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.
Related
The function actually works but the list it returns, if the first element is equal to x, won't be in the correct order.
The function:
myelemIndices :: Eq a => a -> [a] -> [Int]
myelemIndices x [] = []
myelemIndices x l = if posic n l == x
then myReverse (n : myelemIndices x (init l))
else myelemIndices x (init l)
where n = length l - 1
Will return something like:
myelemIndices 1 [1,2,1,2,1]
[2,0,4]
Posic funtion is equal to :
posic :: Int -> [a] -> a
posic 0 (h:t) = h
posic x (a:b) = posic (x-1) b
And myReverse does exactly the same as reverse. I'm not looking for a different funtion that works, just a correction on mine. Thanks in advance!
You're reversing the list every time you find an element. The minimal change here would be to use myElemIndices x (init l) ++ [n] instead of myReverse (n : myelemIndices x (init l))
A first anti-pattern in your code is that you use length. length will usually run in O(n) (with n the number of elements in the list), but furthermore it is troublesome since the list can have infinite length. In that case length will never terminate. In functional programming it is sometimes seen as a sign that something is probably not right.
So the first question is: do we need length. What your code needs to do is return the indices. But say you for instance need to know on what pages in a phone book the name "John" is listed, then you do not need to know in advance how many pages the phone book has: you can simply take a look at the first page. In case it has a person with first name John, then you say that is on page 1, and regardless of that, you move further.
We can use the same approach here. The only thing we need is a parameter that keeps track of the page we are currently looking at. We can introduce this parameter by defining a new function that will do most of the work. So:
myElemIndices :: Eq a => a -> [a] -> [Int]
myElemIndices x l = go 0 l
where go = ...
So we defined a function go, and the first parameter will keep track of the page number. We will need to update that number in case we do recursion. But now of course we still need to define the go function.
The base case is easy: in case we reached the end of the list (phone book), we can say we will not find any occurences anymore. So we can write:
go _ [] = []
This thus means that, regardless of the page number (_), in case there are no pages anymore ([]), we return an empty list as match [].
In case we did not reach the end of the phone book, we can fetch the head h, and the tail t. We will have to check if the head h matches with the queried element x. In case it does, we return the page number, otherwise we don not. Regardless of that, we keep searching for more pages. So we can write:
go i (h:t) | x == h = i : <next-matches>
| otherwise = <next-matches>
the <next-matches> simply is a recursive call where we update the page number (i+1), and we continue with the tail of the list, so:
go i (h:t) | x == h = i : tl
| otherwise = tl
where tl = go (i+1) t
Now we can put that all together into:
myElemIndices :: Eq a => a -> [a] -> [Int]
myElemIndices x l = go 0 l
where go _ [] = []
go i (h:t) | x == h = i : tl
| otherwise = tl
where tl = go (i+1) t
We can still improve the code a bit. First of all, we do not have to write:
myElemIndices x l = go 0 l
Notice that both the head and the body end with l. We can omit this, and turn it into:
myElemIndices x = go 0
Furthermore we do not need the result to be Ints as well. As long as these are Nums, we are fine. So we can generalize this to:
myElemIndices :: (Eq a, Num n) => a -> [a] -> [n]
myElemIndices x l = go 0 l
where go _ [] = []
go i (h:t) | x == h = i : tl
| otherwise = tl
where tl = go (i+1) t
I went through a post for this problem but I do not understand it. Could someone please explain it?
Q: Find every n-th element of the list in the form of a list start from the n-th element itself.
everyNth :: Int -> [t] -> [t]
everyNth elt = map snd . filter (\(lst,y) -> (mod lst elt) == 0) . zip [1..]
