I'm trying to combine 2 lists from input but I am getting an error every time.
Here is my code:
myAppend :: [a] -> [a] -> [a]
myAppend a b = zipWith (+) a b
Getting this error:
"No instance for (Num a) arising from a use of ‘+’"
I was given this solution but it doesn't really make sense to me
myAppend :: [a] -> [a] -> [a]
myAppend [] xs = xs
myAppend (y:ys) xs = y:(myAppend ys xs)
I don't really understand the second and third line.
Can anyone help?
Thanks
Your myAppend does not concatenate two lists, it aims to sum elementwise the two lists, so myAppend [1,4,2,5] [1,3,0,2] will produce [2,7,2,7]. It will require a Num a constraint, since it can only work if the elements of the lists are Numbers:
myAppend :: Num a => [a] -> [a] -> [a]
myAppend a b = zipWith (+) a b
As for the solution here it uses recursion. Lists in Haskell are like linked lists: you have a an empty list ("nil") which is represented by the [] data constructor, and a node ("cons") which is represented with (x:xs) where x points to the first item, and xs points to the list of remaining elements. So [1,4,2,5] is short for (1:(4:(2:(5:[])))).
If we want to append [1,4] and [2,5] we thus want to produce a list (1:(4:(2:(5:[])))) out of (1:(4:[])) and (2:(5:[])). This means we create a linked list with all the elements of the first list, but instead of pointing to the empty list [], we let it point to the second list for the remaining elements. We do this through recursion:
myAppend (y:ys) xs = y : myAppend ys xs
will match if the first list unifies with the (y:ys) pattern. In that case we thus produce a list with y as first element, and the result of myAppend ys xs as as list of remaining elements ("tail"). Eventually we will thus call myAppend ys xs with the empty list [] as first item. In that case, we thus return the second list instead of the empty list, to append the second list to it.
We thus make calls that look like:
myAppend [1, 4] [2, 5]
= myAppend (1:(4:[])) (2:(5:[]))
-> 1 : (myAppend (4:[]) (2:(5:[])))
-> 1 : (4 : (myAppend [] (2:(5:[]))))
-> 1 : (4 : (2:(5:[]))
= [1, 4, 2, 5]
I have implemented a function (!!=) that given a list and a tuple containing an index in the list and a
new value, updates the given list with the new value at the given
index.
(!!=) :: [a] -> (Int,a) -> [a]
(!!=) xs (0, a) = a : tail xs
(!!=) [] (i, a) = error "Index not in the list"
(!!=) (x:xs) (i, a) = x : xs !!= (i-1, a)
Being a beginner with the concept of folding I was wondering if there is a way to achieve the same result using foldl or foldr instead?
Thanks a lot in advance.
I'll give you the foldl version which is easier to understand I think and the easiest / most straight-forward version I can think of.
But please note that you should not use foldl (use foldl': https://wiki.haskell.org/Foldr_Foldl_Foldl') - nor should you use ++ like this (use : and reverse after) ;)
Anway this is the idea:
(!!=) xs (i, a) = snd $ foldl
(\(j, ys) x -> (j+1, if j == i then ys ++ [a] else ys ++ [x]))
(0, [])
xs
as the state/accumulator for the fold I take a tuple of the current index and the accumulated result list (therefore the snd because I only want this in the end)
then the folding function just have to look if we are at the index and exchange the element - returning the next index and the new accumulated list
as an exercise you can try to:
use : instead of ++ and a reverse
rewrite as foldr
look at zipWith and rewrite this using this (zipWith (...) [0..] xs) instead of the fold (this is similar to using a map with index
Neither foldl nor foldr can do this particular job efficiently (unless you "cheat" by pattern matching on the list as you fold over it), though foldr can do it a bit less badly. No, what you really need is a different style of fold, sometimes called para:
para :: (a -> [a] -> b -> b) -> b -> [a] -> b
para _f n [] = n
para f n (a : as) = f a as (para f n as)
para is very similar to foldr. Each of them takes a combining function and, for each element, passes the combining function that element and the result of folding up the rest of the list. But para adds something extra: it also passes in the rest of the list! So there's no need to reconstruct the tail of the list once you've reached the replacement point.
