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I want to fix this code
h :: (a -> b) -> [a] -> [b]
h f = foldr (\x y -> f x : y) []
if i put h (+100) [1,2,3,4,5] in GHCI
it returns to me [101,202,303,404,505]
when i put h (*10) [1,2,3,4,5] then
i want to get [10,200,3000,40000,500000] list
can anyone help me fixing this code?
You here implemented a map, but in order to repeat the same operation multiple times, you need to perform a mapping on the tail y:
h :: (a -> a) -> [a] -> [a]
h f = foldr (\x y -> f x : map f y) []
Solving the general problem, as Willem Van Onsem's answer does, requires O(n^2) time to calculate the first n elements, because the function has to be applied k times to calculate the kth element.
To solve this sort of problem efficiently, you will need to take advantage of some additional structure. Based on your examples, I think the most obvious approach is to think about semigroup actions. That is, instead of applying an arbitrary function repeatedly, look for an efficient way to represent the compositions of the function. For example, (*x) can be represented by x, allowing (*x) . (*y) to be represented by x*y.
To apply this idea, we first need to transform Willem's solution to make the compositions explicit.
h :: (a -> a) -> [a] -> [a]
h f0 as0 = go as0 f0
where
go [] _ = []
go (a:as) f = f a : go as (f0 . f)
If we like, we can write that as a fold:
h :: (a -> a) -> [a] -> [a]
h f0 as = foldr go stop as f0
where
stop _ = []
go a r f = f a : r (f0 . f)
Now we've structured the function using an accumulator (which is a function). As we compose onto the accumulator, it will get slower and slower to apply it. We want to replace that accumulator with one we can "apply" quickly.
{-# language BangPatterns #-}
import Data.Semigroup
repeatedly :: Semigroup s => (s -> a -> a) -> s -> [a] -> [a]
repeatedly act s0 as = foldr go stop as s0
where
stop _ = []
go a r !s = act s a : r (s0 <> s)
Now you can use, for example,
repeatedly (\(Product s) -> (s*)) (Product 10) [1..5]
==> [10,200,3000,40000,500000]
repeatedly (\(Sum s) -> (s+)) (Sum 100) [1..5]
==> [101,202,303,404,505]
In each of these, you accumulate a product/sum which is added to/multiplied by the current list element.
I need some kind of fold which can terminate if I already have the data I want.
For example I need to find first 3 numbers which are greater than 5. I decided to use Either for termination and my code looks like this:
terminatingFold :: ([b] -> a -> Either [b] [b]) -> [a] -> [b]
terminatingFold f l = reverse $ either id id $ fold [] l
where fold acc [] = Right acc
fold acc (x:xs) = f acc x >>= flip fold xs
first3NumsGreater5 acc x =
if length acc >= 3
then Left acc
else Right (if x > 5 then (x : acc) else acc)
Are there some more clever/generic approaches?
The result of your function is a list, and it would be desirable if it were produced lazily, that is, extracting one item from the result should only require evaluating the input list up until the item is found there.
Unfolds are under-appreciated for these kinds of tasks. Instead of focusing on "consuming" the input list, let's think of it as a seed from which (paired with some internal accumulator) we can produce the result, element by element.
Let's define a Seed type that contains a generic accumulator paired with the as-yet unconsumed parts of the input:
{-# LANGUAGE NamedFieldPuns #-}
import Data.List (unfoldr)
data Seed acc input = Seed {acc :: acc, pending :: [input]}
Now let's reformulate first3NumsGreater5 as a function that either produces the next output element from the Seed, of signals that there aren't any more elements:
type Counter = Int
first3NumsGreater5 :: Seed Counter Int -> Maybe (Int, Seed Counter Int)
first3NumsGreater5 (Seed {acc, pending})
| acc >= 3 =
Nothing
| otherwise =
case dropWhile (<= 5) pending of
[] -> Nothing
x : xs -> Just (x, Seed {acc = succ acc, pending = xs})
Now our main function can be written in terms of unfoldr:
unfoldFromList ::
(Seed acc input -> Maybe (output, Seed acc input)) ->
acc ->
[input] ->
[output]
unfoldFromList next acc pending = unfoldr next (Seed {acc, pending})
Putting it to work:
main :: IO ()
main = print $ unfoldFromList first3NumsGreater5 0 [0, 6, 2, 7, 9, 10, 11]
-- [6,7,9]
Normally an early termination-capable fold is foldr with the combining function which is non-strict in its second argument. But, its information flow is right-to-left (if any), while you want it left-to-right.
