How would I define an append function that takes two lists and output 1 list.
So plusplus [1,2,3] [4,5,6] should return [1,2,3,4,5,6].
I have the following so far:
plusplus :: [Int] -> [Int] -> [Int]
plusplus [] [] = []
plusplus [] [x] = [x]
plusplus [x] [] = [x]
plusplus (x:xs) (y:ys) = x: plusplus xs (y:ys)
I want to define my own ++ function, but with recursion. The code above gives an error when I actually execute it with two lists.
This is an example of an error I get with the command:
plusplus [1,2,3] [4,5,6]
[1,2,3*** Exception: baby.hs:(...): Non-exhaustive patterns in function plusplus
ok basically the only problem you have here is a missing case:
pluspluse (x:xs) [] = ...
is missing (and will cause an error)
but you can clean up quite a bit:
The first two cases are basically the same: whatever the second input is will be the result - so you can just it those a name (say ys) and write:
plusplus [] ys = ys
(can you see how this includes your first two cases? - what is ys in those?)
your 3rd line one will be covered by the correct last case (see below) - to see this remember that [x] == x:[]
I'll give you the complete answer in a few minutes - but you should try to figure it out:
plusplus :: [Int] -> [Int] -> [Int]
plusplus [] ys = ys
plusplus (x:xs) ? = ??
try to find the correct things to replace ? and ?? with
you almost had it - and the answer can be:
plusplus :: [Int] -> [Int] -> [Int]
plusplus [] ys = ys
plusplus (x:xs) ys = x: plusplus xs ys
as you can see you never have to touch the second list - but need to recreate all of the first list (to link to the second)
We know that plusplus = flip (foldr (:)). Here's the definition of foldr:
foldr :: (a -> b -> b) -> b -> [a] -> b
foldr step = let fold acc [] = acc
fold acc (x:xs) = step x (fold acc xs)
in fold
On specializing, we get:
foldr :: (a -> b -> b) -> b -> [a] -> b
| | | | | |
| | | | | |
(:) :: (a -> [a] -> [a]) | | |
| | |
foldr (:) :: [a] -> [a] -> [a]
Here's the definition of the specialized function:
fold :: [a] -> [a] -> [a]
fold acc [] = acc
fold acc (x:xs) = x : fold acc xs
Ofcourse, we need to flip the arguments:
plusplus :: [a] -> [a] -> [a]
plusplus [] ys = ys
plusplus (x:xs) ys = x : plusplus xs ys
That's the reason why plusplus is defined the way it is.
Thanks guys! With the help of Carsten spotting my silly mistake.
plusplus :: [Int] -> [Int] -> [Int]
plusplus [] [] = []
plusplus (x:xs) [] = x:xs
plusplus [] (x:xs) = x:xs
plusplus (x:xs) (y:ys) = x: plusplus xs (y:ys)
This works.
So for input:
plusplus [1,2,3] [4,5,6]
Output is:
[1,2,3,4,5,6]
This is really a riff on Carsten's answer, but I think it's worth going into and won't fit into a comment. One of the keys to avoiding non-exhaustive pattern errors in function definitions is to approach writing functions in a more mechanical, formulaic manner. The definition of the list type is something like this:
data [a] = [] | a : [a]
So list types have two data constructors, [] and :. So a good first thing to try when starting your function is to write two equations, one for each of the constructors. Carsten's template (which I'm mostly copying now) is following that pattern:
plusplus :: [Int] -> [Int] -> [Int]
plusplus [] ys = _ -- Equation for empty list
plusplus (x:xs) ys = _ -- Equation for list cells
Now since we've written one pattern for each of the constructors of the type, we're now guaranteed that our patterns are exhaustive. So we stare at what we got a bit, and now we see the solution:
plusplus :: [Int] -> [Int] -> [Int]
plusplus [] ys = ys
plusplus (x:xs) ys = x : plusplus xs ys
Let's try another one! Signature:
-- | Merge two lists by alternating elements of one with the
-- other. After either list runs out, just continue with the
-- remaining elements of the other.
