I'm having some real trouble designing the counterfunction of Haskell's sequence function, which Hoogle tells me doesn't yet exist. This is how it behaves:
ghci> sequence [Just 7, Just 8, Just 9]
Just [7,8,9]
ghci> sequence [getLine, getLine, getLine]
hey
there
stack exchange
["hey","there","stack exchange"] :: IO [String]
My problem is making a function like this:
unsequence :: (Monad m) => m [a] -> [m a]
So that it behaves like this:
ghci> unsequence (Just [7, 8, 9])
[Just 7, Just 8, Just 9]
ghci> sequence getLine
hey
['h','e','y'] :: [IO Char] --(This would actually cause an error, but hey-ho.)
I don't actually know if that's possible, because I'd be escaping the monad at some point, but I've made a start, though I don't know how to set a breakpoint for this recursive function:
unsequence m = (m >>= return . head) : unsequence (m >>= return . tail)
I realise that I need a breakpoint when the m here is equal to return [], but not all monads have Eq instances, so how can I do this? Is this even possible? If so, why and why not? Please tell me that.
You can't have an unsequence :: (Monad m) => m [a] -> [m a]. The problem lies with lists: you can't be sure how may elements you are going to get with a list, and that complicates any reasonable definition of unsequence.
Interestingly, if you were absolutely, 100% sure that the list inside the monad is infinite, you could write something like:
unsequenceInfinite :: (Monad m) => m [a] -> [m a]
unsequenceInfinite x = fmap head x : unsequenceInfinite (fmap tail x)
And it would work!
Also imagine that we have a Pair functor lying around. We can write unsequencePair as
unsequencePair :: (Monad m) => m (Pair a) -> Pair (m a)
unsequencePair x = Pair (fmap firstPairElement x) (fmap secondPairElement x)
In general, it turns out you can only define unsequence for functors with the property that you can always "zip" together two values without losing information. Infinite lists (in Haskell, one possible type for them is Cofree Identity) are an example. The Pair functor is another. But not conventional lists, or functors like Maybe or Either.
In the distributive package, there is a typeclass called Distributive that encapsulates this property. Your unsequence is called distribute there.
It is indeed not possible to create an unsequence function using monads alone. The reason is:
You can safely and easily create a monadic structure from a value using return.
However, it is not safe to remove a value from a monadic structure. For example you can't remove an element from an empty list (i.e. a function of the type [a] -> a is not safe).
Hence we have a special function (i.e. >>=) which safely removes a value from a monadic structure (if one exists), processes it and returns another safe monadic structure.
Hence it is safe to create a monadic structure from a value. However it is not safe to remove a value from a monadic structure.
Suppose we had a function extract :: Monad m => m a -> a which could “safely” remove a value from a monadic structure. We could then implement unsequence as follows:
unsequence :: Monad m => m [a] -> [m a]
unsequence = map return . extract
However, there's no safe way to extract a value from a monadic structure. Hence unsequence [] and unsequence Nothing will return undefined.
You can however create an unsequence function for structures that are both monadic and comonadic. A Comonad is defined as follows:
class Functor w => Comonad w where
extract :: w a -> a
duplicate :: w a -> w (w a)
extend :: (w a -> b) -> w a -> w b
duplicate = extend id
extend f = fmap f . duplicate
A comonadic structure is the opposite of a monadic structure. In particular:
You can safely extract a value from a comonadic structure.
However you can't safely create a new comonadic structure from a value, which is why the duplicate function safely creates a new comonadic structure from a value.
Remember that the definition of unsequence required both return and extract? You can't safely create a new comonadic structure from a value (i.e. comonadic structures don't have return). Hence the unsequence function is defined as follows:
unsequence :: (Comonad m, Monad m) => m [a] -> [m a]
unsequence = map return . extract
Interestingly sequence works on simply monadic structures. So via intuition you might assume that unsequence works on simply comonadic structures. However it not so because you need to first extract the list from the comonadic structure and then put each element of the list into a monadic structure.
The general version of the unsequence function converts a comonadic list structure to a list of monadic structures:
unsequence :: (Comonad w, Monad m) => w [a] -> [m a]
unsequence = map return . extract
On the other hand the sequence function works on simply monadic structures because you are just folding the list of monadic structures into a monadic list structure by chaining all the monads:
import Control.Monad (liftM2)
sequence :: Monad m => [m a] -> m [a]
sequence = foldr (liftM2 (:)) (return [])
Hope that helps.