Also, please explain how pattern matching can be used for this problem. That is using
[]->[]
It's easy to use pattern matching to 'select every nth element' for particular cases of n:
every2nd (first:second:rest) = second : every2nd rest
every2nd _ = []
-- >>> every2nd [1..12]
-- [2,4,6,8,10,12]
every3rd (first:second:third:rest) = third : every3rd rest
every3rd _ = []
-- >>> every3rd [1..13]
-- [3,6,9,12]
every4th (first:second:third:fourth:rest) = fourth : every4th rest
every4th _ = []
-- >>> every4th [1..12]
-- [4,8,12]
For the general case, though, we're out of luck, at least with that particular approach. Patterns like those above will need some definite length to be definite patterns. The composed function you mention starts from the thought that we do know how to find every nth member of [1..], namely if it's a multiple of n
multiple n m = m `mod` n == 0
-- >>> filter (multiple 3) [1..12]
-- [3,6,9,12]
So the solution you are trying to understand zips [1..] with the list
index xs = zip [1..] xs
-- >>> index [1..5]
-- [(1,1),(2,2),(3,3),(4,4),(5,5)]
-- >>> index "hello"
-- [(1,'h'),(2,'e'),(3,'l'),(4,'l'),(5,'o')]
Then it filters out just those pairs whose first element is a multiple of n
every_nth_with_index n xs = filter (\(m,a) -> multiple n m) (index xs)
-- >>> every_nth_with_index 3 [1..12]
-- [(3,3),(6,6),(9,9),(12,12)]
-- >>> every_nth_with_index 3 "stackoverflow.com"
-- [(3,'a'),(6,'o'),(9,'r'),(12,'o'),(15,'c')]
Then it gets rid of the ancillary construction, leaving us with just the second element of each pair:
every_nth n xs = map snd (every_nth_with_index n xs)
-- >>> every_nth 3 [1..12]
-- [3,6,9,12]
-- >>> every_nth 3 "stackoverflow.com"
-- "aoroc"
Retracinging our steps we see that this is the same as
everyNth elt = map snd . filter (\(lst,y) -> (mod lst elt) == 0) . zip [1..]
The notorious fold fan strikes again.
everyNth n xs = foldr go (`seq` []) xs n where
go x r 0 = x : r (n - 1)
go _ r k = r (k - 1)
This is very similar to chepner's approach but it integrates the dropping into the recursion. Rewritten without the fold, it's pure pattern matching:
everyNth n = go n where
go k [] = k `seq` []
go 0 (x : xs) = x : go (n - 1) xs
go k (_ : xs) = go (k - 1) xs
With a little cheating, you can define everyNth using pattern matching. Really, we're abstracting out the part that makes pattern matching difficult, as pointed out in Michael's answer.
everyNth n lst = e (shorten lst)
where shorten = drop (n-1) -- here's the cheat
e [] = []
e (first:rest) = first : e (shorten rest)
If you have never seen Haskell before then this takes a bit of explaining.
everyNth :: Int -> [t] -> [t]
everyNth elt = map snd . filter (\(lst,y) -> (mod lst elt) == 0) . zip [1..]
First, note that the type has two arguments, but the definition has only one. This is because the value returned by everyNth is in fact another function. elt is the Int, and the expression in the second line creates a new function that does the job.
Second, note the "." operators. This is an operator that joins two functions together. It is defined like this:
(f . g) x = f (g x)
Here is an equivalent version of the definition with the second argument made explicit:
everyNth elt xs = map snd (filter (\(lst y) -> (mod lst elt) == 0) (zip xs))
When you see a bunch of functions in a chain linked by "." operators you need to read it from right to left. In my second version pay attention to the bracket nesting. zip [1..] xs is the inner-most expression, so it gets evaluated first. It turns a list like ["foo", "bar"] into [(1, "foo"),(2, "bar")]. Then this is filtered to find entries where the number is a multiple of elt. Finally the map snd strips the numbers back out to return just the required entries.
I need to get the nth element of a list but without using the !! operator. I am extremely new to haskell so I'd appreciate if you can answer in more detail and not just one line of code. This is what I'm trying at the moment:
nthel:: Int -> [Int] -> Int
nthel n xs = 0
let xsxs = take n xs
nthel n xs = last xsxs
But I get: parse error (possibly incorrect indentation)
There's a lot that's a bit off here,
nthel :: Int -> [Int] -> Int
is technically correct, really we want
nthel :: Int -> [a] -> a
So we can use this on lists of anything (Optional)
nthel n xs = 0
What you just said is "No matter what you give to nthel return 0". which is clearly wrong.
let xsxs = ...
This is just not legal haskell. let ... in ... is an expression, it can't be used toplevel.
From there I'm not really sure what that's supposed to do.
Maybe this will help put you on the right track
nthelem n [] = <???> -- error case, empty list
nthelem 0 xs = head xs
nthelem n xs = <???> -- recursive case
Try filling in the <???> with your best guess and I'm happy to help from there.
