I wanted to define the function:
accumulate_list' :: Num a => [a] -> ( a -> a -> a ) -> a
accumulate_list' l f
| [] f = 0
| (x:xs) f = f x (accumulate_list xs f)
but it doesn't compile, complaining that x, xs, x, xs, are Not in scope.
Trying to achieve the same result, I have defined this function
accumulate_list :: Num a => [a] -> ( a -> a -> a ) -> a
accumulate_list [] f = 0
accumulate_list (x:xs) f = f x $ accumulate_list xs f
it compiles well, and does like sum does on a list if the functgion passed in parameter is (+). Yes, I finally found that indeed what I wanted to achieve was already existing in Prelude as the sum function.
However, I don't understand why the first version with mixed guards and pattern matching doesn't compile. What is the problem ?
It is because guards are basically a boolean expression. They have to evaluate to either True or False. Something like this should typecheck:
accumulate_list' :: (Eq a, Num a) => [a] -> ( a -> a -> a ) -> a
accumulate_list' l f
| l == [] = 0
| otherwise = undefined -- fill out undefined
It's also worth mentioning that since pattern guards were added to Haskell 2010, you're allowed to mix patterns and guards like so:
accumulate_list' :: (Eq a, Num a) => [a] -> ( a -> a -> a ) -> a
accumulate_list' l f
| [] <- l = 0 --pattern for the empty list case
| 10 < 5 = 10 --arbitrary regular guard just because
| (x:xs) <- l = undefined --pattern for the non-empty case
Using what #Sibi's answer provided, I've completed it so as to provide entire working code example:
accumulate_list' :: (Eq a ,Num a) => [a] -> ( a -> a -> a ) -> a
accumulate_list' l f
| l == [] = 0
| otherwise = f x $ accumulate_list xs f
where (x:xs) = l
Related
I'm currently trying to write a function that takes as arguments an Int and an array of Ints and for every even value in the array, it concatenates the Int to the final array.
So, something like this:
f 3 [1,2,3,4,5,6] = [1,2,3,3,4,3,5,6,3]
This is the code I imagined would work (I'm just beginning so sorry if it's bad):
f :: Int -> [Int] -> [Int]
f(x,[]) = []
f(x,y)
|even head(y) = (head(y) ++ [x] ++ f(x,drop 1 y)
|otherwise = head(y) ++ f(x,(drop 1 y))
The error I'm getting is "Couldn't match expected type of 'Int' with actual type (a3, [[a3]])'. I understand the parameters types are mismatched, but I'm not sure how a proper syntax would look like here
You use (x, []), so that means the input type would be a tuple, so f :: (Int, [Int]) -> [Int].
I would also use pattern matching instead of head and tail, so:
f :: Int -> [Int] -> [Int]
f _ [] = []
f x (y:ys)
| even y = y : x : f x ys
| otherwise = y : f x ys
You can also generalize the type signature, and work with an inner function to avoid passing the x each time:
f :: Integral a => a -> [a] -> [a]
f x = go
where go [] = []
go (y:ys)
| even y = y : x : go ys
| otherwise = y : go ys
Another way of looking at this would be using a right fold to insert the desired element after even numbers.