interleave :: [a] -> [a] -> [a]
interleave xs ys = _
Let's expand this to two equations by splitting the xs variable according to the two constructors, and fill in the easy parts on the right hand side:
interleave :: [a] -> [a] -> [a]
interleave [] ys = ys -- this one was easy
interleave (x:xs) ys = x : _ -- we know it has to start with `x`
Now we can go two ways. The first one is we could take the second equation and split it into two cases for ys:
interleave :: [a] -> [a] -> [a]
interleave [] ys = ys
interleave (x:xs) [] = x : _
interleave (x:xs) (y:ys) = x : _
And filling in those blanks is easy:
interleave :: [a] -> [a] -> [a]
interleave [] ys = ys
interleave (x:xs) [] = x : xs
interleave (x:xs) (y:ys) = x : y : interleave xs ys
The second way is, instead of splitting ys on the constructors' cases, to notice that this works:
interleave :: [a] -> [a] -> [a]
interleave [] ys = ys
interleave (x:xs) ys = x : interleave ys xs
So by going about it systematically based on the constructors we know for sure that all combinations are covered.
Related
I'm trying to merge:
[[('a',False),('b',False)]]
with
[[('a',True)],[('b',True)]]
and get the following:
[[(a,True),(b,True),(a,False),('b',False)]]
Essiently merging the two lists tuples into one.
I tried to create a function to do this but I'm not getting the output I want. Here's my current function and the output it gives me
mergeFunc :: [[a]] -> [[a]] -> [[a]]
mergeFunc xs [] = xs
mergeFunc [] ys = ys
mergeFunc (x:xs) ys = x : mergeFunc ys xs
[[('a',True),('b',True)],[('a',False)],[('b',False)]]
It seems that I'm merging a level higher than I want to, but I'm not sure how to fix this.
You can concatenate the sublists together in one list with concat :: Foldable f => f [a] -> [a]:
mergeFunc :: [[a]] -> [[a]] -> [[a]]
mergeFunc xs ys = [concat xs ++ concat ys]
and this will produce:
ghci> mergeFunc [[('a',True)],[('b',True)]] [[('a',False),('b',False)]]
[[('a',True),('b',True),('a',False),('b',False)]]
But then it makes not much sense to wrap the list in a singleton list. In that case it makes more sense to define this as:
mergeFunc :: [[a]] -> [[a]] -> [a]
mergeFunc xs ys = concat xs ++ concat ys
since this removes an unnecessary list level.
So I'm trying to write a function that, given two lists of integers, adds the ith even number of each list and returns them in another list. In case one of the list doesn't have an ith even number, a 0 is considered. For example, if the lists are [1,2,1,4,6] and [2,2], it returns [4,6,6] ([2+2,4+2,6+0]). I have the following code:
addEven :: [Int] -> [Int] -> [Int]
addEeven [] [] = []
addEeven (x:xs) [] = filter (\g -> g `mod`2 == 0) (x:xs)
addEven [] (y:ys) = filter (\g -> g `mod` 2 == 0) (y:ys)
addEven (x:xs) (y:ys) = (a + b):(addEven as bs)
where
(a:as) = filter (\g -> g `mod` 2 == 0) (x:xs)
(b:bs) = filter (\g -> g `mod` 2 == 0) (y:ys)
When I run that with the previous example, I get:
[4,6*** Exception: ex.hs:(4,1)-(8,101): Non-exhaustive patterns in function addEven
I really can't see what I'm missing, since it doesn't work with any input I throw at it.
A filter might eliminate elements, hence filter (\g -> gmod2 == 0) is not said to return any elements, and thus the patterns (a:as) and (b:bs) might fail.