Alternatively you can use Haskell's "pattern matching" syntax. I explain how you can do this with lists here.
That changes our above to
nthelem n [] = <???> -- error case, empty list
nthelem 0 (x:xs) = x --bind x to the first element, xs to the rest of the list
nthelem n (x:xs) = <???> -- recursive case
Doing this is handy since it negates the need to use explicit head and tails.
I think you meant this:
nthel n xs = last xsxs
where xsxs = take n xs
... which you can simplify as:
nthel n xs = last (take n xs)
I think you should avoid using last whenever possible - lists are made to be used from the "front end", not from the back. What you want is to get rid of the first n elements, and then get the head of the remaining list (of course you get an error if the rest is empty). You can express this quite directly as:
nthel n xs = head (drop n xs)
Or shorter:
nthel n = head . drop n
Or slightly crazy:
nthel = (head .) . drop
As you know list aren't naturally indexed, but it can be overcome using a common tips.
Try into ghci, zip [0..] "hello", What's about zip [0,1,2] "hello" or zip [0..10] "hello" ?
Starting from this observation, we can now easily obtain a way to index our list.
Moreover is a good illustration of the use of laziness, a good hint for your learning process.
Then based on this and using pattern matching we can provide an efficient algorithm.
Management of bounding cases (empty list, negative index).
Replace the list by an indexed version using zipper.
Call an helper function design to process recursively our indexed list.
Now for the helper function, the list can't be empty then we can pattern match naively, and,
if our index is equal to n we have a winner
else, if our next element is empty it's over
else, call the helper function with the next element.
Additional note, as our function can fail (empty list ...) it could be a good thing to wrap our result using Maybe type.
Putting this all together we end with.
nth :: Int -> [a] -> Maybe a
nth n xs
| null xs || n < 0 = Nothing
| otherwise = helper n zs
where
zs = zip [0..] xs
helper n ((i,c):zs)
| i == n = Just c
| null zs = Nothing
| otherwise = helper n zs
im searching for a solution for my Haskell class.
I have a list of numbers and i need to return SUM for every part of list. Parts are divided by 0. I need to use FOLDL function.
Example:
initial list: [1,2,3,0,3,4,0,5,2,1]
sublist [[1,2,3],[3,4],[5,2,1]]
result [6,7,7]
I have a function for finding 0 in initial list:
findPos list = [index+1 | (index, e) <- zip [0..] list, e == 0]
(returns [4,6] for initial list from example)
and function for making SUM with FOLDL:
sumList list = foldl (+) 0 list
But I completely failed to put it together :/
---- MY SOLUTION
In the end I found something completely different that you guys suggested.
Took me whole day to make it :/
groups :: [Int] -> [Int]
groups list = [sum x | x <- makelist list]
makelist :: [Int] -> [[Int]]
makelist xs = reverse (foldl (\acc x -> zero x acc) [[]] xs)
zero :: Int -> [[Int]] -> [[Int]]
zero x acc | x == 0 = addnewtolist acc
| otherwise = addtolist x acc
addtolist :: Int -> [[Int]] -> [[Int]]
addtolist i listlist = (i : (head listlist)) : (drop 1 listlist)
addnewtolist :: [[Int]] -> [[Int]]
addnewtolist listlist = [] : listlist
I'm going to give you some hints, rather than a complete solution, since this sounds like it may be a homework assignment.
I like the breakdown of steps you've suggested. For the first step (going from a list of numbers with zero markers to a list of lists), I suggest doing an explicit recursion; try this for a template:
splits [] = {- ... -}
splits (0:xs) = {- ... -}
splits (x:xs) = {- ... -}
You can also abuse groupBy if you're careful.
For the second step, it looks like you're almost there; the last step you need is to take a look at the map :: (a -> b) -> ([a] -> [b]) function, which takes a normal function and runs it on each element of a list.
As a bonus exercise, you might want to think about how you might do the whole thing in one shot as a single fold. It's possible -- and even not too difficult, if you track through what the types of the various arguments to foldr/foldl would have to be!