Related
I built a list of this structure:
[(Interger, Double)]
The List was created by using a zip over a list of Integers and a list of Doubles of exactly the same size.
Now I want to filter the list for Doubles that are either <18.5 or >25. The problem I have is I can't access the Doubles to use them in the filter function.
It's probably easy but I'm a bloody noob in this language. I googled around a lot and read some other threads but I didn't find an answer.
I got:
filter (<18.5) listexpression
So what I'm struggling with is that listexpression. It's easy if it's a list of single values. I could filter before zipping but then I can't connect the data from the filtered list to the other unfiltered List anymore.
Edit: I forgot to mention. It's a worksheet. We were asked to build filter and map functions ourselves and are not allowed to use any additions to the basic Haskell. Meaning no imports are allowed.
You can do something like this:
Prelude> filter (\p -> (snd p) < 18.5 || (snd p) > 25) [(1, 2.3), (1, 20.0)]
[(1,2.3)]
The lambda function passed to filter, namely
(\p -> (snd p) < 18.5 || (snd p) > 25)
says that for every p, the second element of p must be less than 18.5 or over 25.
Alternatively, you could write it like this
Prelude> filter (\(_, f) -> f < 18.5 || f > 25) [(1, 2.3), (1, 20.0)]
[(1,2.3)]
Here the function says that for any pair whose first value doesn't matter and the second one is f, f must be less than 18.5 or over 25.
Glad to see Ami Tavory's answer solved your problem.
But under that answer, you commented:
I tried accessing it with a combination of (!!) but that didn't work.
With the insight of a teaching assistant [:D], I guess you confused list with tuple in Haskell.
zip returns a list of tuple, whereas (!!) take a list as (the first) argument (hence (!!1) take a single list argument), so (!!1) can't be applied to elements of the list returned by zip, which are of type tuple.
Prelude> :t zip
zip :: [a] -> [b] -> [(a, b)]
Prelude> :t (!!)
(!!) :: [a] -> Int -> a
Prelude> :t (!!1)
(!!1) :: [a] -> a
And you've known that fst and snd are applied to tuple.
Prelude> :t fst
fst :: (a, b) -> a
Prelude> :t snd
snd :: (a, b) -> b
A compact version using point free style would be
filter ((>18.5).snd) listexpression
This uses the function composition operator ., which reads as: First apply the snd function to a tuple from the list to extract the 2nd value, then apply the comparison to 18.5 to this value.
Just for a variety and some additional information which won't bite...
In Haskell the list type is an instance of Monad class. So a list operation like filter can simply be implemented by a monadic bind operator.
*Main> [(1,2.3),(3,21.2),(5,17.1),(4,24.4)] >>= \t -> if snd t < 25 && snd t > 18.5 then [t] else []
[(3,21.2),(4,24.4)]
Monad is all about handling the contained data in a sequential manner. In the list monad the contained data is the value within the list itself. So the bind operator can be very handy to access to contained values (tuples) of the monadic value (the list of tuples) in a sequential manner.
(>>=) :: Monad m => m a -> (a -> m b) -> m b
The type signature of the monadic bind operator states that it takes a monad type value m a as the first argument (the list of tuples here) and a function as the second argument which takes a pure value and returns a monadic value (takes a tuple and returns a tuple in a list or an empty list in this case).
\t -> if snd t < 25 && snd t > 18.5 then [t] else []
It's critical to understand how and why the list items are applied one by one to the provided function. An entire list is one monadic value and the contained values those accessed by the bind operator are passed to the provided a -> m b (take a pure value and return monadic value) type function. So all of the list items those applied to this function become a monadic value ([t] if condition satisfies or [] if it fails), are then concatenated by the bind operator to form one monadic return value (in this case a list of tuples those satisfy the condition which is implemented in the lambda function).
This monadic operation can also be implemented with the do notation
do
t <- [(1,2.3),(3,21.2),(5,17.1),(4,24.4)]
if snd t < 25 && snd t > 18.5 then return t else []
[(3,21.2),(4,24.4)]
Of course this terribly resembles the list comprehensions which is in fact a syntactical sugar to the monadic list operations. So lets implement it for a final time by using the list comprehensions.
