length xs]]
Digression: I couldn't help myself, sorry, and I started again.
I still don't like all that dropping and taking every single time, I'd rather pop
infinitely many copies of xs next to each other (cycle xs) and take infinitely
many tails of that, chopping them all to the right length, but just give you the first n:
rotationsL' n xs = let l = length xs in
take n . map (take l) . tails . cycle $ xs
Because of lazy evaluation, only a finite amount of cycle xs ever gets calculated,
but this one can run and run: rotationsL' 10 [1..4] gives you:
[[1,2,3,4],[2,3,4,1],[3,4,1,2],[4,1,2,3],[1,2,3,4],[2,3,4,1],[3,4,1,2],[4,1,2,3],[1,2,3,4],[2,3,4,1]]
It would be nice to do the right roations that way too, but it doesn't work because
I'd need to start at the end of an infinite list and work my way back. Let's reuse
your reverse, take what you need, reverse trick again, though:
rotationsR' n xs = let l = length xs in
take n . map (reverse.take l) . tails . cycle . reverse $ xs
Undigression: If you'd rather stick more closely to your original code, you can do
generatingListforRightShifting n xs =
[reverse (take i (reverse xs)) ++ reverse (drop i (reverse xs)) | i <- [1..n]]
I would drop the current approach, which is very convoluted. Instead, focus on abstracting the different components of the operation. If you break the operation into parts, you will notice that there are two symmetric components: rotating the list to the left, and rotating the list to the right. The operation you wish to define iterates the right rotation a specified number of times over some list. This suggests that the desired operation can be defined by taking a specified number of iterations of either the left or right rotation. For example,
left :: [a] -> [a]
left [] = []
left xs = tail xs ++ [head xs]
right :: [a] -> [a]
right [] = []
right xs = last xs : init xs
shiftL :: Int -> [a] -> [[a]]
shiftL n = take n . iterate left
shiftR :: Int -> [a] -> [[a]]
shiftR n = take n . iterate right
Using cycle here seems nice:
shifts n xs = take (n+1) $ shifts' (cycle xs)
where
len = length xs
shifts' ys = take len ys:shifts' (drop (len-1) ys)
I find a left rotation to be very straight forward using splitAt:
import Data.Tuple (swap)
rotateLeft n = uncurry (++) . swap . splitAt n
> rotateLeft 2 "Hello World!"
>>> "llo World!He"
The function I'm trying to write should remove the element at the given index from the given list of any type.
Here is what I have already done:
delAtIdx :: [x] -> Int -> [x]
delAtIdx x y = let g = take y x
in let h = reverse x
in let b = take (((length x) - y) - 1) h
in let j = g ++ (reverse b)
in j
Is this correct? Could anyone suggest another approach?
It's much simpler to define it in terms of splitAt, which splits a list before a given index. Then, you just need to remove the first element from the second part and glue them back together.
reverse and concatenation are things to avoid if you can in haskell. It looks like it would work to me, but I am not entirely sure about that.
However, to answer the "real" question: Yes there is another (easier) way. Basically, you should look in the same direction as you always do when working in haskell: recursion. See if you can make a recursive version of this function.
Super easy(I think):
removeIndex [] 0 = error "Cannot remove from empty array"
removeIndex xs n = fst notGlued ++ snd notGlued
where notGlued = (take (n-1) xs, drop n xs)
I'm a total Haskell noob, so if this is wrong, please explain why.
I figured this out by reading the definition of splitAt. According to Hoogle, "It is equivalent to (take n xs, drop n xs)". This made me think that if we just didn't take one extra number, then it would be basically removed if we rejoined it.
Here is the article I referenced Hoogle link
Here's a test of it running:
*Main> removeIndex [0..10] 4
[0,1,2,4,5,6,7,8,9,10]
deleteAt :: Int -> [a] -> [a]
deleteAt 0 (x:xs) = xs
deleteAt n (x:xs) | n >= 0 = x : (deleteAt (n-1) xs)
deleteAt _ _ = error "index out of range"
Here is my solution:
removeAt xs n | null xs = []
removeAt (x:xs) n | n == 0 = removeAt xs (n-1)
| otherwise = x : removeAt xs (n-1)
remove_temp num l i | elem num (take i l) == True = i
| otherwise = remove_temp num l (i+1)
remove num l = (take (index-1) l) ++ (drop index l)
where index = remove_temp num l 1
Call 'remove' function with a number and a list as parameters. And you'll get a list without that number as output.
In the above code, remove_temp function returns the index at which the number is present in the list. Then remove function takes out the list before the number and after the number using inbuilt 'take' and 'drop' function of the prelude. And finally, concatenation of these two lists is done which gives a list without the input number.
Just starting out with Haskell! As an exercise, the current problem I'm trying to implement is as follows:
We have n squares, print all possible world configurations where :
(1) Each square could have a "P" (pit) or not (2^n possibilities).
(2) There can be at most one "W" (wumpus) in all n squares (n+1 possibilities).
Representing two squares as two strings, here is an output example for n=2. We have (2^n)·(n+1) = (2^2)·(2+1) = 12 configurations.
[[" W"," "],[" "," W"],[" "," "],
[" W","P"],[" ","PW"],[" ","P"],
["PW"," "],["P"," W"],["P"," "],
["PW","P"],["P","PW"],["P","P"]]
Condition (1) is easily implemented. Looking around, I've found a few ways to express it :
p 0 = [[]]
p n = [x:xs | x <- [" ","P"], xs <- p (n-1)]
or
p n = mapM (\x -> [" ","P"]) [1..n]
or
p n = replicateM n [" ","P"]
I cannot claim to understand the last two yet, but here they are for completeness.
Question : How can I add condition (2)? Can it be done with list comprehension?
My not-so-good-looking novice solution involved these functions:
insertw :: Int -> [String] -> [String]
insertw n xs
| n < 0 = xs
| n >= lgth = xs
| otherwise = (take (n) xs) ++ [xs!!n++"W"] ++ (drop (n+1) xs)
where lgth = length xs
duplicate :: Int -> [String] -> [[String]]
duplicate i squares
| i > lgth = []
| otherwise = (insertw i squares) : duplicate (i+1) squares
where lgth = length squares
worlds :: Int -> [[String]]
worlds n = concat . map (duplicate 0) . p $ n
Condition 2 isn't an obvious candidate for a list comprehension, but the working code you have already written can be cleaned up.
The iteration from 0 to lgth in duplicate can be done with a map instead of explicit recursion:
duplicate squares = map (\i -> insertw i squares) [0 .. length squares]
duplicate no longer takes an index parameter, and concat . map is the same as concatMap:
worlds = concatMap duplicate . p
If you do both a drop and a take, then splitAt is often the better operation.
insertw n xs =
case splitAt n xs of
(as, []) -> as
(as, b : bs) -> as ++ ((b ++ "W") : bs)
Note that we got rid of the length xs and xs !! n operations too.
As an exercise, another short duplicate function can be written by zipping over the inits and tails of the squares list.
Seems obvious to me :). In list comprehensions, the later lists can depend on the values generated in the earlier ones. The second function generates your set by calling the first when it adds a wumpus..
p 0 = [[]]
p n = [[x,' ']:xs | x <- [' ','P'], xs <- p (n-1)]
pw 0 = [[]]
pw n = [[x,w]:xs | w <- [' ','W'], x <- [' ','P'], xs <- if w == 'W' then p (n-1) else pw (n-1)]
it isn't as clean as possible, but I always find list comprehensions bring an elegance to the problem :). Totally worth it.