Applying a function that may fail to all values in a list - haskell

I want to apply a function f to a list of values, however function f might randomly fail (it is in effect making a call out to a service in the cloud).
I thought I'd want to use something like map, but I want to apply the function to all elements in the list and afterwards, I want to know which ones failed and which were successful.
Currently I am wrapping the response objects of the function f with an error pair which I could then effectively unzip afterwards
i.e. something like
g : (a->b) -> a -> [ b, errorBoolean]
f : a-> b
and then to run the code ... map g (xs)
Is there a better way to do this? The other alternative approach was to iterate over the values in the array and then return a pair of arrays, one which listed the successful values and one which listed the failures. To me, this seems to be something that ought to be fairly common. Alternatively I could return some special value. What's the best practice in dealing with this??

If f is making a call out to the cloud, than f is undoubtedly using some monad, probably the IO monad or a monad derived from the IO monad. There are monadic versions of map. Here is what you would typically do, as a first attempt:
f :: A -> IO B -- defined elsewhere
g :: [A] -> IO [B]
g xs = mapM f xs
-- or, in points-free style:
g = mapM f
This has the (possibly) undesirable property that g will fail, returning no values, if any call to f fails. We fix that by making it so f returns either an answer or an error message.
type Error = String
f :: A -> IO (Either Error B)
g :: [A] -> IO [Either Error B]
g = mapM f
If you want all of the errors to be returned together, and all of the successes clumped together, you can use the lefts and rights functions from Data.Either.
h :: [A] -> IO ([B], [Error])
h xs = do ys <- g xs
return (rights ys, lefts ys)
If you don't need the error messages, just use Maybe B instead of Either Error B.
The Either data type is the most common way to represent a value which can either result in an error or a correct value. Errors use the Left constructor, correct values use the Right constructor. As a bonus, "right" also means "correct" in English, but the reason that the correct value uses the Right constructor is actually deeper (because this means we can create a functor out of the Either type which modifies correct results, which is not possible over the Left constructor).

You could write your g to return a Maybe monad:
f: a -> b
g: (a -> b) -> a -> Maybe b
If f fails, g returns Nothing, otherwise it returns Just (f x).

Related

Generate injective functions with QuickCheck?

I'm using QuickCheck to generate arbitrary functions, and I'd like to generate arbitrary injective functions (i.e. f a == f b if and only if a == b).
I thought I had it figured out:
newtype Injective = Injective (Fun Word Char) deriving Show
instance Arbitrary Injective where
arbitrary = fmap Injective fun
where
fun :: Gen (Fun Word Char)
fun = do
a <- arbitrary
b <- arbitrary
arbitrary `suchThat` \(Fn f) ->
(f a /= f b) || (a == b)
But I'm seeing cases where the generated function maps distinct inputs to the same output.
What I want:
f such that for all inputs a and b, either f a does not equal f b or a equals b.
What I think I have:
f such that there exist inputs a and b where either f a does not equal f b or a equals b.
How can I fix this?
You've correctly identified the problem: what you're generating is functions with the property ∃ a≠b. f a≠f b (which is readily true for most random functions anyway), whereas what you want is ∀ a≠b. f a≠f b. That is a much more difficult property to ensure, because you need to know about all the other function values for generating each individual one.
I don't think this is possible to ensure for general input types, however for word specifically what you can do is “fake” a function by precomputing all the output values sequentially, making sure that you don't repeat one that has already been done, and then just reading off from that predetermined chart. It requires a bit of laziness fu to actually get this working:
import qualified Data.Set as Set
newtype Injective = Injective ([Char] {- simply a list without duplicates -})
deriving Show
instance Arbitrary Injective where
arbitrary = Injective . lazyNub <$> arbitrary
lazyNub :: Ord a => [a] -> [a]
lazyNub = go Set.empty
where go _ [] = []
go forbidden (x:xs)
| x `Set.member` forbidden = go forbidden xs
| otherwise = x : go (Set.insert x forbidden) xs
This is not very efficient, and may well not be ok for your application, but it's probably the best you can do.
In practice, to actually use Injective as a function, you'll want to wrap the values in a suitable structure that has only O (log n) lookup time. Unfortunately, Data.Map.Lazy is not lazy enough, you may need to hand-bake something like a list of exponentially-growing maps.
There's also the concern that for some insufficiently big result types, it is just not possible to generate injective functions because there aren't enough values available. In fact as Joseph remarked, this is the case here. The lazyNub function will go into an infinite loop in this case. I'd say for a QuickCheck this is probably ok though.

