I want to write a Haskell program that replicates the elements of a list a given number of times.
Here's my code:
repli :: [a] -> a -> [a]
repli xs n = foldl1 (\x -> take n (repeat x)) xs
My problem is I get the following errors when compiling:
'take' is applied to too many arguments
couldn't match expected type '[a]->[a]' with actual type '[[a]]'
The type signature for foldl1 is:
foldl1 :: (a -> a -> a) -> [a] -> a
Hence, the first argument needs to be a function of two arguments. The lambda expression you're passing only takes one argument. You probably meant to do something like this:
repli :: [a] -> Int -> [a]
repli xs n = concat $ map (\x -> take n (repeat x)) xs
Or, to do it better, you can use the replicate function:
repli :: [a] -> Int -> [a]
repli xs n = concat $ map (replicate n) xs
Or, to do it even betterer, you can use the [] monad:
repli :: [a] -> Int -> [a]
repli xs n = xs >>= replicate n
How about this very simple line:
repli :: [a] -> Int -> [a]
repli xs n = concat (replicate n xs)
Consider Prelude replicate before rolling out your own function: http://zvon.org/other/haskell/Outputprelude/replicate_f.html
repli xs n = foldr (++) [] (replicate n xs)
you can use "take" and "repeat". You also need to provide a base case where the list is empty.
repli [] _ = []
repli (x:xs) y = take y (repeat x) ++ repli xs y
A simple solution that worked for me
f :: Int -> [Int] -> [Int]
f n arr = concatMap (\num -> replicate n num) arr
Related
Ok so i want to implement my custom map that receives a replication factor and a target list.
Inputs: Int -> [Int]
Output: [[Int]]
E.g.: 2 [1,2] -----> [[1,1],[2,2]]
f [1,2,3] -> map -> [f(1),f(2),f(3)]
What is supposed to happen with f(1) when map goes to the next element of the list?How should i replace 1 with f(1) ?
P.S: This was my initial solution but it does replicate the initial list,not every element.
replicate::Int->[Int]->[[Int]]
replicate 1 x=x
replicate factor (x:xs)= go factor [] (x:xs) where
go factor ls (x:xs) =go factor (repl factor x):ls xs
repl 1 nr=nr
repl times nr=nr:repl (times-1) nr
There are two issues, that prevent your code from compiling:
null function has the type [a0] -> Bool, but you're applying it
on an element of a list, hence you're expecting it to be Int -> Bool
The result f x shouldn't be pushed into the tail of the input, it
should be pushed into the result of recursive call of the function:
f x: (mymap f xs) instead of f x: xs
As a result the following code should work:
mymap :: (Int -> [Int]) -> [Int]-> [[Int]]
mymap f (x:xs) = if null xs then [] else f x : (mymap f xs)
By the way, the Standard Library provides much readable (and also polymorphic) implementation using pattern-matching:
map :: (a -> b) -> [a] -> [b]
map _ [] = []
map f (x:xs) = f x : map f xs
custom map that receives a replication factor and a target list
It is a little unclear to me what you're asking for.
Does mymap receive the replication factor, or does f?
E.g.: 2 [1,2] -----> [[1,1],[2,2]]
If you want mymap 2 [1,2] to give [[1,1],[2,2]], then:
mymap :: Int -> [a] -> [[a]]
mymap = map . replicate
However,
mymap :: (Int -> [Int]) -> [Int] -> [[Int]]
How is this function any different from the built-in map :: (a -> b) -> [a] -> [b] with a as Int and b as [Int]? Here, mymap does not have any Int argument itself, so you must mean that f's argument is the replication factor; but if f 2 3 == [3,3], then f is replicate and you can use the solution above.
You can write this using your own recursive definitions, if you like:
mymap :: Int -> [a] -> [[a]]
mymap _ [] = []
mymap n (x:xs) = myreplicate n x : mymap n xs
myreplicate :: Int -> a -> [a]
myreplicate 0 _ = []
myreplicate n x = x : myreplicate (n-1) x
Or you can use list comprehension instead of a map:
mymap :: Int -> [a] -> [[a]]
mymap n xs = [ replicate n x | x <- xs ]
I'd probably call mymap for replicateMany or something like that.
Say I wanted to remove all zeros at the end of a list:
removeEndingZeros :: (Num a, Eq a) => [a] -> [a]
removeEndingZeros (xs ++ [0]) = removeEndingZeros xs
removeEndingZeros xs = xs
This does not work because of the (++) operator in the argument. How can I determine the end of a list through pattern-matching?