A possible solution is to make foldr function as a left fold, which can then be made to stop early:
foldlWhile :: Foldable t
=> (a -> Bool) -> (r -> a -> r) -> r
-> t a -> r
foldlWhile t f a xs = foldr cons (\acc -> acc) xs a
where
cons x r acc | t x = r (f acc x)
| otherwise = acc
You will need to tweak this for t to test the acc instead of x, to fit your purposes.
This function is foldlWhile from https://wiki.haskell.org/Foldl_as_foldr_alternative, re-written a little. foldl'Breaking from there might fit the bill a bit better.
foldr with the lazy reducer function can express corecursion perfectly fine just like unfoldr does.
And your code is already lazy: terminatingFold (\acc x -> Left acc) [1..] => []. That's why I'm not sure if this answer is "more clever", as you've requested.
edit: following a comment by #danidiaz, to make it properly lazy you'd have to code it as e.g.
first3above5 :: (Foldable t, Ord a, Num a)
=> t a -> [a]
first3above5 xs = foldr cons (const []) xs 0
where
cons x r i | x > 5 = if i==2 then [x]
else x : r (i+1)
| otherwise = r i
This can be generalized further by abstracting the test and the count.
Of course it's just reimplementing take 3 . filter (> 5), but shows how to do it in general with foldr.
Given a condition, I want to search through a list of elements and return the first element that reaches the condition, and the previous one.
In C/C++ this is easy :
int i = 0;
for(;;i++) if (arr[i] == 0) break;
After we get the index where the condition is met, getting the previous element is easy, through "arr[i-1]"
In Haskell:
dropWhile (/=0) list gives us the last element I want
takeWhile (/=0) list gives us the first element I want
But I don't see a way of getting both in a simple manner. I could enumerate the list and use indexing, but that seems messy. Is there a proper way of doing this, or a way of working around this?
I would zip the list with its tail so that you have pairs of elements
available. Then you can just use find on the list of pairs:
f :: [Int] -> Maybe (Int, Int)
f xs = find ((>3) . snd) (zip xs (tail xs))
> f [1..10]
Just (3,4)
If the first element matches the predicate this will return
Nothing (or the second match if there is one) so you might need to special-case that if you want something
different.
As Robin Zigmond says break can also work:
g :: [Int] -> (Int, Int)
g xs = case break (>3) xs of (_, []) -> error "not found"
([], _) -> error "first element"
(ys, z:_) -> (last ys, z)
(Or have this return a Maybe as well, depending on what you need.)
But this will, I think, keep the whole prefix ys in memory until it
finds the match, whereas f can start garbage-collecting the elements
it has moved past. For small lists it doesn't matter.
I would use a zipper-like search:
type ZipperList a = ([a], [a])
toZipperList :: [a] -> ZipperList a
toZipperList = (,) []
moveUntil' :: (a -> Bool) -> ZipperList a -> ZipperList a
moveUntil' _ (xs, []) = (xs, [])
moveUntil' f (xs, (y:ys))
| f y = (xs, (y:ys))
| otherwise = moveUntil' f (y:xs, ys)
moveUntil :: (a -> Bool) -> [a] -> ZipperList a
moveUntil f = moveUntil' f . toZipperList
example :: [Int]
example = [2,3,5,7,11,13,17,19]
result :: ZipperList Int
result = moveUntil (>10) example -- ([7,5,3,2], [11,13,17,19])
The good thing about zippers is that they are efficient, you can access as many elements near the index you want, and you can move the focus of the zipper forwards and backwards. Learn more about zippers here:
http://learnyouahaskell.com/zippers
Note that my moveUntil function is like break from the Prelude but the initial part of the list is reversed. Hence you can simply get the head of both lists.
A non-awkward way of implementing this as a fold is making it a paramorphism. For general explanatory notes, see this answer by dfeuer (I took foldrWithTails from it):
-- The extra [a] argument f takes with respect to foldr
-- is the tail of the list at each step of the fold.
foldrWithTails :: (a -> [a] -> b -> b) -> b -> [a] -> b
foldrWithTails f n = go
where
go (a : as) = f a as (go as)
go [] = n
boundary :: (a -> Bool) -> [a] -> Maybe (a, a)
boundary p = foldrWithTails findBoundary Nothing
where
findBoundary x (y : _) bnd
| p y = Just (x, y)
| otherwise = bnd
findBoundary _ [] _ = Nothing
Notes:
If p y is true we don't have to look at bnd to get the result. That makes the solution adequately lazy. You can check that by trying out boundary (> 1000000) [0..] in GHCi.