But ... how do you count from the beginning with foldr or para? That brings in a classic trick, sometimes called a "higher-order fold". Instead of para go stop xs producing a list, it's going to produce a function that takes the insertion position as an argument.
(!!=) :: [a] -> (Int, a) -> [a]
xs0 !!= (i0, new) = para go stop xs0 i0
where
-- If the list is empty, then no matter what index
-- you seek, it's not there.
stop = \_ -> error "Index not in the list"
-- We produce a function that takes an index. If the
-- index is 0, we combine the new element with "the rest of the list".
-- Otherwise we apply the function we get from folding up the rest of
-- the list to the predecessor of the index, and tack on the current
-- element.
go x xs r = \i -> case i of
0 -> new : xs
_ -> x : r (i - 1)
Note that para is easily powerful enough to implement foldr:
foldr c = para (\a _ b -> c a b)
What's perhaps less obvious is that foldr can implement a (very inefficient version of) para:
para f n = snd . foldr go ([], n)
where
go x ~(xs, r) = (x : xs, f x xs r)
Lest you get the wrong idea and think that para is "better than" foldr, know that when its extra power isn't needed, foldr is simpler to use and will very often be compiled to more efficient code.
Having spent hours looking for ways to manipulate [[a]] into [[a]], I thought this would be the best solution to my problem. The problem consists of appending a to [a] and returning [[a]] with the new change.
For example: xs = [[a],[b],[c]] and y = d.
I want to append y to xs!!0 . I cannot use xs!!0 ++ y because it will return just [a,d], I know this is because of Haskell's immutability.
How would I go about appending a value to a sublist and returning the list of lists? - [[a,d],[b],[c]] using the example from above to illustrate this.
let { xs = [[1]] ; y = 2 ; zs = [(xs!!0) ++ [y]] } in zs is one example to try at the GHCi prompt.
It returns [[1,2]].
And for the case of e.g. [[1],[2,3],[4]] and the like, we can do
appendToFirst :: [[a]] -> a -> [[a]]
appendToFirst (xs:r) y = (xs ++ [y]) : r
so that
> appendToFirst [[1],[2,3],[4]] 0
[[1,0],[2,3],[4]]
The (xs:r) on the left of the equal sign is a pattern.
The (:) in the ( (...) : r) on the right of the equal sign is a "cons" operation, a data constructor, (:) :: t -> [t] -> [t].
xs is bound to the input list's "head" i.e. its first element, and r is bound to the rest of the input list, in the pattern; and thus xs's value is used in creating the updated version of the list, with the first sublist changed by appending a value to its end, and r remaining as is.
xs ++ [y] creates a new entity, new list, while xs and y continue to refer to the same old values they were defined as. Since Haskell's values and variables are immutable, as you indeed have mentioned.
edit: If you want to add new element at the end of some sublist in the middle, not the first one as shown above, this can be done with e.g. splitAt function, like
appendInTheMiddle :: Int -> a -> [[a]] -> [[a]]
appendInTheMiddle i y xs =
let
(a,b) = splitAt i xs
in
init a ++ [last a ++ [y]] ++ b
Trying it out:
> appendInTheMiddle 2 0 [[1],[2],[3],[4]]
[[1],[2,0],[3],[4]]
Adding the error-handling, bounds checking, and adjusting the indexing if 0-based one is desired (that one would lead to a simpler and faster code, by the way), is left as an exercise for the reader.