Maybe monad and a list

Ok, so I am trying to learn how to use monads, starting out with maybe. I've come up with an example that I can't figure out how to apply it to in a nice way, so I was hoping someone else could:
I have a list containing a bunch of values. Depending on these values, my function should return the list itself, or a Nothing. In other words, I want to do a sort of filter, but with the consequence of a hit being the function failing.
The only way I can think of is to use a filter, then comparing the size of the list I get back to zero. Is there a better way?
This looks like a good fit for traverse:
traverse :: (Traversable t, Applicative f) => (a -> f b) -> t a -> f (t b)
That's a bit of a mouthful, so let's specialise it to your use case, with lists and Maybe:
GHCi> :set -XTypeApplications
GHCi> :t traverse #[] #Maybe
traverse #[] #Maybe :: (a -> Maybe b) -> [a] -> Maybe [b]
It works like this: you give it an a -> Maybe b function, which is applied to all elements of the list, just like fmap does. The twist is that the Maybe b values are then combined in a way that only gives you a modified list if there aren't any Nothings; otherwise, the overall result is Nothing. That fits your requirements like a glove:
noneOrNothing :: (a -> Bool) -> [a] -> Maybe [a]
noneOrNothing p = traverse (\x -> if p x then Nothing else Just x)
(allOrNothing would have been a more euphonic name, but then I'd have to flip the test with respect to your description.)
There are a lot of things we might discuss about the Traversable and Applicative classes. For now, I will talk a bit more about Applicative, in case you haven't met it yet. Applicative is a superclass of Monad with two essential methods: pure, which is the same thing as return, and (<*>), which is not entirely unlike (>>=) but crucially different from it. For the Maybe example...
GHCi> :t (>>=) #Maybe
(>>=) #Maybe :: Maybe a -> (a -> Maybe b) -> Maybe b
GHCi> :t (<*>) #Maybe
(<*>) #Maybe :: Maybe (a -> b) -> Maybe a -> Maybe b
... we can describe the difference like this: in mx >>= f, if mx is a Just-value, (>>=) reaches inside of it to apply f and produce a result, which, depending on what was inside mx, will turn out to be a Just-value or a Nothing. In mf <*> mx, though, if mf and mx are Just-values you are guaranteed to get a Just value, which will hold the result of applying the function from mf to the value from mx. (By the way: what will happen if mf or mx are Nothing?)
traverse involves Applicative because the combining of values I mentioned at the beginning (which, in your example, turns a number of Maybe a values into a Maybe [a]) is done using (<*>). As your question was originally about monads, it is worth noting that it is possible to define traverse using Monad rather than Applicative. This variation goes by the name mapM:
mapM :: (Traversable t, Monad m) => (a -> m b) -> t a -> m (t b)
We prefer traverse to mapM because it is more general -- as mentioned above, Applicative is a superclass of Monad.
On a closing note, your intuition about this being "a sort of filter" makes a lot of sense. In particular, one way to think about Maybe a is that it is what you get when you pick booleans and attach values of type a to True. From that vantage point, (<*>) works as an && for these weird booleans, which combines the attached values if you happen to supply two of them (cf. DarthFennec's suggestion of an implementation using any). Once you get used to Traversable, you might enjoy having a look at the Filterable and Witherable classes, which play with this relationship between Maybe and Bool.
duplode's answer is a good one, but I think it is also helpful to learn to operate within a monad in a more basic way. It can be a challenge to learn every little monad-general function, and see how they could fit together to solve a specific problem. So, here's a DIY solution that shows how to use do notation and recursion, tools which can help you with any monadic question.
forbid :: (a -> Bool) -> [a] -> Maybe [a]
forbid _ [] = Just []
forbid p (x:xs) = if p x
then Nothing
else do
remainder <- forbid p xs
Just (x : remainder)
Compare this to an implementation of remove, the opposite of filter:
remove :: (a -> Bool) -> [a] -> [a]
remove _ [] = []
remove p (x:xs) = if p x
then remove p xs
else
let remainder = remove p xs
in x : remainder
The structure is the same, with just a couple differences: what you want to do when the predicate returns true, and how you get access to the value returned by the recursive call. For remove, the returned value is a list, and so you can just let-bind it and cons to it. With forbid, the returned value is only maybe a list, and so you need to use <- to bind to that monadic value. If the return value was Nothing, bind will short-circuit the computation and return Nothing; if it was Just a list, the do block will continue, and cons a value to the front of that list. Then you wrap it back up in a Just.