*Main> [t | t <- [(1,2.3),(3,21.2),(5,17.1),(4,24.4)], snd t < 25 && snd t > 18.5]
[(3,21.2),(4,24.4)]
Given two lists, [a, b] and [c, d], I'd like to get the following result:
[(a,c), (a,d), (b,c), (b,d)]
How can I do this in Haskell? Is there a built-in function for this, or should I implement one myself?
[ (x,y) | x<-[a,b], y<-[c,d] ]
This doesn't really require any further explanation, does it?
Applicative style all the way!
λ> :m + Control.Applicative
λ> (,) <$> ['a','b'] <*> ['c','d']
[('a','c'),('a','d'),('b','c'),('b','d')]
(I've eschewed any String syntactic sugar above, in order to stay close to your example.)
For information, (,) is special syntax for a function that takes two arguments and makes a pair out of them:
λ> :t (,)
(,) :: a -> b -> (a, b)
Edit: As noted by leftaroundabout in his comment, you can also use liftA2:
λ> :m + Control.Applicative
λ> let combine = liftA2 (,)
λ> combine "ab" "cd"
[('a','c'),('a','d'),('b','c'),('b','d')]
How can you do this in an imperative pseudocode?
for each element x in [a,b]:
for each element y in [c,d]:
produce (x,y)
In Haskell, this is written as
outerProduct xs ys =
do
x <- xs -- for each x drawn from xs:
y <- ys -- for each y drawn from ys:
return (x,y) -- produce the (x,y) pair
(following comments by leftaroundabout) this is of course extremely close to how liftM2 monadic combinator is defined, so in fact
outerProduct = liftM2 (,)
which is the same as liftA2 (,), and its various re-writes in terms of list comprehensions, concatMap function, >>=, <$> and <*> operators.
Conceptually though this is the stuff of the Applicative – which would be better named as Pairing, – because this pairing up of the elements of two "containers" ⁄ "carriers" ⁄ whatever is exactly what Applicative Functor is about. It just so happens that Haskell's do notation works for monads, and not (yet) for applicatives.
In some sense compile-time nested loops are Applicative ⁄ Pairing functors; Monads add the ability to create nested loops on the fly, depending on the values produced by an "outer" enumeration.
The most inuitive would be using list comprehension, other aproaches include using applicative functors:
(,) <$> [1,2,3] <*> [4,5,6]
So what does this do?
Remember that (,) :: a -> b -> (a, b) Takes two arguments and returns a tuple.
<$> is acutally fmap, (<$>) :: Functor f => (a -> b) -> f a -> f b
It takes a function and lift it. In this case it takes (,) and lift it to work on list. So let x = (,) <$> [1,2] would generate x :: [b -> (Integer, b)] which is the the list of functions that takes b and returns tuple with one fixed argument (Integer,b). Finally we apply it using <*> to generate all the combinations.
use List Comprehension:
s = [a,b]
s' = [c,d]
all_combinations = [(x,y) | x <- s, y <- s']
I have a list of tuples, for example:
[(1,2), (3,4), (5,6)]
Now I have to write function which sum up the first an second element of each tuple and create a list of these values.
For the example above it should be:
[3, 7, 11]
This should be done with use of list comprehension. It's not allowed to use functions like map, filter and contact.
Any ideas how I could access the elements of the tuple in the list?
Try this:
[ x + y | (x,y) <- yourlist]
The trick is representing the two elements in a tuple from your input list as x and y and then dealing with them as needed.
Let's do it without list comprehensions, using functions from the Prelude:
map (uncurry (+)) [(1,2), (3,4), (5,6)]
-- Result: [3, 7, 11]
How does this work? Let's consider the types:
(+) :: Num a => a -> a -> a
uncurry :: (a -> b -> c) -> (a, b) -> c
map :: (a -> b) -> [a] -> [b]
As you may already know, in Haskell, the normal way of doing multi-argument functions is by **currying* them. The type of (+) reflects this: conceptually it takes one argument and produces a function that then takes the "second" argument to produce the final result.
uncurry takes such a curried two-argument function and adapts it to work on a pair. It's trivial to implement:
uncurry :: (a -> b -> c) -> (a, b) -> c
uncurry f (a, b) = f a b
Funnily enough, the uncurry function is curried, so its partial application uncurry (+) has type Num a => (a, a) -> a. This would then be a function that takes a pair of numbers and adds them.
