just started learning haskell - love it after a week. At the moment going through the monads pain, not there yet but hopefully it will click.
I am trying to put together a function similar to pythons walk() but simpler.
Given a path I want to generate list of tuples. A tuple for each sub directory (lets just assume that there will be only directories). The tuple would contain path of the directory as its first element and list of files the directory contains as the second element.
I don't know if I explained it correctly but here is the code:
walkDir :: String -> IO [IO (FilePath, [FilePath])]
walkDir path = do
dir <- getDirectoryContents path
let nd = [x | x <- dir, notElem x [".",".."]]
return (map getcont nd)
where
getcont path = do
cont <- getDirectoryContents path
return (path,cont)
My concern is IO inside IO and how to deal with it?? Is it possible to cancel them out?
Is it possible to unwrap at least the internal IO? Is this kind of return normal?
I can not even print this kind of return. Do I have to create an instance of show for this to be printed correctly?
There most likely exist a similar function in some haskell library but this is for educational purpose. I want to learn. So any suggestions welcome.
Thank you very much.
Check out Control.Monad for mapM.
Then this
return (map getcont nd)
becomes
mapM getcont nd
Let's look at the types.
map :: (a -> b) -> [a] -> [b]
getCont :: FilePath -> IO (FilePath, [FilePath])
nd :: [FilePath]
map getCont nd :: [IO (FilePath, FilePath)]
Now, at this point, the structure looks inside-out. We have [IO a] but we want IO [a]. Stop! Hoogle time. Generalizing for any ol' monad, we hoogle [m a] -> m [a]. Lo and behold, sequence has that precise type signature. So instead of return :: a -> m a you should use sequence :: [m a] -> m [a], like this:
sequence (map getCont nd)
Then you'll be all set. Notice that this is essentially identical to Kurt S's solution, because
mapM f xs = sequence (map f xs)
Related
Though it is okay to bind IO [[Char]] and IO () but its not allowed to bind Maybe with IO. Can someone give an example how this relaxation would lead to a bad design? Why freedom in the polymorphic type of Monad is allowed though not the Monad itself?
There are a lot of good theoretical reasons, including "that's not what Monad is." But let's step away from that for a moment and just look at the implementation details.
First off - Monad isn't magic. It's just a standard type class. Instances of Monad only get created when someone writes one.
Writing that instance is what defines how (>>) works. Usually it's done implicitly through the default definition in terms of (>>=), but that just is evidence that (>>=) is the more general operator, and writing it requires making all the same decisions that writing (>>) would take.
If you had a different operator that worked on more general types, you have to answer two questions. First, what would the types be? Second, how would you go about providing implementations? It's really not clear what the desired types would be, from your question. One of the following, I guess:
class Poly1 m n where
(>>) :: m a -> n b -> m b
class Poly2 m n where
(>>) :: m a -> n b -> n b
class Poly3 m n o | m n -> o where
(>>) :: m a -> n b -> o b
All of them could be implemented. But you lose two really important factors for using them practically.
You need to write an instance for every pair of types you plan to use together. This is a massively more complex undertaking than just an instance for each type. Something about n vs n^2.
You lose predictability. What does the operation even do? Here's where theory and practice intersect. The theory behind Monad places a lot of restrictions on the operations. Those restrictions are referred to as the "monad laws". They are beyond the ability to verify in Haskell, but any Monad instance that doesn't obey them is considered to be buggy. The end result is that you quickly can build an intuition for what the Monad operations do and don't do. You can use them without looking up the details of every type involved, because you know properties that they obey. None of those possible classes I suggested give you any kind of assurances like that. You just have no idea what they do.
I’m not sure that I understand your question correctly, but it’s definitely possible to compose Maybe with IO or [] in the same sense that you can compose IO with [].
For example, if you check the types in GHCI using :t,
getContents >>= return . lines
gives you an IO [String]. If you add
>>= return . map Text.Read.readMaybe
you get a type of IO [Maybe a], which is a composition of IO, [] and Maybe. You could then pass it to
>>= return . Data.Maybe.catMaybes
to flatten it to an IO [a]. Then you might pass the list of parsed valid input lines to a function that flattens it again and computes an output.
Putting this together, the program
import Text.Read (readMaybe)
import Data.Maybe (catMaybes)
main :: IO ()
main = getContents >>= -- IO String
return . lines >>= -- IO [String]
return . map readMaybe >>= -- IO [Maybe Int]
return . catMaybes >>= -- IO [Int]
return . (sum :: [Int] -> Int) >>= -- IO Int
print -- IO ()
with the input:
1
2
Ignore this!