This solution gives no special treatment to the edge case of the first element of the list matching the condition. For instance:
GHCi> boundary (<1) [0..9]
Nothing
GHCi> boundary even [0..9]
Just (1,2)
There's several alternatives; either way, you'll have to implement this yourself. You could use explicit recursion:
getLastAndFirst :: (a -> Bool) -> [a] -> Maybe (a, a)
getLastAndFirst p (x : xs#(y:ys))
| p y = Just (x, y)
| otherwise = getLastAndFirst p xs
getLastAndFirst _ [] = Nothing
Alternately, you could use a fold, but that would look fairly similar to the above, except less readable.
A third option is to use break, as suggested in the comments:
getLastAndFirst' :: (a -> Bool) -> [a] -> Maybe (a,a)
getLastAndFirst' p l =
case break p l of
(xs#(_:_), (y:_)) -> Just (last xs, y)
_ -> Nothing
(\(xs, ys) -> [last xs, head ys]) $ break (==0) list
Using break as Robin Zigmond suggested ended up short and simple, not using Maybe to catch edge-cases, but I could replace the lambda with a simple function that used Maybe.
I toyed a bit more with the solution and came up with
breakAround :: Int -> Int -> (a -> Bool) -> [a] -> [a]
breakAround m n cond list = (\(xs, ys) -> (reverse (reverse take m (reverse xs))) ++ take n ys) $ break (cond) list
which takes two integers, a predicate, and a list of a, and returns a single list of m elements before the predicate and n elements after.
Example: breakAround 3 2 (==0) [3,2,1,0,10,20,30] would return [3,2,1,0,10]
I'm trying to convert basic functions into higher order functions (specifically map, filter, or foldr). I was wondering if there are any simple concepts to apply where I could see old functions I've written using guards and turn them into higher order.
I'm working on changing a function called filterFirst that removes the first element from the list (second argument) that does not satisfy a given predicate function (first argument).
filterFirst :: (a -> Bool) -> [a] -> [a]
filterFirst _ [] = []
filterFirst x (y:ys)
| x y = y : filterFirst x ys
| otherwise = ys
For an example:
greaterOne :: Num a=>Ord a=>a->Bool
greaterOne x = x > 1
filterFirst greaterOne [5,-6,-7,9,10]
[5,-7,9,10]
Based on the basic recursion, I was wondering if there might be a way to translate this (and similar functions) to higher order map, filter, or foldr. I'm not very advanced and these functions are new to me.
There is a higher-order function that's appropriate here, but it's not in the base library. What's the trouble with foldr? If you just fold over the list, you'll end up rebuilding the whole thing, including the part after the deletion.
A more appropriate function for the job is para from the recursion-schemes package (I've renamed one of the type variables):
para :: Recursive t => (Base t (t, r) -> r) -> t -> r
In the case of lists, this specializes to
para :: (ListF a ([a], r) -> r) -> [a] -> r
where
data ListF a b = Nil | Cons a b
deriving (Functor, ....)
This is pretty similar to foldr. The recursion-schemes equivalent of foldr is
cata :: Recursive t => (Base t r -> r) -> t -> r
Which specializes to
cata :: (ListF a r -> r) -> [a] -> r
Take a break here and figure out why the type of cata is basically equivalent to that of foldr.
The difference between cata and para is that para passes the folding function not only the result of folding over the tail of the list, but also the tail of the list itself. That gives us an easy and efficient way to produce the rest of the list after we've found the first non-matching element:
filterFirst :: (a -> Bool) -> [a] -> [a]
filterFirst f = para go
where
--go :: ListF a ([a], [a]) -> [a]
go (Cons a (tl, r))
| f a = a : r
| otherwise = tl
go Nil = []
para is a bit awkward for lists, since it's designed to fit into a more general context. But just as cata and foldr are basically equivalent, we could write a slightly less awkward function specifically for lists.