That being said, I think you make the problem too complex here. You can first define a helper function that adds two elements of a list:
addList :: Num a => [a] -> [a] -> [a]
addList (x:xs) (y:ys) = (x+y) : addList xs ys
addList xs [] = xs
addList [] ys = ys
Then we do the filter on the two parameters, and make a function addEven that looks like:
addEven :: Integral a => [a] -> [a] -> [a]
addEven xs ys = addList (filter even xs) (filter even ys)
or with on :: (b -> b -> c) -> (a -> b) -> a -> a -> c:
import Data.Function(on)
addEven :: Integral a => [a] -> [a] -> [a]
addEven = addList `on` filter even
While using filter is very instinctive in this case, perhaps using filter twice and then summing up the results might be slightly ineffficient for large lists. Why don't we do the job all at once for a change..?
addMatches :: [Int] -> [Int] -> [Int]
addMatches [] [] = []
addMatches [] ys = filter even ys
addMatches xs [] = filter even xs
addMatches xs ys = first [] xs ys
where
first :: [Int] -> [Int] -> [Int] -> [Int]
first rs [] ys = rs ++ filter even ys
first rs (x:xs) ys = rs ++ if even x then second [x] xs ys
else first [] xs ys
second :: [Int] -> [Int] -> [Int] -> [Int]
second [r] xs [] = [r] ++ filter even xs
second [r] xs (y:ys) = if even y then first [r+y] xs ys
else second [r] xs ys
λ> addMatches [1,2,1,4,6] [2,2]
[4,6,6]
I am practicing some Haskell exam paper questions, and have come across the following
Define a Haskell function weaveHunks which takes an int and
two lists and weaves them together in hunks of the given size.
Be sure to declare its type signature.
Example:
weaveHunks 3 "abcdefghijklmno" "ABCDEFGHIJKLMNO"
=> "abcABCdefDEFghiGHIjklJKLmnoMNO"
I have found the following on Stack Overflow, which is just too weave two lists together but only in chunks of 1
weaveHunks :: [a] -> [a] -> [a]
weaveHunks xs [] = xs
weaveHunks [] ys = ys
weaveHunks (x:xs) (y:ys) = x : y : weaveHunks xs ys
I am having problems adjusting this to take chunks fo n size, I am very new to Haskell but this is what I have so far
weaveHunks :: Int -> [a] -> [a] -> [a]
weaveHunks n xs [] = xs
weaveHunks n [] ys = ys
weaveHunks n xs ys = (take n xs) : (take n ys) : weaveHunks n (drop n xs) (drop n ys)
I am getting an error on the last line
(Couldn't match type a' with[a]')
Is (drop n xs) not a list?
You're very close!
By using the : operator to prepend the hunks, you're expressing that take n xs is one element of the result list, take n ys the next, and so on. But actually in both cases it's multiple elements you're prepending. That's the [a] that should actually be just a.
The solution is to use the ++ operator instead, which prepends an entire list rather than just a single element.
This is the full solution as I'd write it:
weaveHunks :: Int -> [a] -> [a] -> [a]
weaveHunks _ xs [] = xs
weaveHunks _ [] ys = ys
weaveHunks n xs ys = xHunk ++ yHunk ++ weaveHunks n xRemain yRemain
where [(xHunk, xRemain), (yHunk, yRemain)] = splitAt n <$> [xs,ys]
As #leftaroundabout said, since your appending lists of type [a], you need to use ++ instead of :. With this in mind, your code would then look like this:
weaveHunks :: Int -> [a] -> [a] -> [a]
weaveHunks _ xs [] = xs
weaveHunks _ [] ys = ys
weaveHunks n xs ys = (take n xs) ++ (take n ys) ++ weaveHunks n (drop n xs) (drop n ys)
If your interested, you can also use library functions to do this task:
import Data.List.Split
weaveHunks :: Int -> [a] -> [a] -> [a]
weaveHunks n xs ys = concat $ zipWith (++) (chunksOf n xs) (chunksOf n ys)
Note: chunksOf is from Data.List.Split, which splits the list into sublists of length n, so the type of this function is Int -> [a] -> [[a]]. zipWith zips two lists based on a condition, in this case concatenation ++. concat turns a list of [[a]] into [a].