3
prints 6.
It would also be possible to work with an IO (Maybe [String]), a Maybe [IO String], etc.
You can do this with >> as well. Contrived example: getContents >> (return . Just) False reads the input, ignores it, and gives you back an IO (Maybe Bool).
To illustrate the point with a trivial example, say I have implemented filter:
filter :: (a -> Bool) -> [a] -> [a]
And I have a predicate p that interacts with the real world:
p :: a -> IO Bool
How do it make it work with filter without writing a separate implementation:
filterIO :: (a -> IO Bool) -> [a] -> IO [a]
Presumably if I can turn p into p':
p': IO (a -> Bool)
Then I can do
main :: IO ()
main = do
p'' <- p'
print $ filter p'' [1..100]
But I haven't been able to find the conversion.
Edited:
As people have pointed out in the comment, such a conversion doesn't make sense as it would break the encapsulation of the IO Monad.
Now the question is, can I structure my code so that the pure and IO versions don't completely duplicate the core logic?
How do it make it work with filter without writing a separate implementation
That isn't possible and the fact this sort of thing isn't possible is by design - Haskell places firm limits on its types and you have to obey them. You cannot sprinkle IO all over the place willy-nilly.
Now the question is, can I structure my code so that the pure and IO versions don't completely duplicate the core logic?
You'll be interested in filterM. Then, you can get both the functionality of filterIO by using the IO monad and the pure functionality using the Identity monad. Of course, for the pure case, you now have to pay the extra price of wrapping/unwrapping (or coerceing) the Identity wrapper. (Side remark: since Identity is a newtype this is only a code readability cost, not a runtime one.)
ghci> data Color = Red | Green | Blue deriving (Read, Show, Eq)
Here is a monadic example (note that the lines containing only Red, Blue, and Blue are user-entered at the prompt):
ghci> filterM (\x -> do y<-readLn; pure (x==y)) [Red,Green,Blue]
Red
Blue
Blue
[Red,Blue] :: IO [Color]
Here is a pure example:
ghci> filterM (\x -> Identity (x /= Green)) [Red,Green,Blue]
Identity [Red,Blue] :: Identity [Color]
As already said, you can use filterM for this specific task. However, it is usually better to keep with Haskell's characteristic strict seperation of IO and calculations. In your case, you can just tick off all necessary IO in one go and then do the interesting filtering in nice, reliable, easily testable pure code (i.e. here, simply with the normal filter):
type A = Int
type Annotated = (A, Bool)
p' :: Annotated -> Bool
p' = snd
main :: IO ()
main = do
candidates <- forM [1..100] $ \n -> do
permitted <- p n
return (n, permitted)
print $ fst <$> filter p' candidates
Here, we first annotate each number with a flag indicating what the environment says. This flag can then simply be read out in the actual filtering step, without requiring any further IO.
In short, this would be written:
main :: IO ()
main = do
candidates <- forM [1..100] $ \n -> (n,) <$> p n
print $ fst <$> filter snd candidates
While it is not feasible for this specific task, I'd also add that you can in principle achieve the IO seperation with something like your p'. This requires that the type A is “small enough” that you can evaluate the predicate with all values that are possible at all. For instance,
import qualified Data.Map as Map
type A = Char
p' :: IO (A -> Bool)
p' = (Map.!) . Map.fromList <$> mapM (\c -> (c,) <$> p c) ['\0'..]
This evaluates the predicate once for all of the 1114112 chars there are and stores the results in a lookup table.
What is the intuitive meaning of join for a Monad?
The monads-as-containers analogies make sense to me, and inside these analogies join makes sense. A value is double-wrapped and we unwrap one layer. But as we all know, a monad is not a container.
How might one write sensible, understandable code using join in normal circumstances, say when in IO?
An action :: IO (IO a) is a way of producing a way of producing an a. join action, then, is a way of producing an a by running the outermost producer of action, taking the producer it produced and then running that as well, to finally get to that juicy a.
join collapses consecutive layers of the type constructor.
A valid join must satisfy the property that, for any number of consecutive applications of the type constructor, it shouldn't matter the order in which we collapse the layers.
For example
ghci> let lolol = [[['a'],['b','c']],[['d'],['e']]]
ghci> lolol :: [[[Char]]]
ghci> lolol :: [] ([] ([] Char)) -- the type can also be expressed like this
ghci> join (fmap join lolol) -- collapse inner layers first
"abcde"
ghci> join (join lolol) -- collapse outer layers first
"abcde"
(We used fmap to "get inside" the outer monadic layer so that we could collapse the inner layers first.)