foldrWithTails
:: (a -> [a] -> b -> b)
-> b -> [a] -> b
foldrWithTails f n = go
where
go (a : as) = f a as (go as)
go [] = n
Then
filterFirst :: (a -> Bool) -> [a] -> [a]
filterFirst f = foldrWithTails go []
where
go a tl r
| f a = a : r
| otherwise = tl
First, let's flip the argument order of your function. This will make a few steps easier, and we can flip it back when we're done. (I'll call the flipped version filterFirst'.)
filterFirst' :: [a] -> (a -> Bool) -> [a]
filterFirst' [] _ = []
filterFirst' (y:ys) x
| x y = y : filterFirst' ys x
| otherwise = ys
Note that filterFirst' ys (const True) = ys for all ys. Let's substitute that in place:
filterFirst' :: [a] -> (a -> Bool) -> [a]
filterFirst' [] _ = []
filterFirst' (y:ys) x
| x y = y : filterFirst' ys x
| otherwise = filterFirst' ys (const True)
Use if-else instead of a guard:
filterFirst' :: [a] -> (a -> Bool) -> [a]
filterFirst' [] _ = []
filterFirst' (y:ys) x = if x y then y : filterFirst' ys x else filterFirst' ys (const True)
Move the second argument to a lambda:
filterFirst' :: [a] -> (a -> Bool) -> [a]
filterFirst' [] = \_ -> []
filterFirst' (y:ys) = \x -> if x y then y : filterFirst' ys x else filterFirst' ys (const True)
And now this is something we can turn into a foldr. The pattern we were going for is that filterFirst' (y:ys) can be expressed in terms of filterFirst' ys, without using ys otherwise, and we're now there.
filterFirst' :: Foldable t => t a -> (a -> Bool) -> [a]
filterFirst' = foldr (\y f -> \x -> if x y then y : f x else f (const True)) (\_ -> [])
Now we just need to neaten it up a bit:
filterFirst' :: Foldable t => t a -> (a -> Bool) -> [a]
filterFirst' = foldr go (const [])
where go y f x
| x y = y : f x
| otherwise = f (const True)
And flip the arguments back:
filterFirst :: Foldable t => (a -> Bool) -> t a -> [a]
filterFirst = flip $ foldr go (const [])
where go y f x
| x y = y : f x
| otherwise = f (const True)
And we're done. filterFirst implemented in terms of foldr.
Addendum: Although filter isn't strong enough to build this, filterM is when used with the State monad:
{-# LANGUAGE FlexibleContexts #-}
import Control.Monad.State
filterFirst :: (a -> Bool) -> [a] -> [a]
filterFirst x ys = evalState (filterM go ys) False
where go y = do
alreadyDropped <- get
if alreadyDropped || x y then
return True
else do
put True
return False
If we really want, we can write filterFirst using foldr, since foldr is kind of "universal" -- it allows any list transformation we can perform using recursion. The main downside is that the resulting code is rather counter-intuitive. In my opinion, explicit recursion is far better in this case.
Anyway here's how it is done. This relies on what I consider to be an antipattern, namely "passing four arguments to foldr". I call this an antipattern since foldr is usually called with three arguments only, and the result is not a function taking a fourth argument.
filterFirst :: (a->Bool)->[a]->[a]
filterFirst p xs = foldr go (\_ -> []) xs True
where
go y ys True
| p y = y : ys True
| otherwise = ys False
go y ys False = y : ys False
Clear? Not very much. The trick here is to exploit foldr to build a function Bool -> [a] which returns the original list if called with False, and the filtered-first list if called with True. If we craft that function using
foldr go baseCase xs
the result is then obviously
foldr go baseCase xs True
Now, the base case must handle the empty list, and in such case we must return a function returning the empty list, whatever the boolean argument is. Hence, we arrive at
foldr go (\_ -> []) xs True
Now, we need to define go. This takes as arguments:
a list element y
the result of the "recursion" ys (a function Bool->[a] for the rest of the list)
and must return a function Bool->[a] for the larger list. So let's also consider
a boolean argument
and finally make go return a list. Well, if the boolean is False we must return the list unchanged, so
go y ys False = y : ys False
Note that ys False means "the tail unchanged", so we are really rebuilding the whole list unchanged.
If instead the boolean is true, we query the predicate as in p y. If that is false, we discard y, and return the list tail unchanged
go y ys True
| p y = -- TODO
| otherwise = ys False
If p y is true, we keep y and we return the list tail filtered.
go y ys True
| p y = y : ys True
| otherwise = ys False
As a final note, we cold have used a pair ([a], [a]) instead of a function Bool -> [a], but that approach does not generalize as well to more complex cases.