f :: Int -> [Int] -> [Int]
f x lst = foldr (\y i -> if even y then y:x:i else y:i) [] lst
Which we can simplify to:
f :: Int -> [Int] -> [Int]
f x = foldr (\y i -> if even y then y:x:i else y:i) []
Note that without specifying the type, the more general inferred type of f would be:
f :: (Foldable t, Integral a) => a -> t a -> [a]
I am trying to create a function that will sort a list of strings in ascending order bases on the number of occurrences of a certain character without using any library functions, but i do want to know how i can use my version of inssort to do this i am bit confused any help appreciated
My code so far returns wrong order,
say i give it
sortwords 'c' ["abcc", "abc", "bbc", "aa"]
i want it to return
["aa", "abc", "bbc", abcc"]
but i get
["abc","bbc","aa","abcc"]
Here is my code:
insert :: Ord b => (a -> b) -> a -> [a] -> [a]
insert f a [] = [a]
insert f a (x:xs) = if f a <= f x then a:x:xs else x : insert f a xs
inssort :: Ord b => (a -> b) -> [a] -> [a]
inssort f [] = []
inssort f [x] = [x]
inssort f (x:xs) = insert f x (inssort f xs)
countocc :: Eq a => a -> [a] -> Int
countocc x [] = 0
countocc x xs = length [x' | x' <- xs, x == x']
sortwords :: Char -> [String] -> [String]
sortwords c [] = []
sortwords c [x]= [x]
sortwords c (x:y:ys) = if countocc c x <= countocc c y then x : sortwords c (y:ys) else y : sortwords c (x:ys)
I won't spoil the whole solution, but here's a hint: it looks like you are trying to reimplement your sorting procedure in sortwords. Don't do that; just reuse inssort directly! Try filling in this replacement function skeleton:
sortwords :: Char -> [String] -> [String]
sortwords c strings = inssort f strings where
f s = ...
What do you think the ... should be?
I would like to define a greaters function, which selects from a list items that are larger than the one before it.
For instance:
greaters [1,3,2,4,3,4,5] == [3,4,4,5]
greaters [5,10,6,11,7,12] == [10,11,12]
The definition I came up with is this :
greaters :: Ord a => [a] -> [a]
Things I tried so far:
greaters (x:xs) = group [ d | d <- xs, x < xs ]
Any tips?
We can derive a foldr-based solution by a series of re-writes starting from the hand-rolled recursive solution in the accepted answer:
greaters :: Ord a => [a] -> [a]
greaters [] = []
greaters (x:xs) = go x xs -- let's re-write this clause
where
go _ [] = []
go last (act:xs)
| last < act = act : go act xs
| otherwise = go act xs
greaters (x:xs) = go xs x -- swap the arguments
where
go [] _ = []
go (act:xs) last
| last < act = act : go xs act
| otherwise = go xs act
greaters (x:xs) = foldr g z xs x -- go ==> foldr g z
where
foldr g z [] _ = []
foldr g z (act:xs) last
| last < act = act : foldr g z xs act
| otherwise = foldr g z xs act
greaters (x:xs) = foldr g z xs x
where -- simplify according to
z _ = [] -- foldr's definition
g act (foldr g z xs) last
| last < act = act : foldr g z xs act
| otherwise = foldr g z xs act
Thus, with one last re-write of foldr g z xs ==> r,
greaters (x:xs) = foldr g z xs x
where
z = const []
g act r last
| last < act = act : r act
| otherwise = r act
The extra parameter serves as a state being passed forward as we go along the input list, the state being the previous element; thus avoiding the construction by zip of the shifted-pairs list serving the same purpose.
I would start from here:
greaters :: Ord a => [a] -> [a]
greaters [] = []
greaters (x:xs) = greatersImpl x xs
where
greatersImpl last [] = <fill this out>
greatersImpl last (x:xs) = <fill this out>
The following functions are everything you’d need for one possible solution :)
zip :: [a] -> [b] -> [(a, b)]
drop 1 :: [a] -> [a]
filter :: (a -> Bool) -> [a] -> [a]
(<) :: Ord a => a -> a -> Bool
uncurry :: (a -> b -> c) -> (a, b) -> c
map :: (a -> b) -> [a] -> [b]
snd :: (a, b) -> b
Note: drop 1 can be used when you’d prefer a “safe” version of tail.
If you like over-generalization like me, you can use the witherable package.