And map simply applies a function to every element of a list, collecting the individual results into a list. Plug them all together and that's a solution.
In The Little Schemer there is a function to check, whether the list is flat:
(define lat?
(lambda (l)
(cond
((null? l) #t)
((atom? (car l)) (lat? (cdr l)))
(else #f))))
I'm trying to write the same recursive function in Haskell, but have no success:
is_lat :: [a] -> Bool
is_lat [] = True
is_lat ???
How do i check that the parameter is not in the form [[a]]? In other words, [1,2,3] is a valid input, but [[1,3], [2,4]] and [[[1,2,3]]] aren't.
I want to use this further in recursive functions that accept lists to make sure that i deal with flat lists only.
EDIT: I see that people are confused because of the is_lat :: [a] -> Bool type signature. I agree now, that i shouldn't check type on runtime. However, is it possible to check the type on compile-time? How can i make the function work only for flat lists? Or should i completely change my way of thinking?
You can't really think of nested lists the same way in Haskell as in Scheme, because they're not the same data structure. A Haskell list is homogenous, where as a Lisp "list" is actually closer to a rose tree (as pointed out by C.A.McCann below). As an illustrative example, take a look at how the WYAS48 parsing section defines LispVal.
If you really, really, really want to do runtime type checking, even though it's usually a bad idea and very unconventional in Haskell, look into Data.Typeable. This response might be useful too.
The real answer to this question is "You need to think about your arguments differently in Haskell than in Lisp, which results in never needing to perform this check yourself at runtime" (and I say this as a Common Lisper, so I understand how frustrating that is to start with).
Addendum: In response to your edit, Haskell's type system automatically ensures this. If you have a function of type foo :: [Int] -> Int, for example, and you pass it ["One", "Two", "Three"] or [[1, 2, 3]], you'll get a compile-time error telling you what just exploded and why. If you want to specialize a function, just declare a more specific type.
For instance (don't write code like this, it's just for illustrative purposes), say you have a simple function like
myLookup index map = lookup index map
If you load this into GHCi and run :t myLookup, it'll tell you that the functions' type is myLookup :: Eq a => a -> [(a, b)] -> Maybe b which means that it can take a key of any type that derives Eq (anything you can run == on). Now, say that for whatever reason you want to ensure that you only use numbers as keys. You'd ensure that by adding a more specific type declaration
myLookup :: Int -> [(Int, a)] -> Maybe a
myLookup index map = lookup index map
Now, even though there's nothing in the body of the function preventing it from dealing with other key types, you'll get a type error at compile time if you try to pass it something other than an Int index or something other than an [(Int, a)] map. As a result, this
myLookup :: Int -> [(Int, a)] -> Maybe a
myLookup ix lst = lookup ix lst
main :: IO ()
main = putStrLn . show $ myLookup 1 [(1, "Foo")]
will compile and run fine, but this
myLookup :: Int -> [(Int, a)] -> Maybe a
myLookup ix lst = lookup ix lst
main :: IO ()
main = putStrLn . show $ myLookup "Nope.jpg" [("Foo", 1)]
will do neither. On my machine it errors at compile time with
/home/inaimathi/test.hs:5:35:
Couldn't match expected type `Int' with actual type `[Char]'
In the first argument of `myLookup', namely `"Nope.jpg"'
In the second argument of `($)', namely
`myLookup "Nope.jpg" [("Foo", 1)]'
In the expression:
putStrLn . show $ myLookup "Nope.jpg" [("Foo", 1)]
Failed, modules loaded: none.
I really hope that didn't confuse you further.
This is both impossible and unnecessary with standard Haskell lists because Haskell is strongly typed; either all elements of a list are themselves lists (in which case the type is [a] = [[b]] for some b), or they are not.
E.g. if you try to construct a mixed list, you will get an error from the compiler:
Prelude> ["hello", ["world!"]]
<interactive>:3:12:
Couldn't match expected type `Char' with actual type `[Char]'
In the expression: "world!"
In the expression: ["world!"]
In the expression: ["hello", ["world!"]]
The function type [a] -> Bool implicitly means forall a. [a] -> Bool, in other words it's defined identically for lists of all possible element types. That does include types like [[Int]] or [[[String]]] or any depth of nesting you can think of. But it doesn't--and can't--matter to your function what the element type is.