There is a function in Data.List to do this:
dropWhileEnd :: (a -> Bool) -> [a] -> [a]
dropWhileEnd p = foldr (\x xs -> if p x && null xs then [] else x : xs) []
So you can drop the trailing zeros with
dropWhileEnd (== 0)
Another, very similar, function can be implemented like this:
dropWhileEnd2 :: (a -> Bool) -> [a] -> [a]
dropWhileEnd2 p = foldr (\x xs -> if null xs && p x then [] else x : xs) []
dropWhileEnd2 p has exactly the same semantics as reverse . dropWhile p . reverse, but can reasonably be expected to be faster by a constant factor. dropWhileEnd has different, non-comparable strictness properties than the others (it's stricter in some ways and less strict in others).
Can you figure out circumstances under which each can be expected to be faster?
I am trying to create a function that accepts a function as parameter, and applies that function on every pair of elements in a list. For example, if I call my function foldPairs, then I would use it as so:
foldPairs (+) [1..10]
[3,7,11,15,19]
I tried using foldl in my attempt...
foldPairs :: (a->a->a) -> [a] -> [a]
foldPairs func lis = foldl func lis
However this clearly does not work. I think I might have to use curried and uncurried but I am not quite sure how. Could someone help me out?
Assuming, that for an odd-numbered input list we just discard the last element, the following will do the required:
foldPairs :: (a->a->a) -> [a] -> [a]
foldPairs _ [] = []
foldPairs _ [_] = []
foldPairs f (x:y:xs) = f x y : foldPairs f xs
The solution I would go with is to turn [1..10] into [[1,2],[3,4],[5,6],[7,8],[9,10]], then filter out any lists of length 1, convert to tuples, then map your function:
chunks :: Int -> [a] -> [[a]]
chunks n = reverse . go []
where
go acc [] = acc
go acc xs =
let (h, t) = splitAt n xs
in go (h:acc) t
Then simply
foldpairs :: (a -> a -> b) -> [a] -> [b]
foldpairs f
= map (uncurry f)
. map (\[x, y] -> (x, y))
. filter ((== 2) . length)
. chunks 2
For a function that maps a function to every nth element in a list:
mapEvery :: Int -> (a -> a) -> [a] -> [a]
mapEvery n f = zipWith ($) (drop 1 . cycle . take n $ f : repeat id)
Is it possible to implement this with foldr like ordinary map?
EDIT: In the title, changed 'folder' to 'foldr'. Autocorrect...
Here's one solution
mapEvery :: Int -> (a -> a) -> [a] -> [a]
mapEvery n f as = foldr go (const []) as 1 where
go a as m
| m == n = f a : as 1
| otherwise = a : as (m+1)
This uses the "foldl as foldr" trick to pass state from the left to the right along the list as you fold. Essentially, if we read the type of foldr as (a -> r -> r) -> r -> [a] -> r then we instantiate r as Int -> [a] where the passed integer is the current number of elements we've passed without calling the function.
Yes, it can:
mapEvery :: Int -> (a -> a) -> [a] -> [a]
mapEvery n f xs
= foldr (\y ys -> g y : ys) []
$ zip [1..] xs
where
g (i, y) = if i `mod` n == 0 then f y else y
And since it's possible to implement zip in terms of foldr, you could get even more fold-y if you really wanted. This even works on infinite lists:
> take 20 $ mapEvery 5 (+1) $ repeat 1
[1,1,1,1,2,1,1,1,1,2,1,1,1,1,2,1,1,1,1,2]
This is what it looks like with even more foldr and inlining g:
mapEvery :: Int -> (a -> a) -> [a] -> [a]
mapEvery _ _ [] = []
mapEvery n f xs
= foldr (\(i, y) ys -> (if i `mod` n == 0 then f y else y) : ys) []
$ foldr step (const []) [1..] xs
where
step _ _ [] = []
step x zipsfn (y:ys) = (x, y) : zipsfn ys
Now, would I recommend writing it this way? Absolutely not. This is about as obfuscated as you can get while still writing "readable" code. But it does demonstrate that this is possible to use the very powerful foldr to implement relatively complex functions.
I'm doing a bit of self study on functional languages (currently using Haskell). I came across a Haskell based assignment which requires defining map and filter in terms of foldr. For the life of me I'm not fully understanding how to go about this.