{-# language ScopedTypeVariables #-}
import Control.Monad.State.Lazy
import Data.Witherable
{-
class (Traversable t, Filterable t) => Witherable t where
-- `wither` is an effectful version of mapMaybe.
wither :: Applicative f => (a -> f (Maybe b)) -> t a -> f (t b)
-}
greaters
:: forall t a. (Ord a, Witherable t)
=> t a -> t a
greaters xs = evalState (wither go xs) Nothing
where
go :: a -> State (Maybe a) (Maybe a)
go curr = do
st <- get
put (Just curr)
pure $ case st of
Nothing -> Nothing
Just prev ->
if curr > prev
then Just curr
else Nothing
The state is the previous element, if there is one. Everything is about as lazy as it can be. In particular:
If the container is a Haskell list, then it can be an infinite one and everything will still work. The beginning of the list can be produced without withering the rest.
If the container extends infinitely to the left (e.g., an infinite snoc list), then everything will still work. How can that be? We only need to know what was in the previous element to work out the state for the current element.
"Roll your own recursive function" is certainly an option here, but it can also be accomplished with a fold. filter can't do it because we need some sort of state being passed, but fold can nicely accumulate the result while keeping that state at the same time.
Of course the key idea is that we keep track of last element add the next one to the result set if it's greater than the last one.
greaters :: [Int] -> [Int]
greaters [] = []
greaters (h:t) = reverse . snd $ foldl (\(a, r) x -> (x, if x > a then x:r else r)) (h, []) t
I'd really love to eta-reduce it but since we're dropping the first element and seeding the accumulator with it it kinda becomes awkward with the empty list; still, this is effectively an one-liner.
So i have come up with a foldr solution. It should be similar to what #Will Ness has demonstrated but not quite i suppose as we don't need a separate empty list check in this one.
The thing is, while folding we need to encapsulate the previous element and also the state (the result) in a function type. So in the go helper function f is the state (the result) c is the current element of interest and p is the previous one (next since we are folding right). While folding from right to left we are nesting up these functions only to run it by applyying the head of the input list to it.
go :: Ord a => a -> (a -> [a]) -> (a -> [a])
go c f = \p -> let r = f c
in if c > p then c:r else r
greaters :: Ord a => [a] -> [a]
greaters = foldr go (const []) <*> head
*Main> greaters [1,3,2,4,3,4,5]
[3,4,4,5]
*Main> greaters [5,10,6,11,7,12]
[10,11,12]
*Main> greaters [651,151,1651,21,651,1231,4,1,16,135,87]
[1651,651,1231,16,135]
*Main> greaters [1]
[]
*Main> greaters []
[]
As per rightful comments of #Will Ness here is a modified slightly more general code which hopefully doesn't break suddenly when the comparison changes. Note that const [] :: b -> [a] is the initial function and [] is the terminator applied to the result of foldr. We don't need Maybe since [] can easily do the job of Nothing here.
gs :: Ord a => [a] -> [a]
gs xs = foldr go (const []) xs $ []
where
go :: Ord a => a -> ([a] -> [a]) -> ([a] -> [a])
go c f = \ps -> let r = f [c]
in case ps of
[] -> r
[p] -> if c > p then c:r else r
So I'm learning about Haskell at the moment, and I came across this question:
The type of a function f is supposed to be a->[a]->a. The
following definitions of f are incorrect because their types are all
different from a->[a]->a:
i. f x xs = xs
ii. f x xs = x+1
iii. f x xs = x ++ xs
iv. f x (y:ys) = y
My answers as I see it are:
i) f :: a -> a -> a
This is because x or xs can be of any value and is not a list as it does not contain the ':' operator.
ii) f :: Int -> a -> Int
This is because the + operator is used on x, meaning x is of type Int.
iii) f :: Eq a => a -> a -> a
The ++ operators are used, therefore in order to concatenate they must be of the same type..?
iv) f :: a -> [a] -> a
f returns an element from the list.
The last one is definitely wrong, because it can't be of type a -> [a] -> a. Are there any others I did wrong, and why? I'm hoping I can fully understand types and how to find out the types of functions.
i) f :: a -> a -> a
f x xs = xs
This is because x or xs can be of any value and is not a list as it does not contain the ':' operator.
True, but it also does not have to be of the same type!
So, it's actually f :: a -> b -> b.
ii) f :: Int -> a -> Int
f x xs = x+1
This is because the + operator is used on x, meaning x is of type Int.