I want to write a function that takes a list of sorted lists, then merges everything together and sorts them again.
I managed to write this so far:
merge_:: Ord a => [[a]] -> [a] --takes in the list and merges it
merge_ [] = []
merge_ (x:xs) = x ++ merge_ xs
isort:: Ord a => [a] -> [a] --Sorts a list
isort [] = []
isort (a:x) = ins a (isort x)
where
ins a [] = [a]
ins a (b:y) | a<= b = a:(b:y)
| otherwise = b: (ins a y)
I haven't been able to find a way to combine these two in one function in a way that makes sense. Note that I'm not allowed to use things such as ('.', '$'..etc) (homework)
We start simple. How do we merge two sorted lists?
mergeTwo :: Ord a => [a] -> [a] -> [a]
mergeTwo [] ys = ys
mergeTwo xs [] = xs
mergeTwo (x:xs) (y:ys)
| x <= y = x : mergeTwo xs (y:ys)
| otherwise = y : mergeTwo (x:xs) ys
How do we merge multiple? Well, we start with the first and the second and merge them together. Then we merge the new one and the third together:
mergeAll :: Ord a => [[a]] -> [a]
mergeAll (x:y:xs) = mergeAll ((mergeTwo x y) : xs)
mergeAll [x] = x
mergeAll _ = []
Allright. Now, to sort all elements, we need to create a list from every element, and then merge them back. Let's write a function that creates a list for a single item:
toList :: a -> [a]
toList x = -- exercise
And now a function to wrap all elements in lists:
allToList :: [a] -> [[a]]
allToList = -- exercise
And now we're done. We simply need to use allToList and then mergeAll:
isort :: Ord a => [a] -> [a]
isort xs = mergeAll (allToList xs)
Note that this exercise got a lot easier since we've split it into four functions.
Exercises (which might not be possible for you(r homework))
Write toList and allToList.
Try a list comprehension for allToList. Try a higher order function for allToList.
Write isort point-free (with (.)).
Check whether there is already a toList function with the same type. Use that one.
Rewrite mergeAll using foldr
Try this (not tested):
merge :: Ord a => [a] -> [a] -> [a]
merge [] l1 = l1
merge l1 [] = l1
merge (e1:l1) (e2:l2)
| e1<e2 = e1:merge l1 (e2:l2)
| otherwise = e2:merge (e1:l1) l2
I've been tasked to create a recursive function that takes in a list and gives the reverse of the list.
I have been able to create the a function named rev1 which does this:
rev1 :: [a] -> [a]
rev1 [] = []
rev1 (x:xs) = reverse xs ++ [x]
But I have been asked to create a another function 'rev2' which should use an additional argument serving as an accumulator.
Could somebody please help me compose the rev2 function.
Thanks in advance,
Sam
First of all, your rev1 should be this instead:
rev1 :: [a] -> [a]
rev1 [] = []
rev1 (x:xs) = rev1 xs ++ [x]
The point of rev2 would be to achieve tail recursion by means of passing the intermediate result inside an accumulator argument:
rev2 :: [a] -> [a] -> [a]
-- if applied to an empty list, just return the result
-- so far, i.e. whatever is inside the accumulator:
rev2 acc [] = acc
-- otherwise, take the head of the list and append it to
-- the accumulator, and carry on with the rest of the list
rev2 acc (x:xs) = rev2 (x:acc) xs
this, however, obviously has the downside of exposing the acc argument to users of rev2, so a typical approach is hiding the accumulator based implementation behind a façade that looks exactly like rev1:
rev2 :: [a] -> [a]
rev2 xs = go [] xs where
go :: [a] -> [a] -> [a] -- signature included for clarity
go acc [] = acc
go acc (x:xs) = go (x:acc) xs
Let me start things off:
rev2 :: [a] -> [a]
rev2 xs = rev2' xs ???
rev2' :: [a] -> [a] -> [a]
rev2' [] ys = ???
rev2' (x : xs) ys = ???