For example when I define a map function like:
map' :: (a -> b) -> [a] -> [b]
map' f [] = []
map' f (x:xs) = foldr (\x xs -> (f x):xs) [] xs
I don't know why the first element of the list is always ignored. Meaning that:
map' (*2) [1,2,3,4]
results in [4,6,8] instead of [2,4,6,8]
Similarly, my filter' function:
filter' :: (a -> Bool) -> [a] -> [a]
filter' p [] = []
filter' p (x:xs) = foldr (\x xs -> if p x then x:xs else xs ) [] xs
when run as:
filter' even [2,3,4,5,6]
results in [4,6] instead of [2,4,6]
Why would this be the case? And how SHOULD I have defined these functions to get the expected results? I'm assuming something is wrong with my lambda expressions...
I wish I could just comment, but alas, I don't have enough karma.
The other answers are all good ones, but I think the biggest confusion seems to be stemming from your use of x and xs.
If you rewrote it as
map' :: (a -> b) -> [a] -> [b]
map' f [] = []
map' f (x:xs) = foldr (\y ys -> (f y):ys) [] xs
you would clearly see that x is not even mentioned on the right-hand side, so there's no way that it could be in the solution.
Cheers
For your first question, foldr already has a case for the empty list, so you need not and should not provide a case for it in your own map.
map' f = foldr (\x xs -> f x : xs) []
The same holds for filter'
filter' p = foldr (\x xs -> if p x then x : xs else xs) []
Nothing is wrong with your lambda expressions, but there is something wrong with your definitions of filter' and map'. In the cons case (x:xs) you eat the head (x) away and then pass the tail to foldr. The foldr function can never see the first element you already ate. :)
Alse note that:
filter' p = foldr (\x xs -> if p x then x : xs else xs) []
is equivalent (η-equivalent) to:
filter' p xs = foldr (\x xs -> if p x then x : xs else xs) [] xs
I would define map using foldr and function composition as follows:
map :: (a -> b) -> [a] -> [b]
map f = foldr ((:).f) []
And for the case of filter:
filter :: (a -> Bool) -> [a] -> [a]
filter p = foldr (\x xs -> if p x then x:xs else xs) []
Note that it is not necessary to pass the list itself when defining functions over lists using foldr or foldl.
The problem with your solution is that you drop the head of the list and then apply the map over the list and
this is why the head of the list is missing when the result is shown.
In your definitions, you are doing pattern matching for x:xs, which means, when your argument is [1,2,3,4], x is bound to 1 and xs is bound to the rest of the list: [2,3,4].
What you should not do is simply throw away x: part. Then your foldr will be working on whole list.
So your definitions should look as follows:
map' :: (a -> b) -> [a] -> [b]
map' f [] = []
map' f xs = foldr (\x xs -> (f x):xs) [] xs
and
filter' :: (a -> Bool) -> [a] -> [a]
filter' p [] = []
filter' p xs = foldr (\x xs -> if p x then x:xs else xs ) [] xs
I am new to Haskell (in fact I've found this page asking the same question) but this is my understanding of lists and foldr so far:
lists are elements that are linked to the next element with the cons (:) operator. they terminate with the empty list []. (think of it as a binary operator just like addition (+) 1+2+3+4 = 10, 1:2:3:4:[] = [1,2,3,4]
foldr function takes a function that takes two parameters. this will replace the cons operator, which will define how each item is linked to the next.
it also takes the terminal value for the operation, which can be tought as the initial value that will be assigned to the empty list. for cons it is empty list []. if you link an empty list to any list the result is the list itself. so for a sumfunction it is 0. for a multiply function it is 1, etc.
and it takes the list itself
So my solution is as follows:
filter' p = foldr (\x n -> if p x then x : n else n) []
the lambda expression is our link function, which will be used instead of the cons (:) operator. Empty list is our default value for an empty list. If predicate is satisfied we link to the next item using (:) as normal, else we simply don't link at all.
map' f = foldr (\x n -> f x : n) []
here we link f x to the next item instead of just x, which would simply duplicate the list.
Also, note that you don't need to use pattern matching, since we already tell foldr what to do in case of an empty list.
I know this question is really old but I just wanted to answer it anyway. I hope it is not against the rules.