How to implement delete with foldr in Haskell

I've been studying folds for the past few days. I can implement simple functions with them, like length, concat and filter. What I'm stuck at is trying to implement with foldr functions like delete, take and find. I have implemented these with explicit recursion but it doesn't seem obvious to me how to convert these types of functions to right folds.
I have studied the tutorials by Graham Hutton and Bernie Pope. Imitating Hutton's dropWhile, I was able to implement delete with foldr but it fails on infinite lists.
From reading Implement insert in haskell with foldr, How can this function be written using foldr? and Implementing take using foldr, it would seem that I need to use foldr to generate a function which then does something. But I don't really understand these solutions and don't have an idea how to implement for example delete this way.
Could you explain to me a general strategy for implementing with foldr lazy versions of functions like the ones I mentioned. Maybe you could also implement delete as an example since this probably is one of the easiest.
I'm looking for a detailed explanation that a beginner can understand. I'm not interested in just solutions, I want to develop an understanding so I can come up with solutions to similar problems myself.
Thanks.
Edit: At the moment of writing there is one useful answer but it's not quite what I was looking for. I'm more interested in an approach that uses foldr to generate a function, which then does something. The links in my question have examples of this. I don't quite understand those solutions so I would like to have more information on this approach.
delete is a modal search. It has two different modes of operation - whether it's already found the result or not. You can use foldr to construct a function that passes the state down the line as each element is checked. So in the case of delete, the state can be a simple Bool. It's not exactly the best type, but it will do.
Once you have identified the state type, you can start working on the foldr construction. I'm going to walk through figuring it out the way I did. I'll be enabling ScopedTypeVariables just so I can annotate the type of subexpressions better. One you know the state type, you know you want foldr to generate a function taking a value of that type, and returning a value of the desired final type. That's enough to start sketching things.
{-# LANGUAGE ScopedTypeVariables #-}
delete :: forall a. Eq a => a -> [a] -> [a]
delete a xs = foldr f undefined xs undefined
where
f :: a -> (Bool -> [a]) -> (Bool -> [a])
f x g = undefined
It's a start. The exact meaning of g is a little bit tricky here. It's actually the function for processing the rest of the list. It's accurate to look at it as a continuation, in fact. It absolutely represents performing the rest of the folding, with your whatever state you choose to pass along. Given that, it's time to figure out what to put in some of those undefined places.
{-# LANGUAGE ScopedTypeVariables #-}
delete :: forall a. Eq a => a -> [a] -> [a]
delete a xs = foldr f undefined xs undefined
where
f :: a -> (Bool -> [a]) -> (Bool -> [a])
f x g found | x == a && not found = g True
| otherwise = x : g found
That seems relatively straightforward. If the current element is the one being searched for, and it hasn't yet been found, don't output it, and continue with the state set to True, indicating it's been found. otherwise, output the current value and continue with the current state. This just leaves the rest of the arguments to foldr. The last one is the initial state. The other one is the state function for an empty list. Ok, those aren't too bad either.
{-# LANGUAGE ScopedTypeVariables #-}
delete :: forall a. Eq a => a -> [a] -> [a]
delete a xs = foldr f (const []) xs False
where
f :: a -> (Bool -> [a]) -> (Bool -> [a])
f x g found | x == a && not found = g True
| otherwise = x : g found
No matter what the state is, produce an empty list when an empty list is encountered. And the initial state is that the element being searched for has not yet been found.
This technique is also applicable in other cases. For instance, foldl can be written as a foldr this way. If you look at foldl as a function that repeatedly transforms an initial accumulator, you can guess that's the function being produced - how to transform the initial value.
{-# LANGUAGE ScopedTypeVariables #-}
foldl :: forall a b. (a -> b -> a) -> a -> [b] -> a
foldl f z xs = foldr g id xs z
where
g :: b -> (a -> a) -> (a -> a)
g x cont acc = undefined
The base cases aren't too tricky to find when the problem is defined as manipulating the initial accumulator, named z there. The empty list is the identity transformation, id, and the value passed to the created function is z.
