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I'm writing a function that compares two vectors in haskell using comprehension lists. The thing is that I want to add booleans to my final list, but Haskell interprets this code as if x == y, add the element to the list (that's how comprehensive lists works I know). What I want is a list with booleans if the coordinates I'm comparing are true or false.
Is it possible to do this with comprehension lists?
igualdad :: Vector -> Vector -> [Bool]
igualdad v1 v2 = [ x == y | x <- xs, y <- ys]
where xs = vectorToFloatList v1
ys = vectorToFloatList v2
PD: I'm going to use foldr (&&) True with the list that returns igualdad, in order to get the final result that I want.
Thanks.
What I want is a list with booleans if the coordinates I'm comparing are True or False. Is it possible to do this with comprehension lists?
You get such a list. For two Vectors v and w with lengths m and n respectively, you will get a list with m×n elements, such that the item vi and wj will be compared in the result list in the element with index i×m + j.
If you hwever want a list of length min(m, n), such that the item at index i checks if vi and wi are the same, then we can make use of zip :: [a] -> [b] -> [(a, b)]:
igualdad :: Vector -> Vector -> [Bool]
igualdad v1 v2 = [ x == y | (x, y) <- zip (vectorToFloatList xs) (vectorToFloatList ys)]
or with zipWith :: (a -> b -> c) -> [a] -> [b] -> [c] and on :: (b -> b -> c) -> (a -> b) -> a -> a -> c:
import Data.Function(on)
igualdad :: Vector -> Vector -> [Bool]
igualdad = on (zipWith (==)) vectorToFloatList
or we can make use of the ParallelListComp extension [ghc-doc] and run this with:
{-# LANGUAGE ParallelListComp #-}
igualdad :: Vector -> Vector -> [Bool]
igualdad v1 v2 = [ x == y | x <- vectorToFloatList xs | y <- vectorToFloatList ys]
PD: I'm going to use foldr (&&) True with the list that returns igualdad.
There exists a function for that already: that is and :: Foldable f => f Bool -> Bool. If you however want to check if all the items are the same, you can just use all :: Foldable f => (a -> Bool) -> f a -> Bool here:
import Data.Function(on)
sameVec :: Vector -> Vector -> Bool
sameVec = on (all (uncurry (==) .) . zip) vectorToFloatList
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 practicing some Haskell to understand the \, case.. of and Maybe better.
I've got this little function here which should return Nothing if the array is empty, Just y if y is equal to the head of the array xs and Just (tail xs) if y is not equal to the head of the array xs.
I set the return type of the function to Maybe a because in one case it should return an Int and in the other an [Int].
funct :: Int -> [Int] -> Maybe a
funct = \y xs -> case xs of
[] -> Nothing
xs -> if ((head xs) == y)
then Just y
else Just (tail xs)
What am I missing? I am getting the error that it couldn't match type a with [Int]. Isn't the a in Maybe a generic or is it influenced by the fact that I "used" the a as an Int in the Just y part?
EDIT: Ok my suggestion was bs, I tested it with Just (tail xs) in the then and else part and I'm still getting the same error.
set the return type of the function to Maybe a because in one case it should return an Int and in the other an [Int].
Haskell is statically typed. Meaning it can not - at runtime - have a different return type. It can only have one return type. a is not an ad hoc type (in the sense that it can be any type at runtime). It means that a will be determined - at compile time - based on the types of other parameters.
For instance you can write: foo :: a -> a -> a to specify that if foo takes two Ints (again known at compile time), the result will be an Int.
You can however use Either a b to say that you will either return a Left a, or a Right b. So you can rewrite it to:
funct :: Int -> [Int] -> Maybe (Either Int [Int])
funct = \y xs -> case xs of
[] -> Nothing
xs -> if ((head xs) == y)
then Just (Left y)
else Just (Right (tail xs))
Your function however is quite verbose, you can make it more clear and compact as follows:
funct :: Int -> [Int] -> Maybe (Either Int [Int])
funct _ [] = Nothing
funct y (h:t) | h == y = Just (Left y)
| otherwise = Just (Right t)
Furthermore we can generalize it to:
funct :: Eq a => a -> [a] -> Maybe (Either a [a])
funct _ [] = Nothing
funct y (h:t) | h == y = Just (Left y)
| otherwise = Just (Right t)
Here Eq is a typeclass that specifies that there exists a function (==) :: a -> a -> Bool that we can use. Otherwise using == in the body of the function would not be possible.