A small non container example where join is useful: for the function monad (->) a, join is equivalent to \f x -> f x x, a function of type (a -> a -> b) -> a -> b that applies two times the same argument to another function.
For the List monad, join is simply concat, and concatMap is join . fmap.
So join implicitly appears in any list expression which uses concat
or concatMap.
Suppose you were asked to find all of the numbers which are divisors of any
number in an input list. If you have a divisors function:
divisors :: Int -> [Int]
divisors n = [ d | d <- [1..n], mod n d == 0 ]
you might solve the problem like this:
foo xs = concat $ (map divisors xs)
Here we are thinking of solving the problem by first mapping the
divisors function over all of the input elements and then concatenating
all of the resulting lists. You might even think that this is a very
"functional" way of solving the problem.
Another approch would be to write a list comprehension:
bar xs = [ d | x <- xs, d <- divisors x ]
or using do-notation:
bar xs = do x <- xs
d <- divisors
return d
Here it might be said we're thinking a little more
imperatively - first draw a number from the list xs; then draw
a divisors from the divisors of the number and yield it.
It turns out, though, that foo and bar are exactly the same function.
Morever, these two approaches are exactly the same in any monad.
That is, for any monad, and appropriate monadic functions f and g:
do x <- f
y <- g x is the same as: (join . fmap g) f
return y
For instance, in the IO monad if we set f = getLine and g = readFile,
we have:
do x <- getLine
y <- readFile x is the same as: (join . fmap readFile) getLine
return y
The do-block is a more imperative way of expressing the action: first read a
line of input; then treat returned string as a file name, read the contents
of the file and finally return the result.
The equivalent join expression seems a little unnatural in the IO-monad.
However it shouldn't be as we are using it in exactly the same way as we
used concatMap in the first example.
Given an action that produces another action, run the action and then run the action that it produces.
If you imagine some kind of Parser x monad that parses an x, then Parser (Parser x) is a parser that does some parsing, and then returns another parser. So join would flatten this into a Parser x that just runs both actions and returns the final x.
Why would you even have a Parser (Parser x) in the first place? Basically, because fmap. If you have a parser, you can fmap a function that changes the result over it. But if you fmap a function that itself returns a parser, you end up with a Parser (Parser x), where you probably want to just run both actions. join implements "just run both actions".
I like the parsing example because a parser typically has a runParser function. And it's clear that a Parser Int is not an integer. It's something that can parse an integer, after you give it some input to parse from. I think a lot of people end up thinking of an IO Int as being just a normal integer but with this annoying IO bit that you can't get rid of. It isn't. It's an unexecuted I/O operation. There's no integer "inside" it; the integer doesn't exist until you actually perform the I/O.
I find these things easier to interpret by writing out the types and refactoring them a bit to reveal what the functions do.
Reader monad
The Reader type is defined thus, and its join function has the type shown:
newtype Reader r a = Reader { runReader :: r -> a }
join :: Reader r (Reader r a) -> Reader r a
Since this is a newtype, this means that the type Reader r a is isomorphic to r -> a. So we can refactor the type definition to give us this type that, albeit it's not the same, it's really "the same" with scare quotes:
In the (->) r monad, which is isomorphic to Reader r, join is the function:
join :: (r -> r -> a) -> r -> a
So the Reader join is the function that takes a two-place function (r -> r -> a) and applies to the same value at both its argument positions.
Writer monad
Since the Writer type has this definition:
newtype Writer w a = Writer { runWriter :: (a, w) }
...then when we remove the newtype, its join function has a type isomorphic to:
join :: Monoid w => ((a, w), w) -> (a, w)
The Monoid constraint needs to be there because the Monad instance for Writer requires it, and it lets us guess right away what the function does:
join ((a, w0), w1) = (a, w0 <> w1)
State monad
Similarly, since State has this definition:
newtype State s a = State { runState :: s -> (a, s) }
...then its join is like this:
join :: (s -> (s -> (a, s), s)) -> s -> (a, s)
...and you can also venture just writing it directly:
join f s0 = (a, s2)
where
(g, s1) = f s0
(a, s2) = g s1
{- Here's the "map" to the variable names in the function:
f g s2 s1 s0 s2
join :: (s -> (s -> (a, s ), s )) -> s -> (a, s )
-}
If you stare at this type a bit, you might think that it bears some resemblance to both the Reader and Writer's types for their join operations. And you'd be right! The Reader, Writer and State monads are all instances of a more general pattern called update monads.