Correct. (Actually, in Haskell we get Num b => b -> a -> b which generalized the Int to any numeric type, but it's not that important.)
iii) f :: Eq a => a -> a -> a
f x xs = x ++ xs
The ++ operators are used, therefore in order to concatenate they must be of the same type..?
True, but they must be lists. Also, Eq is only needed if you use == or something which compares values.
Here, f :: [a] -> [a] -> [a].
iv) f :: a -> [a] -> a
f x (y:ys) = y
f returns an element from the list.
The type of x does not have to be the same. We get f :: b -> [a] -> a.
i. f x xs = xs
(...)
i) f :: a -> a -> a
Although this can be a type signature, you make it too restrictive. The function takes two parameters x and xs. Initially we can reason that x and xs can have different types, so we say that x :: a, and xs :: b. Since the function returns xs, the return type is b as well, so the type is:
f :: a -> b -> b
f x xs = xs
ii. f x xs = x+1
(...)
ii) f :: Int -> a -> Int
Again you make the function too restrictive. Let us again assume that x :: a and xs :: b have different types. We see that we return x + 1 (or in more canonical form (+) x 1. Since (+) has signature (+) :: Num c => c -> c -> c (we here use c since a is already used), and 1 has signature 1 :: Num d => d, we thus see that we call (+) with x and 1, as a result we know that a ~ c (a and c are the same type), and c ~ d, so as a result we obtain the signature:
f :: Num c => c -> b -> c
f x xs = x+1
iii. f x xs = x ++ xs
(...)
iii) f :: Eq a => a -> a -> a
This is wrong: we here see that f has two parameters, x :: a and xs :: b. We see that we return (++) x xs. Since (++) has signature (++) :: [c] -> [c] -> [c], we thus know that a ~ [c] and b ~ [c], so the type is:
f :: [c] -> [c] -> [c]
f x xs = x ++ xs
iv. f x (y:ys) = y
(...)
iv) f :: a -> [a] -> a
This is again too restrictive. Here we see again two parameters: x and (y:ys). We first generate a type a for x :: a, and (y:ys) :: b, since the pattern of the second parameter is (y:ys), this is a list constructor with as parameters (:) :: c -> [c] -> [c]. As a result we can derive that y :: c, and ys :: [c], furthermore the pattern (y:ys) has type [c]. Since the function returns y, we know that the return type is c, so:
f :: a -> [c] -> c
f x (y:ys) = y
Note: you can let Haskell derive the type of the function itself. In GHCi you can use the :t command to query the type of an expression. For example:
Prelude> f x (y:ys) = y
Prelude> :t f
f :: t1 -> [t] -> t
I'm learning Haskell and I'm doing now an exercise with the Maybe Class. I have to create a function which applies f("Maybe function") to a(and its following results) repeatedly until f a returns Nothing.
For example f a0 = Just a1,f a1= Just a2,...,f an = Nothing. Then
unfold f a0 = [a0,a1,...,an]
I've tried to do it and and I've got:
unfold :: (a- > Maybe a) -> a -> [a]
unfold f a = case f a of
Just n -> n: unfold f a
Nothing -> []
The problem is that the solution is:
unfold' :: ( a -> Maybe a) -> a -> [a]
unfold' f a = a : rest ( f a )
where rest Nothing = []
rest ( Just x ) = unfold' f x
And my program doesn't work like the solution. Maybe I'm using wrong "case of" but I'm not sure.
Your use of case is fine, but have a look where you cons the new value on the list, and where the solution does.
testFunc = const Nothing
unfold testFunc 1 == [] -- your version prepends only if f a isn't Nothing
unfold' testFunc 1 == [1] -- the solution _always_ prepends the current value
Also, you're using the same value all the time.
unfold :: (a -> Maybe a) ->a -> [a]
unfold f a = a : case f a of -- cons before the case
Just n -> unfold f n -- use n as parameter for f
Nothing -> []