Syntactically, this can be streamlined with "view patterns", as
{-# LANGUAGE ViewPatterns #-}
appendInTheMiddle :: Int -> a -> [[a]] -> [[a]]
appendInTheMiddle i y (splitAt i -> (a,b)) =
init a ++ [last a ++ [y]] ++ b
I have been working with Haskell for a little over a week now so I am practicing some functions that might be useful for something. I want to compare two lists recursively. When the first list appears in the second list, I simply want to return the index at where the list starts to match. The index would begin at 0. Here is an example of what I want to execute for clarification:
subList [1,2,3] [4,4,1,2,3,5,6]
the result should be 2
I have attempted to code it:
subList :: [a] -> [a] -> a
subList [] = []
subList (x:xs) = x + 1 (subList xs)
subList xs = [ y:zs | (y,ys) <- select xs, zs <- subList ys]
where select [] = []
select (x:xs) = x
I am receiving an "error on input" and I cannot figure out why my syntax is not working. Any suggestions?
Let's first look at the function signature. You want to take in two lists whose contents can be compared for equality and return an index like so
subList :: Eq a => [a] -> [a] -> Int
So now we go through pattern matching on the arguments. First off, when the second list is empty then there is nothing we can do, so we'll return -1 as an error condition
subList _ [] = -1
Then we look at the recursive step
subList as xxs#(x:xs)
| all (uncurry (==)) $ zip as xxs = 0
| otherwise = 1 + subList as xs
You should be familiar with the guard syntax I've used, although you may not be familiar with the # syntax. Essentially it means that xxs is just a sub-in for if we had used (x:xs).
You may not be familiar with all, uncurry, and possibly zip so let me elaborate on those more. zip has the function signature zip :: [a] -> [b] -> [(a,b)], so it takes two lists and pairs up their elements (and if one list is longer than the other, it just chops off the excess). uncurry is weird so lets just look at (uncurry (==)), its signature is (uncurry (==)) :: Eq a => (a, a) -> Bool, it essentially checks if both the first and second element in the pair are equal. Finally, all will walk over the list and see if the first and second of each pair is equal and return true if that is the case.
Say I have any list like this:
[4,5,6,7,1,2,3,4,5,6,1,2]
I need a Haskell function that will transform this list into a list of lists which are composed of the segments of the original list which form a series in ascending order. So the result should look like this:
[[4,5,6,7],[1,2,3,4,5,6],[1,2]]
Any suggestions?
You can do this by resorting to manual recursion, but I like to believe Haskell is a more evolved language. Let's see if we can develop a solution that uses existing recursion strategies. First some preliminaries.
{-# LANGUAGE NoMonomorphismRestriction #-}
-- because who wants to write type signatures, amirite?
import Data.List.Split -- from package split on Hackage
Step one is to observe that we want to split the list based on a criteria that looks at two elements of the list at once. So we'll need a new list with elements representing a "previous" and "next" value. There's a very standard trick for this:
previousAndNext xs = zip xs (drop 1 xs)
However, for our purposes, this won't quite work: this function always outputs a list that's shorter than the input, and we will always want a list of the same length as the input (and in particular we want some output even when the input is a list of length one). So we'll modify the standard trick just a bit with a "null terminator".
pan xs = zip xs (map Just (drop 1 xs) ++ [Nothing])
Now we're going to look through this list for places where the previous element is bigger than the next element (or the next element doesn't exist). Let's write a predicate that does that check.
bigger (x, y) = maybe False (x >) y
Now let's write the function that actually does the split. Our "delimiters" will be values that satisfy bigger; and we never want to throw them away, so let's keep them.
ascendingTuples = split . keepDelimsR $ whenElt bigger
The final step is just to throw together the bit that constructs the tuples, the bit that splits the tuples, and a last bit of munging to throw away the bits of the tuples we don't care about:
ascending = map (map fst) . ascendingTuples . pan
Let's try it out in ghci:
*Main> ascending [4,5,6,7,1,2,3,4,5,6,1,2]
[[4,5,6,7],[1,2,3,4,5,6],[1,2]]
*Main> ascending [7,6..1]
[[7],[6],[5],[4],[3],[2],[1]]
*Main> ascending []
[[]]
*Main> ascending [1]
[[1]]
P.S. In the current release of split, keepDelimsR is slightly stricter than it needs to be, and as a result ascending currently doesn't work with infinite lists. I've submitted a patch that makes it lazier, though.