List monad
join :: [[a]] -> [a]
As other people have pointed out, this is the type of the concat function.
Parsing monads
Here comes a really neat thing to realize. Very often, "fancy" monads turn out to be combinations or variants of "basic" ones like Reader, Writer, State or lists. So often what I do when confronted with a novel monad is ask: which of the basic monads does it resemble, and how?
Take for example parsing monads, which have been brought up in other answers here. A simplistic parser monad (with no support for important things like error reporting) looks like this:
newtype Parser a = Parser { runParser :: String -> [(a, String)] }
A Parser is a function that takes a string as input, and returns a list of candidate parses, where each candidate parse is a pair of:
A parse result of type a;
The leftovers (the suffix of the input string that was not consumed in that parse).
But notice that this type looks very much like the state monad:
newtype Parser a = Parser { runParser :: String -> [(a, String)] }
newtype State s a = State { runState :: s -> (a, s) }
And this is no accident! Parser monads are nondeterministic state monads, where the state is the unconsumed portion of the input string, and parse steps generate alternatives that may be later rejected in light of further input. List monads are often called "nondeterminism" monads, so it's no surprise that a parser resembles a mix of the state and list monads.
And this intuition can be systematized by using monad transfomers. The state monad transformer is defined like this:
newtype StateT s m a = StateT { runStateT :: s -> m (a, s) }
Which means that the Parser type from above can be written like this as well:
type Parser a = StateT String [] a
...and its Monad instance follows mechanically from those of StateT and [].
The IO monad
Imagine we could enumerate all of the possible primitive IO actions, somewhat like this:
{-# LANGUAGE GADTs #-}
data Command a where
-- An action that writes a char to stdout
putChar :: Char -> Command ()
-- An action that reads a char from stdin
getChar :: Command Char
-- ...
Then we could think of the IO type as this (which I've adapted from the highly-recommended Operational monad tutorial):
data IO a where
-- An `IO` action that just returns a constant value.
Return :: a -> IO a
-- An action that binds the result of a `Command` to
-- a function that computes the next step after it.
Bind :: Command x -> (x -> IO a) -> IO a
instance Monad IO where ...
Then join action would then look like this:
join :: IO (IO a) -> IO a
-- If the action is just `Return`, then its payload already
-- is what we need to return.
join (Return ioa) = ioa
-- If the action is a `Bind`, then its "next step" function
-- `f` produces `IO (IO a)`, so we can just recursively stick
-- a `join` to its result end.
join (Bind cmd f) = Bind cmd (join . f)
So all that the join does here is "chase down" the IO action until it sees a result that fits the pattern Return (ma :: IO a), and strip out the outer Return.
So what did I do here? Just like for parser monads, I just defined (or rather copied) a toy model of the IO type that has the virtue of being transparent. Then I work out the behavior of join from the toy model.
Consider this:
ruleset = [rule0, rule1, rule2, rule3, rule4, rule5]
where rule0, rule1, etc. are boolean functions that take one argument. What is the cleanest way to find if all elements of a particular list satisfy all the rules in the ruleset?
Obviously, a loop would work, but Haskell folks always seem to have clever one-liners for these types of problems.
The all function seems appropriate (eg. all (== check_one_element) ruleset) or nested maps. Also, map ($ anElement) ruleset is roughly what I want, but for all elements.
I'm a novice at Haskell and the many ways one could approach this problem are overwhelming.
If you require all the functions to be true for each argument, then it's just
and (ruleset <*> list)
(You'll need to import Control.Applicative to use <*>.)
Explanation:
When <*> is given a pair of lists, it applies each function from the list on the left to each argument from the list on the right, and gives back a list containing all the results.
A one-liner:
import Control.Monad.Reader
-- sample data
rulesetL = [ (== 1), (>= 2), (<= 3) ]
list = [1..10]
result = and $ concatMap (sequence rulesetL) list
(The type we're working on here is Integer, but it could be anything else.)
Let me explain what's happening: rulesetL is of type [Integer -> Bool]. By realizing that (->) e is a monad, we can use
sequence :: Monad m => [m a] -> m [a]
which in our case will get specialized to type [Integer -> Bool] -> (Integer -> [Bool]). So
sequence rulesetL :: Integer -> [Bool]
will pass a value to all the rules in the list. Next, we use concatMap to apply this function to list and collect all results into a single list. Finally, calling
and :: [Bool] -> Bool
will check that all combinations returned True.