The implementation of g is trickier. It can't just be done blindly on types, because there are two different implementations that use all the expected values and type-check. This is a case where types aren't enough, and you need to consider the meanings of the functions available.
Let's start with an inventory of the values that seem like they should be used, and their types. The things that seem like they must need to be used in the body of g are f :: a -> b -> a, x :: b, cont :: (a -> a), and acc :: a. f will obviously take x as its second argument, but there's a question of the appropriate place to use cont. To figure out where it goes, remember that it represents the transformation function returned by processing the rest of the list, and that foldl processes the current element and then passes the result of that processing to the rest of the list.
{-# LANGUAGE ScopedTypeVariables #-}
foldl :: forall a b. (a -> b -> a) -> a -> [b] -> a
foldl f z xs = foldr g id xs z
where
g :: b -> (a -> a) -> (a -> a)
g x cont acc = cont $ f acc x
This also suggests that foldl' can be written this way with only one tiny change:
{-# LANGUAGE ScopedTypeVariables #-}
foldl' :: forall a b. (a -> b -> a) -> a -> [b] -> a
foldl' f z xs = foldr g id xs z
where
g :: b -> (a -> a) -> (a -> a)
g x cont acc = cont $! f acc x
The difference is that ($!) is used to suggest evaluation of f acc x before it's passed to cont. (I say "suggest" because there are some edge cases where ($!) doesn't force evaluation even as far as WHNF.)
delete doesn't operate on the entire list evenly. The structure of the computation isn't just considering the whole list one element at a time. It differs after it hits the element it's looking for. This tells you it can't be implemented as just a foldr. There will have to be some sort of post-processing involved.
When that happens, the general pattern is that you build a pair of values and just take one of them at completion of the foldr. That's probably what you did when you imitated Hutton's dropWhile, though I'm not sure since you didn't include code. Something like this?
delete :: Eq a => a -> [a] -> [a]
delete a = snd . foldr (\x (xs1, xs2) -> if x == a then (x:xs1, xs1) else (x:xs1, x:xs2)) ([], [])
The main idea is that xs1 is always going to be the full tail of the list, and xs2 is the result of the delete over the tail of the list. Since you only want to remove the first element that matches, you don't want to use the result of delete over the tail when you do match the value you're searching for, you just want to return the rest of the list unchanged - which fortunately is what's always going to be in xs1.
And yeah, that doesn't work on infinite lists - but only for one very specific reason. The lambda is too strict. foldr only works on infinite lists when the function it is provided doesn't always force evaluation of its second argument, and that lambda does always force evaluation of its second argument in the pattern match on the pair. Switching to an irrefutable pattern match fixes that, by allowing the lambda to produce a constructor before ever examining its second argument.
delete :: Eq a => a -> [a] -> [a]
delete a = snd . foldr (\x ~(xs1, xs2) -> if x == a then (x:xs1, xs1) else (x:xs1, x:xs2)) ([], [])
That's not the only way to get that result. Using a let-binding or fst and snd as accessors on the tuple would also do the job. But it is the change with the smallest diff.
The most important takeaway here is to be very careful with handling the second argument to the reducing function you pass to foldr. You want to defer examining the second argument whenever possible, so that the foldr can stream lazily in as many cases as possible.
If you look at that lambda, you see that the branch taken is chosen before doing anything with the second argument to the reducing function. Furthermore, you'll see that most of the time, the reducing function produces a list constructor in both halves of the result tuple before it ever needs to evaluate the second argument. Since those list constructors are what make it out of delete, they are what matter for streaming - so long as you don't let the pair get in the way. And making the pattern-match on the pair irrefutable is what keeps it out of the way.
As a bonus example of the streaming properties of foldr, consider my favorite example:
dropWhileEnd :: (a -> Bool) -> [a] -> [a]
dropWhileEnd p = foldr (\x xs -> if p x && null xs then [] else x:xs) []
It streams - as much as it can. If you figure out exactly when and why it does and doesn't stream, you'll understand pretty much every detail of the streaming structure of foldr.
here is a simple delete, implemented with foldr:
delete :: (Eq a) => a -> [a] -> [a]
delete a xs = foldr (\x xs -> if x == a then (xs) else (x:xs)) [] xs