ascend :: Ord a => [a] -> [[a]]
ascend xs = foldr f [] xs
where
f a [] = [[a]]
f a xs'#(y:ys) | a < head y = (a:y):ys
| otherwise = [a]:xs'
In ghci
*Main> ascend [4,5,6,7,1,2,3,4,5,6,1,2]
[[4,5,6,7],[1,2,3,4,5,6],[1,2]]
This problem is a natural fit for a paramorphism-based solution. Having (as defined in that post)
para :: (a -> [a] -> b -> b) -> b -> [a] -> b
foldr :: (a -> b -> b) -> b -> [a] -> b
para c n (x : xs) = c x xs (para c n xs)
foldr c n (x : xs) = c x (foldr c n xs)
para c n [] = n
foldr c n [] = n
we can write
partition_asc xs = para c [] xs where
c x (y:_) ~(a:b) | x<y = (x:a):b
c x _ r = [x]:r
Trivial, since the abstraction fits.
BTW they have two kinds of map in Common Lisp - mapcar
(processing elements of an input list one by one)
and maplist (processing "tails" of a list). With this idea we get
import Data.List (tails)
partition_asc2 xs = foldr c [] . init . tails $ xs where
c (x:y:_) ~(a:b) | x<y = (x:a):b
c (x:_) r = [x]:r
Lazy patterns in both versions make it work with infinite input lists
in a productive manner (as first shown in Daniel Fischer's answer).
update 2020-05-08: not so trivial after all. Both head . head . partition_asc $ [4] ++ undefined and the same for partition_asc2 fail with *** Exception: Prelude.undefined. The combining function g forces the next element y prematurely. It needs to be more carefully written to be productive right away before ever looking at the next element, as e.g. for the second version,
partition_asc2' xs = foldr c [] . init . tails $ xs where
c (x:ys) r#(~(a:b)) = (x:g):gs
where
(g,gs) | not (null ys)
&& x < head ys = (a,b)
| otherwise = ([],r)
(again, as first shown in Daniel's answer).
You can use a right fold to break up the list at down-steps:
foldr foo [] xs
where
foo x yss = (x:zs) : ws
where
(zs, ws) = case yss of
(ys#(y:_)) : rest
| x < y -> (ys,rest)
| otherwise -> ([],yss)
_ -> ([],[])
(It's a bit complicated in order to have the combining function lazy in the second argument, so that it works well for infinite lists too.)
One other way of approaching this task (which, in fact lays the fundamentals of a very efficient sorting algorithm) is using the Continuation Passing Style a.k.a CPS which, in this particular case applied to folding from right; foldr.
As is, this answer would only chunk up the ascending chunks however, it would be nice to chunk up the descending ones at the same time... preferably in reverse order all in O(n) which would leave us with only binary merging of the obtained chunks for a perfectly sorted output. Yet that's another answer for another question.
chunks :: Ord a => [a] -> [[a]]
chunks xs = foldr go return xs $ []
where
go :: Ord a => a -> ([a] -> [[a]]) -> ([a] -> [[a]])
go c f = \ps -> let (r:rs) = f [c]
in case ps of
[] -> r:rs
[p] -> if c > p then (p:r):rs else [p]:(r:rs)
*Main> chunks [4,5,6,7,1,2,3,4,5,6,1,2]
[[4,5,6,7],[1,2,3,4,5,6],[1,2]]
*Main> chunks [4,5,6,7,1,2,3,4,5,4,3,2,6,1,2]
[[4,5,6,7],[1,2,3,4,5],[4],[3],[2,6],[1,2]]
In the above code c stands for current and p is for previous and again, remember we are folding from right so previous, is actually the next item to process.