Additions since the question changed:
Since it looks like you've worked out a solution, I now feel comfortable giving some spoilers. =)
I suggested two possible implementations; one that goes step-by-step, as you suggested, and another that goes all at once. The step-by-step one could look like this:
splits [] = []
splits (0:xs) = [] : splits xs
splits (x:xs) = case splits xs of
[] -> [[x]]
(ys:yss) -> ((x:ys):yss)
groups' = map sum . splits
Or like this:
splits' = groupBy (\x y -> y /= 0)
groups'' = map sum . splits'
The all-at-once version might look like this:
accumulate 0 xs = 0:xs
accumulate n (x:xs) = (n+x):xs
groups''' = foldr accumulate [0]
To check that you understand these, here are a few exercises you might like to try:
What do splits and splits' do with [1,2,3,0,4,5]? [1,2,0,3,4,0]? [0]? []? Check your predictions in ghci.
Predict what each of the four versions of groups (including yours) output for inputs like [] or [1,2,0,3,4,0], and then test your prediction in ghci.
Modify groups''' to exhibit the behavior of one of the other implementations.
Modify groups''' to use foldl instead of foldr.
Now that you've completed the problem on your own, I am showing you a slightly less verbose version. Foldr seems better in my opinion to this problem*, but because you asked for foldl I will show you my solution using both functions.
Also, your example appears to be incorrect, the sum of [5,2,1] is 8, not 7.
The foldr version.
makelist' l = foldr (\x (n:ns) -> if x == 0 then 0:(n:ns) else (x + n):ns) [0] l
In this version, we traverse the list, if the current element (x) is a 0, we add a new element to the accumulator list (n:ns). Otherwise, we add the value of the current element to the value of the front element of the accumulator, and replace the front value of the accumulator with this value.
Step by step:
acc = [0], x = 1. Result is [0+1]
acc = [1], x = 2. Result is [1+2]
acc = [3], x = 5. Result is [3+5]
acc = [8], x = 0. Result is 0:[8]
acc = [0,8], x = 4. Result is [0+4,8]
acc = [4,8], x = 3. Result is [4+3,8]
acc = [7,8], x = 0. Result is 0:[7,8]
acc = [0,7,8], x = 3. Result is [0+3,7,8]
acc = [3,7,8], x = 2. Result is [3+2,7,8]
acc = [5,7,8], x = 1. Result is [5+1,7,8] = [6,7,8]
There you have it!
And the foldl version. Works similarly as above, but produces a reversed list, hence the use of reverse at the beginning of this function to unreverse the list.
makelist l = reverse $ foldl (\(n:ns) x -> if x == 0 then 0:(n:ns) else (x + n):ns) [0] l
*Folding the list from the right allows the cons (:) function to be used naturally, using my method with a left fold produces a reversed list. (There is likely a simpler way to do the left fold version that I did not think of that eliminates this triviality.)
As you already solved it, another version:
subListSums list = reverse $ foldl subSum [0] list where
subSum xs 0 = 0 : xs
subSum (x:xs) n = (x+n) : xs
(Assuming that you have only non-negative numbers in the list)
i'm trying to create an infinte reverse add then sort list in haskell
r = map head (iterate rlist [2..])
rlist (x:xs) = [x : xs | x <- xs , x quick $ p+x ]
where p = reverse x
quick [] = []
quick (x:xs) = quick [u |u <- xs, u < x] ++ [x] ++ quick [u | u <- xs , u >= x]
but its not working, any suggestions?
thanks.
I'm not sure how you expect your code to work (perhaps there was a problem when you posted the code). Namely, part of your list comprehension ... x quick $ p + x makes no sense to me - x isn't a function and it isn't a list so reverse x also makes no sense. For that matter you have shadowed x - notice the x in your list comprehension isn't the same as the x in ratslist (x:xs).
A simple solution does exist using read and show to convert the numbers to lists of digits (well, characters, but it works) and back:
import Data.List
myRats = 1 : map ratify myRats
-- Alternatively: myRats = iterate ratify 1
ratify :: Integer -> Integer
ratify n = sortRat (n + rev n)
where
rev = read . reverse . show
sortRat = read . sort . show
And in GHCi:
*Main Data.List> take 10 myRats
[1,2,4,8,16,77,145,668,1345,6677]
I don't quite get your problem, but according to your example, I'd try
rats x = sort (zipWith (+) x (reverse x))
like in rats [1,4,5] which equals [6,6,8].