A different way to think about it - foldr exists because the following recursive pattern is used often:
-- Example 1: Sum up numbers
summa :: Num a => [a] -> a
summa [] = 0
summa (x:xs) = x + suma xs
Taking the product of numbers or even reversing a list looks structurally very similar to the previous recursive function:
-- Example 2: Reverse numbers
reverso :: [a] -> [a]
reverso [] = []
reverso (x:xs) = x `op` reverso xs
where
op = (\curr acc -> acc ++ [curr])
The structure in the above examples only differs in the initial value (0 for summa and [] for reverso) along with the operator between the first value and the recursive call (+ for summa and (\q qs -> qs ++ [q]) for reverso). So the function structure for the above examples can be generally seen as
-- Generic function structure
foo :: (a -> [a] -> [a]) -> [a] -> [a] -> [a]
foo op init_val [] = init_val
foo op init_val (x:xs) = x `op` foo op init_val xs
To see that this "generic" foo works, we could now rewrite reverso by using foo and passing it the operator, initial value, and the list itself:
-- Test: reverso using foo
foo (\curr acc -> acc ++ [curr]) [] [1,2,3,4]
Let's give foo a more generic type signature so that it works for other problems as well:
foo :: (a -> b -> b) -> b -> [a] -> b
Now, getting back to your question - we could write filter like so:
-- Example 3: filter
filtero :: (a -> Bool) -> [a] -> [a]
filtero p [] = []
filtero p (x:xs) = x `filterLogic` (filtero p xs)
where
filterLogic = (\curr acc -> if (p curr) then curr:acc else acc)
This again has a very similar structure to summa and reverso. Hence, we should be able to use foo to rewrite it. Let's say we want to filter the even numbers from the list [1,2,3,4]. Then again we pass foo the operator (in this case filterLogic), initial value, and the list itself. filterLogic in this example takes a p function, called a predicate, which we'll have to define for the call:
let p = even in foo (\curr acc -> if (p curr) then curr:acc else acc) [] [1,2,3,4]
foo in Haskell is called foldr. So, we've rewritten filter using foldr.
let p = even in foldr (\curr acc -> if (p curr) then curr:acc else acc) [] [1,2,3,4]
So, filter can be written with foldr as we've seen:
-- Solution 1: filter using foldr
filtero' :: (a -> Bool) -> [a] -> [a]
filtero' p xs = foldr (\curr acc -> if (p curr) then curr:acc else acc) [] xs
As for map, we could also write it as
-- Example 4: map
mapo :: (a -> b) -> [a] -> [b]
mapo f [] = []
mapo f (x:xs) = x `op` (mapo f xs)
where
op = (\curr acc -> (f curr) : acc)
which therefore can be rewritten using foldr. For example, to multiply every number in a list by two:
let f = (* 2) in foldr (\curr acc -> (f curr) : acc) [] [1,2,3,4]
So, map can be written with foldr as we've seen:
-- Solution 2: map using foldr
mapo' :: (a -> b) -> [a] -> [b]
mapo' f xs = foldr (\curr acc -> (f curr) : acc) [] xs
Your solution almost works .)
The problem is that you've got two differend bindings for x in both your functions (Inside the patternmatching and inside your lambda expression), therefore you loose track of the first Element.
map' :: (a -> b) -> [a] -> [b]
map' f [] = []
map' f (x:xs) = foldr (\x xs -> (f x):xs) [] (x:xs)
filter' :: (a -> Bool) -> [a] -> [a]
filter' p [] = []
filter' p (x:xs) = foldr (\x xs -> if p x then x:xs else xs ) [] (x:xs)
This should to the trick :). Also: you can write your functions pointfree style easily.
*Main> :{
*Main| map' :: (a -> b) -> [a] -> [b]
*Main| map' = \f -> \ys -> (foldr (\x -> \acc -> f x:acc) [] ys)
*Main| :}
*Main> map' (^2) [1..10]
[1,4,9,16,25,36,49,64,81,100]
*Main> :{
*Main| filter' :: (a -> Bool) -> [a] -> [a]
*Main| filter' = \p -> \ys -> (foldr (\x -> \acc -> if p x then x:acc else acc) [] ys)
*Main| :}
*Main> filter' (>10) [1..100]
In the above snippets acc refers to accumulator and x refers to the last element.
Everything is correct in your lambda expressions. The problem is you are missing the first element in the list. If you try,
map' f (x:xs) = foldr (\x xs -> f x:xs) [] (x:xs)
then you shouldn't miss the first element anymore. The same logic applies to filter.
filter' p (x:xs) = foldr(\ y xs -> if p y then y:xs else xs) [] (x:xs)