As far as this function is concerned, the input is always a list whose elements are some opaque, unknown type. It will never receive nested lists containing that same opaque type.
Well, I guess, in Haskell you are limited to http://ideone.com/sPhRCP:
main = do -- your code goes here
print $isFlat [Node 1, Node 2, Node 3]
print $isFlat [Node 1, Node 2, Branch [Node 3, Node 4, Node 5], Node 6]
data Tree a = Node a | Branch [Tree a]
isFlat :: [Tree a] -> Bool
isFlat = all isNode where
isNode (Node _) = True
isNode _ = False
In non strictly-typed languages every object has run-time type information and thus can be polymorphic. This might be a complicated network of coercions, like in Scala (less complicated if you're in C++), or just "everything is an object, and object is everything" like in purely dynamic languages (Lisp, JS, ...).
Haskell is strictly-typed.
I have a custom list type:
data NNList a = Sing a | Append ( NNList a) ( NNList a) deriving (Eq)
data CList a = Nil | NotNil ( NNList a) deriving (Eq)
I'm trying to implement a function that returns the head and tail of a list:
cListGet :: CList a -> Maybe (a, CList a)
My attempt:
cListGet :: CList a -> Maybe (a, CList a)
cListGet Nil = Nothing
cListGet xs#(NotNil nxs) =
case nxs of
Sing x -> (x, Nil)
Append l r -> ((fst $ cListGet (NotNil l)), (Append (snd $ cListGet (NotNil l)), r))
Which to me means keep going leftwards until I get a single. Once I get the single element (head), return the element and a Nil list. This Nil list is then combined with the list before it's returned as the final result.
I'm not even sure if the logic is 100% correct.
Well, people would normally refer to the data structure you have as a kind of tree, not as a list. But anyway...
Problem #1: Haskell is indentation sensitive, and your case expression is not indented. This leads to a parse error.
Problem #2, and the bigger one: you haven't understood how the Maybe type works yet. I get the impression that you think it works like nulls in more common languages, and this is throwing you off.
In a language like, say, Java, null is a value that can occur where most any other value can. If we have a method with the following signature:
public Foo makeAFoo(Bar someBar)
...then it is legal to call it either of these ways:
// Way #1: pass in an actual value
Bar theBar = getMeABar();
Foo result = makeAFoo(theBar);
// Way #2: pass in a null
Foo result2 = makeAFoo(null)
theBar and null are "parallel" in a sense, or said more precisely, they have the same type—you can replace one with the other in a program and it will compile in both cases.
In Haskell, on the other hand, the string "hello" and Nothing do not have the same type, and you cannot use one where the other goes. Haskell distinguishes between these three things:
A string that's required to be there: "hello" :: String
The absence of an optional string: Nothing :: Maybe String
The presence of an optional string: Just "hello" :: Maybe String
The difference between #1 and #3 is what you're systematically missing in your function. With Maybe a, in the cases where you do have a value you must use Just, which acts like a wrapper to signify "this isn't just an a, it's a Maybe a."
First place you're missing Just is the right hand sides of the case expressions, which we can fix like this:
-- This still fails to compile!
cListGet :: CList a -> Maybe (a, CList a)
cListGet Nil = Nothing
cListGet xs#(NotNil nxs) =
case nxs of
-- I added 'Just' here and in the next line:
Sing x -> Just (x, Nil)
Append l r -> Just (fst $ cListGet (NotNil l), (Append (snd $ cListGet (NotNil l)), r))
But this isn't the end of it, because you're doing fst $ cListGet (NotNil l), which suffers from the converse problem: cListGet returns Maybe (a, CList a), but fst works on (a, b), not on Maybe (a, b). You need to pattern match on the result of cListGet to test whether it's Nothing or Just (x, l'). (This same problem occurs also in your snd $ cListGet (NotNil l).)
Third, you're using your Append constructor wrong. You have it in the form of (Append foo, bar), which should have no comma between foo and bar. In Haskell this sort of thing will give you more confusing error messages than most other languages, because when Haskell sees this, it doesn't tell you "you made a syntax error"; Haskell is rather more literal than most languages, so it figures you're trying to make a pair with Append foo as the first element, and bar as the second one, so it concludes that (Append foo, bar) must have type (NNList a -> NNList a, NNList a).