Furthermore we use patterns in the head of every clause. [] is a pattern that describes the empty list. (h:t) on the other hand is a pattern describing a list containing at least one element: the head h, followed by a (possibly empty tail t).
Need to create a list of tuples from a tuple with a static element and a list. Such as:
(Int, [String]) -> [(Int, String)]
Feel like this should be a simple map call but am having trouble actually getting it to output a tuple as zip would need a list input, not a constant.
I think this is the most direct and easy to understand solution (you already seem to be acquainted with map anyway):
f :: (Int, [String]) -> [(Int, String)]
f (i, xs) = map (\x -> (i, x)) xs
(which also happens to be the desugared version of [(i, x) | x < xs], which Landei proposed)
then
Prelude> f (3, ["a", "b", "c"])
[(3,"a"),(3,"b"),(3,"c")]
This solution uses pattern matching to "unpack" the tuple argument, so that the first tuple element is i and the second element is xs. It then does a simple map over the elements of xs to convert each element x to the tuple (i, x), which I think is what you're after. Without pattern matching it would be slightly more verbose:
f pair = let i = fst pair -- get the FIRST element
xs = snd pair -- get the SECOND element
in map (\x -> (i, x)) xs
Furthermore:
The algorithm is no way specific to (Int, [String]), so you can safely generalize the function by replacing Int and String with type parameters a and b:
f :: (a, [b]) -> [(a, b)]
f (i, xs) = map (\x -> (i, x)) xs
this way you can do
Prelude> f (True, [1.2, 2.3, 3.4])
[(True,1.2),(True,2.3),(True,3.4)]
and of course if you simply get rid of the type annotation altogether, the type (a, [b]) -> [(a, b)] is exactly the type that Haskell infers (only with different names):
Prelude> let f (i, xs) = map (\x -> (i, x)) xs
Prelude> :t f
f :: (t, [t1]) -> [(t, t1)]
Bonus: you can also shorten \x -> (i, x) to just (i,) using the TupleSections language extension:
{-# LANGUAGE TupleSections #-}
f :: (a, [b]) -> [(a, b)]
f (i, xs) = map (i,) xs
Also, as Ørjan Johansen has pointed out, the function sequence does indeed generalize this even further, but the mechanisms thereof are a bit beyond the scope.
For completeness, consider also cycle,
f i = zip (cycle [i])
Using foldl,
f i = foldl (\a v -> (i,v) : a ) []
Using a recursive function that illustrates how to divide the problem,
f :: Int -> [a] -> [(Int,a)]
f _ [] = []
f i (x:xs) = (i,x) : f i xs
A list comprehension would be quite intuitive and readable:
f (i,xs) = [(i,x) | x <- xs]
Do you want the Int to always be the same, just feed zip with an infinite list. You can use repeat for that.
f i xs = zip (repeat i) xs
Completely new to Haskell and learning through Learn Haskell the greater good.
I am looking at the map function
map :: (a -> b) -> [a] -> [b]
map _ [] = []
map f (x:xs) = f x : map f xs
is it possible to add a predicate to this? for example, to only map to every other element in the list?
You can code your own version of map to apply f only to even (or odd) positions as follows. (Below indices start from 0)
mapEven :: (a->a) -> [a] -> [a]
mapEven f [] = []
mapEven f (x:xs) = f x : mapOdd f xs
mapOdd :: (a->a) -> [a] -> [a]
mapOdd f [] = []
mapOdd f (x:xs) = x : mapEven f xs
If instead you want to exploit the library functions, you can do something like
mapEven :: (a->a) -> [a] -> [a]
mapEven f = map (\(flag,x) -> if flag then f x else x) . zip (cycle [True,False])
or even
mapEven :: (a->a) -> [a] -> [a]
mapEven f = map (uncurry (\flag -> if flag then f else id)) . zip (cycle [True,False])
If you want to filter using an arbitrary predicate on the index, then:
mapPred :: (Int -> Bool) -> (a->a) -> [a] -> [a]
mapPred p f = map (\(i,x) -> if p i then f x else x) . zip [0..]