Why does bind (>>=) exist? What are typical cases where a solution without bind is ugly?

This is a type declaration of a bind method:
(>>=) :: (Monad m) => m a -> (a -> m b) -> m b
I read this as follows: apply a function that returns a wrapped value, to a wrapped value.
This method was included to Prelude as part of Monad typeclass. That means there are a lot of cases where it's needed.
OK, but I don't understand why it's a typical solution of a typical case at all.
If you already created a function which returns a wrapped value, why that function doesn't already take a wrapped value?
In other words, what are typical cases where there are many functions which take a normal value, but return a wrapped value? (instead of taking a wrapped value and return a wrapped value)
The 'unwrapping' of values is exactly what you want to keep hidden when dealing with monads, since it is this that causes a lot of boilerplate.
For example, if you have a sequence of operations which return Maybe values that you want to combine, you have to manually propagate Nothing if you receive one:
nested :: a -> Maybe b
nested x = case f x of
Nothing -> Nothing
Just r ->
case g r of
Nothing -> Nothing
Just r' ->
case h r' of
Nothing -> Nothing
r'' -> i r''
This is what bind does for you:
Nothing >>= _ = Nothing
Just a >>= f = f a
so you can just write:
nested x = f x >>= g >>= h >>= i
Some monads don't allow you to manually unpack the values at all - the most common example is IO. The only way to get the value from an IO is to map or >>= and both of these require you to propagate IO in the output.
Everyone focuses on IO monad and inability to "unwrap".
But a Monad is not always a container, so you can't unwrap.
Reader r a == r->a such that (Reader r) is a Monad
to my mind is the simplest best example of a Monad that is not a container.
You can easily write a function that can produce m b given a: a->(r->b). But you can't easily "unwrap" the value from m a, because a is not wrapped in it. Monad is a type-level concept.
Also, notice that if you have m a->m b, you don't have a Monad. What Monad gives you, is a way to build a function m a->m b from a->m b (compare: Functor gives you a way to build a function m a->m b from a->b; ApplicativeFunctor gives you a way to build a function m a->m b from m (a->b))
If you already created a function which returns a wrapped value, why that function doesn't already take a wrapped value?
Because that function would have to unwrap its argument in order to do something with it.
But for many choices of m, you can only unwrap a value if you will eventually rewrap your own result. This idea of "unwrap, do something, then rewrap" is embodied in the (>>=) function which unwraps for you, let's you do something, and forces you to rewrap by the type a -> m b.
To understand why you cannot unwrap without eventually rewrapping, we can look at some examples:
If m a = Maybe a, unwrapping for Just x would be easy: just return x. But how can we unwrap Nothing? We cannot. But if we know that we will eventually rewrap, we can skip the "do something" step and return Nothing for the overall operation.
If m a = [a], unwrapping for [x] would be easy: just return x. But for unwrapping [], we need the same trick as for Maybe a. And what about unwrapping [x, y, z]? If we know that we will eventually rewrap, we can execute the "do something" three times, for x, y and z and concat the results into a single list.
If m a = IO a, no unwrapping is easy because we only know the result sometimes in the future, when we actually run the IO action. But if we know that we will eventually rewrap, we can store the "do something" inside the IO action and perform it later, when we execute the IO action.
I hope these examples make it clear that for many interesting choices of m, we can only implement unwrapping if we know that we are going to rewrap. The type of (>>=) allows precisely this assumption, so it is cleverly chosen to make things work.
While (>>=) can sometimes be useful when used directly, its main purpose is to implement the <- bind syntax in do notation. It has the type m a -> (a -> m b) -> m b mainly because, when used in a do notation block, the right hand side of the <- is of type m a, the left hand side "binds" an a to the given identifier and, when combined with remainder of the do block, is of type a -> m b, the resulting monadic action is of type m b, and this is the only type it possibly could have to make this work.
For example:
echo = do
input <- getLine
putStrLn input
The right hand side of the <- is of type IO String
The left hands side of the <- with the remainder of the do block are of type String -> IO (). Compare with the desugared version using >>=:
echo = getLine >>= (\input -> putStrLn input)
The left hand side of the >>= is of type IO String. The right hand side is of type String -> IO (). Now, by applying an eta reduction to the lambda we can instead get:
echo = getLine >>= putStrLn
which shows why >>= is sometimes used directly rather than as the "engine" that powers do notation along with >>.
I'd also like to provide what I think is an important correction to the concept of "unwrapping" a monadic value, which is that it doesn't happen. The Monad class does not provide a generic function of type Monad m => m a -> a. Some particular instances do but this is not a feature of monads in general. Monads, generally speaking, cannot be "unwrapped".
Remember that m >>= k = join (fmap k m) is a law that must be true for any monad. Any particular implementation of >>= must satisfy this law and so must be equivalent to this general implementation.
What this means is that what really happens is that the monadic "computation" a -> m b is "lifted" to become an m a -> m (m b) using fmap and then applied the m a, giving an m (m b); and then join :: m (m a) -> m a is used to squish the two ms together to yield a m b. So the a never gets "out" of the monad. The monad is never "unwrapped". This is an incorrect way to think about monads and I would strongly recommend that you not get in the habit.
I will focus on your point
If you already created a function which returns a wrapped value, why
that function doesn't already take a wrapped value?
and the IO monad. Suppose you had
getLine :: IO String
putStrLn :: IO String -> IO () -- "already takes a wrapped value"
how one could write a program which reads a line and print it twice? An attempt would be
let line = getLine
in putStrLn line >> putStrLn line
but equational reasoning dictates that this is equivalent to
putStrLn getLine >> putStrLn getLine
which reads two lines instead.
What we lack is a way to "unwrap" the getLine once, and use it twice. The same issue would apply to reading a line, printing "hello", and then printing a line:
let line = getLine in putStrLn "hello" >> putStrLn line
-- equivalent to
putStrLn "hello" >> putStrLn getLine
So, we also lack a way to specify "when to unwrap" the getLine. The bind >>= operator provides a way to do this.
A more advanced theoretical note
If you swap the arguments around the (>>=) bind operator becomes (=<<)
(=<<) :: (a -> m b) -> (m a -> m b)
which turns any function f taking an unwrapped value into a function g taking a wrapped
value. Such g is known as the Kleisli extension of f. The bind operator guarantees
such an extension always exists, and provides a convenient way to use it.
Because we like to be able to apply functions like a -> b to our m as. Lifting such a function to m a -> m b is trivial (liftM, liftA, >>= return ., fmap) but the opposite is not necessarily possible.
You want some typical examples? How about putStrLn :: String -> IO ()? It would make no sense for this function to have the type IO String -> IO () because the origin of the string doesn't matter.
Anyway: You might have the wrong idea because of your "wrapped value" metaphor; I use it myself quite often, but it has its limitations. There isn't necessarily a pure way to get an a out of an m a - for example, if you have a getLine :: IO String, there's not a great deal of interesting things you can do with it - you can put it in a list, chain it in a row and other neat things, but you can't get any useful information out of it because you can't look inside an IO action. What you can do is use >>= which gives you a way to use the result of the action.
Similar things apply to monads where the "wrapping" metaphor applies too; For example the point Maybe monad is to avoid manually wrapping and unwrapping values with and from Just all the time.
My two most common examples:
1) I have a series of functions that generate a list of lists, but I finally need a flat list:
f :: a -> [a]
fAppliedThrice :: [a] -> [a]
fAppliedThrice aList = concat (map f (concat (map f (concat (map f a)))))
fAppliedThrice' :: [a] -> [a]
fAppliedThrice' aList = aList >>= f >>= f >>= f
A practical example of using this was when my functions fetched attributes of a foreign key relationship. I could just chain them together to finally obtain a flat list of attributes. Eg: Product hasMany Review hasMany Tag type relationship, and I finally want a list of all the tag names for a product. (I added some template-haskell and got a very good generic attribute fetcher for my purposes).
2) Say you have a series of filter-like functions to apply to some data. And they return Maybe values.
case (val >>= filter >>= filter2 >>= filter3) of
Nothing -> putStrLn "Bad data"
Just x -> putStrLn "Good data"