The fourth and final problem: the problem you've set yourself is not clearly stated, and thus has no good answer. You say you want to find the "head" and "tail" of a CList a. What does that mean? In the case of the Haskell [a] type, with constructors [] and :, this is clear: the head is the x in x:xs, and the tail is the xs.
As I understand you, what you mean by "head" seems to be the leftmost element of the recursive structure. We could get that this way:
cListHead :: CList a -> Maybe a
cListHead Nil = Nothing
-- No need to cram everything together into one definition; deal with
-- the NNList case in an auxiliary function, it's easier...
cListGet (NotNil nxs) = Just (nnListHead nxs)
-- Note how much easier this function is to write, because since 'NNList'
-- doesn't have a 'Nil' case, there's no need to mess around with 'Maybe'
-- here. Basically, by splitting the problem into two functions, only
-- 'cListHead' needs to care about 'Maybe' and 'Just'.
nnListHead :: NNList a -> a
nnListHead (Sing a) = a
nnListHead (Append l _) = nnListHead l
So you might think that "the tail" is everything else. Well, the problem is that "everything else" is not a subpart of your CList or NNList. Take this example:
example :: CList Int
example = NotNil (Append (Append (Sing 1) (Sing 2)) (Sing 3))
The "head" is 1. But there is no subpart of the structure defined in example that contains 2 and 3 without containing 1 as well. You'd have to construct a new CList with a different shape than the original to get that. That's possible to do, but I don't see the value of it as a beginner's exercise, frankly.
In case it's not clear what I mean by a "subpart," think of the example as a tree:
NotNil
|
v
Append
/ \
v v
Sing Append
| / \
v v v
1 Sing Sing
| |
v v
2 3
Subpart = subtree.
Hint: try to rewrite this using only pattern matching and not equality-checking (==).
Edit:
First off, it's crucial that you understand what pattern matching is and how it works. I'd recommend going here and reading up; there are also plenty of other resources about this on the web (Google is your friend).
Once you've done that, here's another hint: First write a function nnListGet :: NNList a -> (a, CList a), then use it to implement cListGet.
Just to add to the other (very thorough) answers: It's good to realize that your custom list is a foldable structure. This means, it represents a sequence of values that can be combined together. Such datatypes can implement Foldable type class. In your case, it would be:
import Prelude hiding (foldr)
import Data.Foldable
data NNList a = Sing a | Append (NNList a) (NNList a) deriving (Eq)
data CList a = Nil | NotNil (NNList a) deriving (Eq)
instance Foldable NNList where
foldr f z (Sing x) = f x z
foldr f z (Append xs ys) = foldr f (foldr f z ys) xs
instance Foldable CList where
foldr _ z Nil = z
foldr f z (NotNil xs) = foldr f z xs
From that you'll get all functions defined in Data.Foldable for free, such as maximum/minimum, searching for an element etc.
For any Foldable, you can implement headMaybe that returns its first element by using First monoid. It's a very simple monoid that returns the left-most non-empty element. So if you fold all elements of a Foldable using this monoid, you'll get its first one:
import Data.Monoid
headMaybe :: (Foldable f) => f a -> Maybe a
headMaybe = getFirst . foldMap (First . Just)
(Alternatively, you can use foldr directly, using Maybe's instance of Alternative, which again returns the left-most non-empty element:
import Control.Applicative
headMaybe = foldr (\x y -> pure x <|> y) Nothing
.)
However, this doesn't solve the second part of your question - computing tailMaybe. This can't be defined in a generic way like headMaybe, and you'll need your custom function for that, as you did.
See also:
Fold on Wikipedia.
Foldable and Traversable on Haskell wiki.
Fold on Haskell wiki.
List processing on Haskell wikibook.
Why did you declare that in terms of two types? Here's a seemingly more appropriate type declaration with a correct function:
data CList a
= Nil
| Sing a
| Append (CList a) (CList a)
deriving (Eq)
headAndTail :: CList a -> Maybe (a, CList a)
headAndTail Nil = Nothing
headAndTail (Sing a) = Just (a, Nil)
headAndTail (Append a b) =
case headAndTail a of
Nothing -> headAndTail b
Just (head, tail) -> Just (head, Append tail b)