So, that's all. This technique is something nice to know, but I do not recommend it in real code which is meant to be understood by others.
Joseph and chi's answers already show how to derive a foldr implementation, so I'll try to aid intuition.
map is length-preserving, filterFirst is not, so trivially map must be unsuited for implementing filterFirst.
filter (and indeed map) are memoryless - the same predicate/function is applied to each element of the list, regardless of the result on other elements. In filterFirst, behaviour changes once we see the first non-satisfactory element and remove it, so filter (and map) are unsuited.
foldr is used to reduce a structure to a summary value. It's very general, and it might not be immediately obvious without experience what sorts of things this may cover. filterFirst is in fact such an operation, though. The intuition is something like, "can we build it in a single pass through the structure, building it up as we go(, with additional state stored as required)?". I fear Joseph's answer obfuscates a little, as foldr with 4 parameters, it may not be immediately obvious what's going on, so let's try it a little differently.
filterFirst p xs = snd $ foldr (\a (deleted,acc) -> if not deleted && not (p a) then (True,acc) else (deleted,a:acc) ) (False,[]) xs
Here's a first attempt. The "extra state" here is obviously the bool indicating whether or not we've deleted an element yet, and the list accumulates in the second element of the tuple. At the end we call snd to obtain just the list. This implementation has the problem, however, that we delete the rightmost element not satisfying the predicate, because foldr first combines the rightmost element with the neutral element, then the second-rightmost, and so on.
filterFirst p xs = snd $ foldl (\(deleted,acc) a -> if not deleted && not (p a) then (True,acc) else (deleted,a:acc) ) (False,[]) xs
Here, we try using foldl. This does delete the leftmost non-satisfactory element, but has the side-effect of reversing the list. We can stick a reverse at the front, and this would solve the problem, but is somewhat unsatisfactory due to the double-traversal.
Then, if you go back to foldr, having realized that (basically) if you want transform a list whilst preserving order that foldr is the correct variant, you play with it for a while and end up writing what Joseph suggested. I do however agree with chi that straightforward recursion is the best solution here.
Your function can also be expressed as an unfold, or, more specifically, as an apomorphism. Allow me to begin with a brief explanatory note, before the solution itself.
The apomorphism is the recursion scheme dual to the paramorphism (see dfeuer's answer for more about the latter). Apomorphisms are examples of unfolds, which generate a structure from a seed. For instance, Data.List offers unfoldr, a list unfold.
unfoldr :: (b -> Maybe (a, b)) -> b -> [a]
The function given to unfoldr takes a seed and either produces a list element and a new seed (if the maybe-value is a Just) or terminates the list generation (if it is Nothing). Unfolds are more generally expressed by the ana function from recursion-schemes ("ana" is short for "anamorphism").
ana :: Corecursive t => (a -> Base t a) -> a -> t
Specialised to lists, this becomes...
ana #[_] :: (b -> ListF a b) -> b -> [a]
... which is unfoldr in different clothing.
An apomorphism is an unfold in which the generation of the structure can be short-circuited at any point of the process, by producing, instead of a new seed, the rest of the structure in a fell swoop. In the case of lists, we have:
apo #[_] :: (b -> ListF a (Either [a] b)) -> b -> [a]
Either is used to trigger the short-circuit: with a Left result, the unfold short-circuits, while with a Right it proceeds normally.
The solution in terms of apo is fairly direct:
{-# LANGUAGE LambdaCase #-}
import Data.Functor.Foldable
filterFirst :: (a -> Bool) -> [a] -> [a]
filterFirst p = apo go
where
go = \case
[] -> Nil
a : as
| p a -> Cons a (Right as)
| otherwise -> case as of
[] -> Nil
b : bs -> Cons b (Left bs)
It is somewhat more awkward than dfeuer's para-based solution, because if we want to short-circuit without an empty list for a tail we are compelled to emit one extra element (the b in the short-circuiting case), and so we have to look one position ahead. This awkwardness would grow by orders of magnitude if, rather than filterFirst, we were to impĺement plain old filter with an unfold, as beautifully explained in List filter using an anamorphism.