What are the alternatives to prelude's iterate if the "output" values are not the same as those being iterated on?

I have come across a pattern where, I start with a seed value x and at each step generate a new seed value and a value to be output. My desired final result is a list of the output values. This can be represented by the following function:
my_iter :: (a -> (a, b)) -> a -> [b]
my_iter f x = y : my_iter f x'
where (x',y) = f x
And a contrived example of using this would be generating the Fibonacci numbers:
fibs:: [Integer]
fibs = my_iter (\(a,b) -> let c = a+b in ((b, c), c)) (0,1)
-- [1, 2, 3, 5, 8...
My problem is that I have this feeling that there is very likely a more idiomatic way to do this kind of stuff. What are the idiomatic alternatives to my function?
The only ones I can think of right now involve iterate from the Prelude, but they have some shortcomings.
One way is to iterate first and map after
my_iter f x = map f2 $ iterate f1 x
where f1 = fst . f
f2 = snd . f
However, this can look ugly if there is no natural way to split f into the separate f1 and f2 functions. (In the contrived Fibonacci case this is easy to do, but there are some situations where the generated value is not an "independent" function of the seed so its not so simple to split things)
The other way is to tuple the "output" values together with the seeds, and use a separate step to separate them (kind of like the "Schwartzian transform" for sorting things):
my_iter f x = map snd . tail $ iterate (f.fst) (x, undefined)
But this seems wierd, since we have to remember to ignore the generated values in order to get to the seed (the (f.fst) bit) and add we need an "undefined" value for the first, dummy generated value.
As already noted, the function you want is unfoldr. As the name suggests, it's the opposite of foldr, but it might be instructive to see exactly why that's true. Here's the type of foldr:
(a -> b -> b) -> b -> [a] -> b
The first two arguments are ways of obtaining something of type b, and correspond to the two data constructors for lists:
[] :: [a]
(:) :: a -> [a] -> [a]
...where each occurrence of [a] is replaced by b. Noting that the [] case produces a b with no input, we can consolidate the two as a function taking Maybe (a, b) as input.
(Maybe (a, b) -> b) -> ([a] -> b)
The extra parentheses show that this is essentially a function that turns one kind of transformation into another.
Now, simply reverse the direction of both transformations:
(b -> Maybe (a, b)) -> (b -> [a])
The result is exactly the type of unfoldr.
The underlying idea this demonstrates can be applied similarly to other recursive data types, as well.
The standard function you're looking for is called unfoldr.
Hoogle is a very useful tool in this case, since it doesn't only support searching functions by name, but also by type.
In your case, you came up with the desired type (a -> (a, b)) -> a -> [b]. Entering it yields no results - hmm.
Well, maybe there's a standard function with a slightly different syntax. For example, the standard function might have its arguments flipped; let's look for something with (a -> (a, b)) in its type signature somewhere. This time we're lucky as there are plenty of results, but all of them are in exotic packages and none of them seems very helpful.
Maybe the second part of your function is a better match, you want to generate a list out of some initial element after all - so type in a -> [b] and hit search. First result: unfoldr - bingo!
Another possibility is iterateM in State monad:
iterateM :: Monad m => m a -> m [a]
iterateM = sequence . repeat
It is not in standard library but it's easy to build.
So your my_iter is
evalState . sequence . repeat :: State s a -> s -> [a]

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