A more direct solution can be reached using zipWith (as #amalloy suggests).
mapEven :: (a->a) -> [a] -> [a]
mapEven f = zipWith (\flag x -> if flag then f x else x) (cycle [True,False])
This can be further refined as follows
mapEven :: (a->a) -> [a] -> [a]
mapEven f = zipWith ($) (cycle [f,id])
The "canonical" way to perform filtering based on positions is to zip the sequence with the naturals, so as to append an index to each element:
> zip [1, 1, 2, 3, 5, 8, 13] [0..]
[(1,0),(1,1),(2,2),(3,3),(5,4),(8,5),(13,6)]
This way you can filter the whole thing using the second part of the tuples, and then map a function which discards the indices:
indexedFilterMap p f xs = (map (\(x,_) -> f x)) . (filter (\(_,y) -> p y)) $ (zip xs [0..])
oddFibsPlusOne = indexedFilterMap odd (+1) [1, 1, 2, 3, 5, 8, 13]
To be specific to you question, one might simply put
mapEveryOther f = indexedFilterMap odd f
You can map with a function (a lambda is also possible):
plusIfOdd :: Int -> Int
plusIfOdd a
| odd a = a
| otherwise = a + 100
map plusIfOdd [1..5]
As a first step, write the function for what you want to do to the individual element of the list:
applytoOdd :: Integral a => (a -> a) -> a -> a
applytoOdd f x = if odd x
then (f x)
else x
So applytoOdd function will apply the function f to the element if the element is odd or else return the same element if it is even. Now you can apply map to that like this:
λ> let a = [1,2,3,4,5]
λ> map (applytoOdd (+ 100)) a
[101,2,103,4,105]
Or if you want to add 200 to it, then:
λ> map (applytoOdd (+ 200)) a
[201,2,203,4,205]
Looking on the comments, it seems you want to map based on the index position. You can modify your applytoOdd method appropriately for that:
applytoOdd :: Integral a => (b -> b) -> (a, b) -> b
applytoOdd f (x,y) = if odd x
then (f y)
else y
Here, the type variable a corresponds to the index element. If it's odd you are applying the function to the actual element of the list. And then in ghci:
λ> map (applytoOdd (+ 100)) (zip [1..5] [1..])
[101,2,103,4,105]
λ> map (applytoOdd (+ 200)) (zip [1..5] [1..])
[201,2,203,4,205]
Or use a list comprehension:
mapOdd f x = if odd x then f x else x
[ mapOdd (+100) x | x <- [1,2,3,4,5]]
I'm glad that you're taking the time to learn about Haskell. It's an amazing language. However it does require you to develop a certain mindset. So here's what I do when I face a problem in Haskell. Let's start with your problem statement:
Is it possible to add a predicate to the map function? For example, to only map to every other element in the list?
So you have two questions:
Is it possible to add a predicate to the map function?
How to map to every other element in the list?
So the way people think in Haskell is via type signatures. For example, when an engineer is designing a building she visualizes how the building should look for the top (top view), the front (front view) and the side (side view). Similarly when functional programmers write code they visualize their code in terms of type signatures.
Let's start with what we know (i.e. the type signature of the map function):
map :: (a -> b) -> [a] -> [b]
Now you want to add a predicate to the map function. A predicate is a function of the type a -> Bool. Hence a map function with a predicate will be of the type:
mapP :: (a -> Bool) -> (a -> b) -> [a] -> [b]
However, in your case, you also want to keep the unmapped values. For example mapP odd (+100) [1,2,3,4,5] should result in [101,2,103,4,105] and not [101,103,105]. Hence it follows that the type of the input list should match the type of the output list (i.e. a and b must be of the same type). Hence mapP should be of the type:
mapP :: (a -> Bool) -> (a -> a) -> [a] -> [a]
It's easy to implement a function like this:
map :: (a -> Bool) -> (a -> a) -> [a] -> [a]
mapP p f = map (\x -> if p x then f x else x)
Now to answer your second question (i.e. how to map to every other element in the list). You could use zip and unzip as follows:
snd . unzip . mapP (odd . fst) (fmap (+100)) $ zip [1..] [1,2,3,4,5]
Here's what's happening:
We first zip the index of each element with the element itself. Hence zip [1..] [1,2,3,4,5] results in [(1,1),(2,2),(3,3),(4,4),(5,5)] where the fst value of each pair is the index.
