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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]
How can I apply a function to only a single element of a list?
Any suggestion?
Example:
let list = [1,2,3,4,3,6]
function x = x * 2
in ...
I want to apply function only to the first occurance of 3 and stop there.
Output:
List = [1,2,6,4,3,6] -- [1, 2, function 3, 4, 3, 6]
To map or not to map, that is the question.
Better not to map.
Why? Because map id == id anyway, and you only want to map through one element, the first one found to be equal to the argument given.
Thus, split the list in two, change the found element, and glue them all back together. Simple.
See: span :: (a -> Bool) -> [a] -> ([a], [a]).
Write: revappend (xs :: [a]) (ys :: [a]) == append (reverse xs) ys, only efficient.
Or fuse all the pieces together into one function. You can code it directly with manual recursion, or using foldr. Remember,
map f xs = foldr (\x r -> f x : r) [] xs
takeWhile p xs = foldr (\x r -> if p x then x : r else []) [] xs
takeUntil p xs = foldr (\x r -> if p x then [x] else x : r) [] xs
filter p xs = foldr (\x r -> if p x then x : r else r) [] xs
duplicate xs = foldr (\x r -> x : x : r) [] xs
mapFirstThat p f xs = -- ... your function
etc. Although, foldr won't be a direct fit, as you need the combining function of the (\x xs r -> ...) variety. That is known as paramorphism, and can be faked by feeding tails xs to the foldr, instead.
you need to maintain some type of state to indicate the first instance of the value, since map will apply the function to all values.
Perhaps something like this
map (\(b,x) -> if (b) then f x else x) $ markFirst 3 [1,2,3,4,3,6]
and
markFirst :: a -> [a] -> [(Boolean,a)]
markFirst a [] = []
markFirst a (x:xs) | x==a = (True,x): zip (repeat False) xs
| otherwise = (False,x): markFirst a xs
I'm sure there is an easier way, but that's the best I came up with at this time on the day before Thanksgiving.
Here is another approach based on the comment below
> let leftap f (x,y) = f x ++ y
leftap (map (\x -> if(x==3) then f x else x)) $ splitAt 3 [1,2,3,4,3,6]
You can just create a simple function which multiples a number by two:
times_two :: (Num a) => a -> a
times_two x = x * 2
Then simply search for the specified element in the list, and apply times_two to it. Something like this could work:
map_one_element :: (Eq a, Num a) => a -> (a -> a) -> [a] -> [a]
-- base case
map_one_element _ _ [] = []
-- recursive case
map_one_element x f (y:ys)
-- ff element is found, apply f to it and add rest of the list normally
| x == y = f y : ys
-- first occurence hasnt been found, keep recursing
| otherwise = y : map_one_element x f ys
Which works as follows:
*Main> map_one_element 3 times_two [1,2,3,4,3,6]
[1,2,6,4,3,6]
I'm trying to reverse a list in Haskell, and then use the reversed list and do some work on it, but I don't know how, since Haskell picks up the third line and ignores my last line of code
Note: I know I have pattern match overlap but I don't know how to fix it.
This is what I have :
test :: (a -> Bool) -> [a] -> [a]
test _ [] = []
test _ (x:xs) = reverse xs ++ [x]
test func (x:xs)
I tried to use the where clause like this:
test :: (a -> Bool) -> [a] -> [a]
test _ [] = []
test func (x:xs)
| some code here
| some code here
where test _ (x:xs) = reverse xs ++ [x]
But it does not look like it accepts it.
The easiest way to reverse the list before further processing is to use a helper function:
func f xs = func' f (reverse xs)
where
func' _ [] = ...
func' f (x:xs) = ...
(Note that reversing the empty list is a no-op, so you don't need to handle it specially in func.)
remark:
as #chi just pointed out there is something fishy with the types/function
So you either want to filter or use the resulting map - I assumed the later (as you talked about it) but your signature suggests the former.
I'll edit the post and put on the first version as well
When you do your pattern matching like this:
test _ (x:xs)
test func (x:xs)
then both lines will match the same thing - the only difference is that in
the later case you named the first parameter (func) while in the first
you say I don't need the function
And as Haskell matches from top to bottom you will never see the last line
and so you will never get to the right-side of it
so
test :: ([a] -> b) -> [a] -> b
test f [] = f []
test f (x:xs) = f (reverse xs ++ [x])
or (the filter version)
test :: (a -> Bool) -> [a] -> [a]
test _ [] = []
test f (x:xs) = filter f (reverse xs ++ [x])
would have worked (although the deconstruction into (x:xs) and even the
first case are not really needed as #foo remarked)
The idea with the where was a good step but you try to mix it with pattern matching the list (you don't have to here at all - reverse will do it for you):
test :: ([a] -> b) -> [a] -> b
test f xs = f xs'
where xs' = reverse xs
or filter version:
test :: (a -> Bool) -> [a] -> [a]
test f xs = filter f xs'
where xs' = reverse xs
of course once you have this you might see that you can use
composition too:
test f xs = f (reverse xs) = (f . reverse) xs
or filter version
test f xs = filter f (reverse xs) = (filter f . reverse) xs
so that you can write
test f = f . reverse
or filtered
test f = filter f . reverse
instead too
btw: here you can see what I meant by
just put your guards in whatever f is
in my comment as if you implement the first version then you get the second by just replacing f with filter g where g is your guard
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)
I have recently been teaching myself Haskell, and one of my exercises was to re-implement the filter function. However, of all the exercises I have performed, my answer for this one seems to me the most ugly and long. How could I improve it? Are there any Haskell tricks I don't yet know?