This answer is inspired by a comment from luqui on a now-deleted question.
filterFirst can be implemented in a fairly direct way in terms of span:
filterFirst :: (a -> Bool) -> [a] -> [a]
filterFirst p = (\(yeas, rest) -> yeas ++ drop 1 rest) . span p
span :: (a -> Bool) -> [a] -> ([a], [a]) splits the list in two at the first element for which the condition doesn't hold. After span, we drop the first element of the second part of the list (with drop 1 rather than tail so that we don't have to add a special case for []), and reassemble the list with (++).
As an aside, there is a near-pointfree spelling of this implementation which I find too pretty not to mention:
filterFirst :: (a -> Bool) -> [a] -> [a]
filterFirst p = uncurry (++) . second (drop 1) . span p
While span is a higher order function, it would be perfectly understandable if you found this implementation disappointing in the context of your question. After all, span is not much more fundamental than filterFirst itself. Shouldn't we try going a little deeper, to see if we can capture the spirit of this solution while expressing it as a fold, or as some other recursion scheme?
I believe functions like filterFirst can be fine demonstrations of hylomorphisms. A hylomorphism is an unfold (see my other answer for more on that) that generates an intermediate data structure followed by a fold which turns this data structure into something else. Though it might look like that would require two passes to get a result (one through the input structure, and another through the intermediate one), if the hylomorphism implemented properly (as done in the hylo function of recursion-schemes) it can be done in a single pass, with the fold consuming pieces of the intermediate structure as they are generated by the unfold (so that we don't have to actually build it all only to tear it down).
Before we start, here is the boilerplate needed to run what follows:
{-# LANGUAGE LambdaCase #-}
{-# LANGUAGE DeriveFunctor #-}
{-# LANGUAGE DeriveFoldable #-}
{-# LANGUAGE DeriveTraversable #-}
{-# LANGUAGE TypeFamilies #-}
{-# LANGUAGE TemplateHaskell #-}
import Data.Functor.Foldable
import Data.Functor.Foldable.TH
The strategy here is picking an intermediate data structure for the hylomorphism that expresses the essence of what we want to achieve. In this case, we will use this cute thing:
data BrokenList a = Broken [a] | Unbroken a (BrokenList a)
-- I won't actually use those instances here,
-- but they are nice to have if you want to play with the type.
deriving (Eq, Show, Functor, Foldable, Traversable)
makeBaseFunctor ''BrokenList
BrokenList is very much like a list (Broken and Unbroken mirror [] and (:), while the makeBaseFunctor incantation generates a BrokenListF base functor analogous to ListF, with BrokenF and UnbrokenF constructors), except that it has another list attached at its end (the Broken constructor). It expresses, in a quite literal way, the idea of a list being divided in two parts.
With BrokenList at hand, we can write the hylomorphism. coalgSpan is the operation used for the unfold, and algWeld, the one used for the fold.
filterFirst p = hylo algWeld coalgSpan
where
coalgSpan = \case
[] -> BrokenF []
x : xs
| p x -> UnbrokenF x xs
| otherwise -> BrokenF xs
algWeld = \case
UnbrokenF x yeas -> x : yeas
BrokenF rest -> rest
coalgSpan breaks the list upon hitting a x element such that p x doesn't hold. Not adding that element to the second part of the list (BrokenF xs rather than BrokenF (x : xs)) takes care of the filtering. As for algWeld, it is used to concatenate the two parts (it is very much like what we would use to implement (++) using cata).
(For a similar example of BrokenList in action, see the breakOn implementation in Note 5 of this older answer of mine. It suggests what it would take to implement span using this strategy.)
There are at least two good things about this hylo-based implementation. Firstly, it has good performance (casual testing suggests that, if compiled with optimisations, it is at least as good as, and possibly slightly faster than, the most efficient implementations in other answers here). Secondly, it reflects very closely your original, explicitly recursive implementation of filterFirst (or, at any rate, more closely than the fold-only and unfold-only implementations).
The code below retains, for a given integer n, the first n items from a list, drops the following n items, keeps the following n and so on. It works correctly for any finite list.
In order to make it usable with infinite lists, I used the 'seq' operator to force the accumulator evaluation before the recursive step as in foldl' as example.
I tested by tracing the accumulator's value and it seems that it is effectively computed as desired with finite lists.
Nevertheless, it doesn't work when applied to an infinite list. The "take" in the main function is only executed once the inner calculation is terminated, what, of course, never happens with an infinite list.