For every odd index element we apply the (+100) function to the element. Hence the resulting list is [(1,101),(2,2),(3,103),(4,4),(5,105)].
We unzip the list resulting in two separate lists ([1,2,3,4,5],[101,2,103,4,105]).
We discard the list of indices and keep the list of mapped results using snd.
We can make this function more general. The type signature of the resulting function would be:
mapI :: ((Int, a) -> Bool) -> (a -> a) -> [a] -> [a]
The definition of the mapI function is simple enough:
mapI :: ((Int, a) -> Bool) -> (a -> a) -> [a] -> [a]
mapI p f = snd . unzip . mapP p (fmap f) . zip [1..]
You can use it as follows:
mapI (odd . fst) (+100) [1,2,3,4,5]
Hope that helps.
Is it possible to add a predicate to this? for example, to only map to every other element in the list?
Yes, but functions should ideally do one relatively simple thing only. If you need to do something more complicated, ideally you should try doing it by composing two or more functions.
I'm not 100% sure I understand your question, so I'll show a few examples. First: if what you mean is that you only want to map in cases where a supplied predicate returns true of the input element, but otherwise just leave it alone, then you can do that by reusing the map function:
mapIfTrue :: (a -> Bool) -> (a -> a) -> [a] -> [a]
mapIfTrue pred f xs = map step xs
where step x | pred x = f x
| otherwise = x
If what you mean is that you want to discard list elements that don't satisfy the predicate, and apply the function to the remaining ones, then you can do that by combining map and filter:
filterMap :: (a -> Bool) -> (a -> b) -> [a] -> [b]
filterMap pred f xs = map f (filter pred xs)
Mapping the function over every other element of the list is different from these two, because it's not a predicate over the elements of the list; it's either a structural transformation of the list of a stateful traversal of it.
Also, I'm not clear whether you mean to discard or keep the elements you're not applying the function to, which would imply different answers. If you're discarding them, then you can do it by just discarding alternate list elements and then mapping the function over the remaining ones:
keepEven :: [a] -> [a]
keepEven xs = step True xs
where step _ [] = []
step True (x:xs) = x : step False xs
step False (_:xs) = step True xs
mapEven :: (a -> b) -> [a] -> [b]
mapEven f xs = map f (keepEven xs)
If you're keeping them, one way you could do it is by tagging each list element with its position, filtering the list to keep only the ones in even positions, discard the tags and then map the function:
-- Note: I'm calling the first element of a list index 0, and thus even.
mapEven :: (a -> a) -> [a] -> [a]
mapEven f xs = map aux (filter evenIndex (zip [0..] xs))
where evenIndex (i, _) = even i
aux (_, x) = f x
As another answer mentioned, zip :: [a] -> [b] -> [(a, b)] combines two lists pairwise by position.
But this is the general philosophy: to do a complex thing, use a combination of general-purpose generic functions. If you're familiar with Unix, it's similar to that.
Another simple way to write the last one. It's longer, but keep in mind that evens, odds and interleave all are generic and reusable:
evens, odds :: [a] -> [a]
evens = alternate True
odds = alternate False
alternate :: Bool -> [a] -> [a]
alternate _ [] = []
alternate True (x:xs) = x : alternate False xs
alternate False (_:xs) = alternate True xs
interleave :: [a] -> [a] -> [a]
interleave [] ys = ys
interleave (x:xs) ys = x : interleave ys xs
mapEven :: (a -> a) -> [a] -> [a]
mapEven f xs = interleave (map f (evens xs)) (odds xs)
You can't use a predicate because predicates operate on list values, not their indices.
I quite like this format for what you're trying to do, since it makes the case handling quite clear for the function:
newMap :: (t -> t) -> [t] -> [t]
newMap f [] = [] -- no items in list
newMap f [x] = [f x] -- one item in list
newMap f (x:y:xs) = (f x) : y : newMap f xs -- 2 or more items in list
For example, running:
newMap (\x -> x + 1) [1,2,3,4]
Yields:
[2,2,4,4]