myfilter :: (a -> Bool) -> [a] -> [a]
myfilter f (x:xs) = if f x
then x : myfilter f xs
else myfilter f xs
myfilter _ [] = []
Thank You
The simplest way to neaten your implementation is to use guards. Instead of pattern = value, you can write write pattern | boolean = value; this will only match when boolean is true. Thus, we can get
filter1 :: (a -> Bool) -> [a] -> [a]
filter1 p (x:xs) | p x = x : filter1 p xs
| otherwise = filter1 p xs
filter1 _ [] = []
(Note that otherwise is just a synonym for True.) Now, we have filter p xs in two places, so we can move it out into a where clause; these are shared by everything which shares a common pattern, even if it has a different guard:
filter2 :: (a -> Bool) -> [a] -> [a]
filter2 p (x:xs) | p x = x : xs'
| otherwise = xs'
where xs' = filter2 p xs
filter2 _ [] = []
(This implementation is the one used by GHCs Prelude.)
Now, neither of these are tail-recursive. This can be disadvantageous, but it does make the function lazy. If we want a tail-recursive version, we could write
filter3 :: (a -> Bool) -> [a] -> [a]
filter3 p xs = let filter3' p (x:xs) ys | p x = next $! x:ys
| otherwise = next $! ys
where next = filter3' p xs
filter3' _ [] ys = reverse ys
in filter3' p xs []
Note, however, that this would fail on infinite lists (though all the other implementations will work), thanks to the reverse, so we make it strict with $!. (I think I did this right—I could have forced the wrong variable. I think I got this one right, though.)
Those implementations all look like yours. There are, of course, others. One is based on foldr:
filter4 :: (a -> Bool) -> [a] -> [a]
filter4 p = let check x | p x = (x :)
| otherwise = id
in foldr check []
We take advantage of point-free style here; since xs would be the last argument to both filter4 and foldr check [], we can elide it, and similarly with the last argument of check.
You could also take advantage of the list monad:
import Control.Monad
filter5 :: MonadPlus m => (a -> Bool) -> m a -> m a
filter5 p xs = do x <- xs
guard $ p x
return x
The list monad represents nondeterminism. You pick an element x from xs, make sure that it satisfies p, and then return it if it does. All of these results are then collected together. But note that this is now more general; this works for any MonadPlus (a monad which is also a monoid; that is, which has an associative binary operation mplus—++ for lists—and an identity element mzero—[] for lists), such as [] or Maybe. For instance, filter5 even $ Just 1 == Nothing, and filter5 even $ Just 2 == Just 2.
We can also adapt the foldr-based version to get a different generic type signature:
import Control.Monad
import qualified Data.Foldable as F
import qualified Data.Monoid as M
filter6 :: (F.Foldable f, MonadPlus m, M.Monoid (m a))
=> (a -> Bool) -> f a -> m a
filter6 p = let check x | p x = return x
| otherwise = mzero
in F.foldMap check
The Data.Foldable module provides the Foldable type class, which represents any structure which can be folded like a list (placing the result in a generic Monoid instead.) Our filter requires a MonadPlus constraint on the result as well so that we can write return x. The foldMap function requires a function which converts everything to elements of a Monoid, and then concatenates all of them together. The mismatch between the f a on the left and the m a on the right means you could, for instance, filter6 a Maybe and get back a list.
I'm sure that there are (many!) other implementations of filter, but these are the 6 that I could think of relatively quickly. Now, which of these do I actually like best? It's a tossup between the straightforward filter2 and the foldr-based filter4. And filter5 is nice for its generic type signature. (I don't think I've ever needed a type signature like filter6's.) The fact that filter2 is used by GHC is a plus, but GHC also uses some funky rewrite rules, so it's not obvious to me that it's superior without those. Personally, I would probably go with filter4 (or filter5 if I needed the genericity), but filter2 is definitely fine.
How about a list comprehension?
myfilter f xs = [x | x <- xs, f x]
You could at least DRY it up a bit by pulling out that common myfilter f xs code:
myfilter :: (a -> Bool) -> [a] -> [a]
myfilter f (x:xs) = if f x
then x : rest
else rest
where rest = myfilter f xs
myfilter _ [] = []
For comparison, here's Wikipedia's implementation:
myfilter :: (a -> Bool) -> [a] -> [a]
myfilter _ [] = []
myfilter f (x:xs) | f x = x : myfilter f xs
| otherwise = myfilter f xs
In Haskell, most of the time you can (and should) use guards instead of if-then-else:
myfilter :: (a -> Bool) -> [a] -> [a]
myfilter f (x:xs)
| f x = x : myfilter f xs
| otherwise = myfilter f xs
myfilter _ [] = []
This ends up being basically the same definition as used in the standard library.