Edit: Check out dave4420's answer, it's nicer and more concise. Mine answer could help if you'd need to combine rules and apply them later on some lists. In particular
liftM and . sequence :: [a -> Bool] -> (a -> Bool)
combines several rules into one. You can also extend it to other similar combinators like using or etc. Realizing that rules are values of (->) a monad can give you other useful combinators, such as:
andRules = liftM2 (&&) :: (a -> Bool) -> (a -> Bool) -> (a -> Bool)
orRules = liftM2 (||) :: (a -> Bool) -> (a -> Bool) -> (a -> Bool)
notRule = liftM not :: (a -> Bool) -> (a -> Bool)
-- or just (not .)
etc. (don't forget to import Control.Monad.Reader).
An easier-to-understand version (without using Control.Applicative):
satisfyAll elems ruleset = and $ map (\x -> all ($ x) ruleset) elems
Personally, I like this way of writing the function, as the only combinator it uses explicitly is and:
allOkay ruleset items = and [rule item | rule <- ruleset, item <- items]
I am doing a haskell exercise, regarding define a function accumulate :: [IO a] -> IO [a]
which performs a sequence of interactions and accumulates their result in a list.
What makes me confused is how to express a list of IO a ? (action:actions)??
how to write recursive codes using IO??
This is my code, but these exists some problem...
accumulate :: [IO a] -> IO [a]
accumulate (action:actions) = do
value <- action
list <- accumulate (action:actions)
return (convert_to_list value list)
convert_to_list:: Num a =>a -> [a]-> [a]
convert_to_list a [] = a:[]
convert_to_list x xs = x:xs
What you are trying to implement is sequence from Control.Monad.
Just to let you find the answer instead of giving it, try searching for [IO a] -> IO [a] on hoogle (there's a Source link on the right hand side of the page when you've chosen a function).
Try to see in your code what happens when list of actions is empty list and see what does sequence do to take care of that.
There is already such function in Control.Monad and it called sequence (no you shouldn't look at it). You should denote the important decision taken during naming of it. Technically [IO a] says nothing about in which order those Monads should be attached to each other, but name sequence puts a meaning of sequential attaching.
As for the solving you problem. I'd suggest to look more at types and took advice of #sacundim. In GHCi (interpreter from Glasgow Haskell Compiler) there is pretty nice way to check type and thus understand expression (:t (:) will return (:) :: a -> [a] -> [a] which should remind you one of you own function but with less restrictive types).
First of all I'd try to see at what you have showed with more simple example.
data MyWrap a = MyWrap a
accumulate :: [MyWrap a] -> MyWrap [a]
accumulate (action:actions) = MyWrap (convert_to_list value values) where
MyWrap value = action -- use the pattern matching to unwrap value from action
-- other variant is:
-- value = case action of
-- MyWrap x -> x
MyWrap values = accumulate (action:actions)
I've made the same mistake that you did on purpose but with small difference (values is a hint). As you probably already have been told you could try to interpret any of you program by trying to inline appropriate functions definitions. I.e. match definitions on the left side of equality sign (=) and replace it with its right side. In your case you have infinite cycle. Try to solve it on this sample or your and I think you'll understand (btw your problem might be just a typo).
Update: Don't be scary when your program will fall in runtime with message about pattern match. Just think of case when you call your function as accumulate []
Possibly you looking for sequence function that maps [m a] -> m [a]?
So the short version of the answer to your question is, there's (almost) nothing wrong with your code.
First of all, it typechecks:
Prelude> let accumulate (action:actions) = do { value <- action ;
list <- accumulate (action:actions) ; return (value:list) }
Prelude> :t accumulate
accumulate :: (Monad m) => [m t] -> m [t]
Why did I use return (value:list) there? Look at your second function, it's just (:). Calling g
g a [] = a:[]
g a xs = a:xs
is the same as calling (:) with the same arguments. This is what's known as "eta reduction": (\x-> g x) === g (read === as "is equivalent").
So now just one problem remains with your code. You've already taken a value value <- action out of the action, so why do you reuse that action in list <- accumulate (action:actions)? Do you really have to? Right now you have, e.g.,
accumulate [a,b,c] ===
do { v1<-a; ls<-accumulate [a,b,c]; return (v1:ls) } ===
do { v1<-a; v2<-a; ls<-accumulate [a,b,c]; return (v1:v2:ls) } ===
do { v1<-a; v2<-a; v3<-a; ls<-accumulate [a,b,c]; return (v1:v2:v3:ls) } ===
.....
One simple fix and you're there.