Please, can someone tell me where is my mistake?
main :: IO ()
main = print (take 2 (foo 2 [1..100]))
foo :: Show a => Int -> [a] -> [a]
foo l lst = inFoo keepOrNot 1 l lst []
inFoo :: Show a => (Bool -> Int -> [a] -> [a] -> [a]) -> Int -> Int -> [a] -> [a] -> [a]
inFoo keepOrNot i l [] lstOut = lstOut
inFoo keepOrNot i l lstIn lstOut = let lstOut2 = (keepOrNot (odd i) l lstIn lstOut) in
stOut2 `seq` (inFoo keepOrNot (i+1) l (drop l lstIn) lstOut2)
keepOrNot :: Bool -> Int -> [a] -> [a] -> [a]
keepOrNot b n lst1 lst2 = case b of
True -> lst2 ++ (take n lst1)
False -> lst2
Here's how list concatenation is implemented:
(++) :: [a] -> [a] -> [a]
(++) [] ys = ys
(++) (x:xs) ys = x : xs ++ ys
Note that
the right hand list structure is reused as is (even if it's not been evaluated yet, so lazily)
the left hand list structure is rewritten (copied)
This means that if you're using ++ to build up a list, you want the accumulator to be on the right hand side. (For finite lists, merely for efficiency reasons --- if the accumulator is on the left hand side, it will be repeatedly copied and this is inefficient. For infinite lists, the caller can't look at the first element of the result until it's been copied for the last time, and there won't be a last time because there's always something else to concatenate onto the right of the accumulator.)
The True case of keepOrNot has the accumulator on the left of the ++. You need to use a different data structure.
The usual idiom in this case is to use difference lists. Instead of using type [a] for your accumulator, use [a] -> [a]. Your accumulator is now a function that prepends a list to the list it's given as input. This avoids repeated copying, and the list can be built lazily.
keepOrNot :: Bool -> Int -> [a] -> ([a] -> [a]) -> ([a] -> [a])
keepOrNot b n lst1 acc = case b of
True -> acc . (take n lst1 ++)
False -> acc
The initial value of the accumulator should be id. When you want to convert it to a conventional list, call it with [] (i.e., acc []).
seq is a red herring here. seq does not force the entire list. seq only determines whether it is of the form [] or x : xs.
You're learning Haskell, yes? So it would be a good idea as an exercise to modify your code to use a difference list accumulator. Possibly the use of infinite lists will burn you in a different part of your code; I don't know.
But there is a better approach to writing foo.
foo c xs = map snd . filter fst . zipWith f [0..] $ xs
where f i x = (even (i `div` c), x)
So you want to group a list into groups of n elements, and drop every other group. We can write this down directly:
import Data.List (unfoldr)
groups n xs = takeWhile (not.null) $ unfoldr (Just . splitAt n) xs
foo c xs = concatMap head . groups 2 . groups c $ xs
dave4420 already explained what is wrong with your code, but I'd like to comment on how you got there, IMO. Your keepOrNot :: Bool -> Int -> [a] -> [a] -> [a] function is too general. It works according to the received Bool, any Bool; but you know that you will feed it a succession of alternating True and False values. Programming with functions is like plugging a pipe into a funnel - output of one function serves as input to the next - and the funnel is too wide here, so the contact is loose.
A minimal re-write of your code along these lines could be
foo n lst = go lst
where
go lst = let (a,b) = splitAt n lst
(c,d) = splitAt n b
in
a ++ go d
The contact is "tight", there's no "information leakage" here. We just do the work twice (*) ourselves, and "connect the pipes" explicitly, in this code, grabbing one result (a) and dropping the other (c).
--
(*) twice, reflecting the two Boolean values, True and False, alternating in a simple fashion one after another. Thus this is captured frozen in the code's structure, not hanging loose as a parameter able to accommodate an arbitrary Boolean value.
Like dava4420 said, you shouldn't be using (++) to accumulate from the left. But perhaps you shouldn't be accumulating at all! In Haskell, lazyness makes straighforward "head-construction" often more efficient than the tail recursions you'd need to use in e.g. Lisp. For example:
foo :: Int -> [a] -> [a] -- why would you give this a Show constraint?
foo ℓ = foo' True
where foo' _ [] = []
foo' keep lst
| keep = firstℓ ++ foo' False other
| otherwise = foo' True other
where (firstℓ, other) = splitAt ℓ lst