How does the following work? Given a list of items [a] and a function mapping a to g b, how do I preserve the input in the resulting list?
[a] -> (a -> g b) -> g [(a,b)]
Concretely, I have a list of contract ids. I want to map every contract id to the tuple of (id, payload), where payload is some contract variable. And g b in the above example is fetch.
It's a bit easier if you flip the argument order, then you can do:
mapAWithArgs : Applicative m => (a -> m b) -> [a] -> m [(a, b)]
mapAWithArgs f = mapA (withArg f)
where
withArg f x = (x,) <$> f x
Then you can do withIds <- mapAWithArgs fetch <list of contract IDs> in your Update
Related
I want to write a function that returns the longest prefix of a list, where applying a function to every item in that prefix produces a strictly ascending list.
For example:
longestAscendingPrefix (`mod` 5) [1..10] == [1,2,3,4]
longestAscendingPrefix odd [1,4,2,6,8,9,3,2,1] == [1]
longestAscendingPrefix :: Ord b => (a -> b) -> [a] -> [a]
longestAscendingPrefix _ [] = []
longestAscendingPrefix f (x:xs) = takeWhile (\y z -> f y <= f z) (x:xs)
This code snippet produces the error message in the title. It seems the problem lies within that lambda function.
takeWhile has type takeWhile :: (a -> Bool) -> [a] -> [a]. The first parameter is thus a function that maps an element of the list to a Bool. Your lambda expression has type Ord b => a -> a -> Bool, which does not make much sense.
You can work with explicit recursion with:
longestAscendingPrefix :: Ord b => (a -> b) -> [a] -> [a]
longestAscendingPrefix f = go
where go [] = []
go [x] = …
go (x1:x2:xs) = …
where you need to fill in the … parts the last one makes a recursive call to go.
nestedApply :: (Applicative f, Applicative g) => g (f (a -> b)) -> f a -> g (f b)
As the type indicates, how to get that (a->b) applied to that a in the context f?
Thanks for help.
This one of those cases where it's helpful to focus on types. I will try to keep it simple and explain the reasoning.
Let's start with describing the task. We have gfab :: g(f(a->b)) and fa :: f a, and we want to have g(f b).
gfab :: g (f (a -> b))
fa :: f a
??1 :: g (f b)
Since g is a functor, to obtain type g T we can start with a value ??2 of type g U and apply fmap to ??3 :: U -> T. In our case, we have T = f b, so we are looking for:
gfab :: g (f (a -> b))
fa :: f a
??2 :: g U
??3 :: U -> f b
??1 = fmap ??3 ??2 :: g (f b)
Now, it looks like we should pick ??2 = gfab. After all,that's the only value of type g Something we have. We obtain U = f (a -> b).
gfab :: g (f (a -> b))
fa :: f a
??3 :: f (a -> b) -> f b
??1 = fmap ??3 gfab :: g (f b)
Let's make ??3 into a lambda, \ (x :: f (a->b)) -> ??4 with ??4 :: f b. (The type of x can be omitted, but I decided to add it to explain what's going on)
gfab :: g (f (a -> b))
fa :: f a
??4 :: f b
??1 = fmap (\ (x :: f (a->b)) -> ??4) gfab :: g (f b)
How to craft ??4. Well, we have values of types f (a->b) and f a, so we can <*> those to get f b. We finally obtain:
gfab :: g (f (a -> b))
fa :: f a
??1 = fmap (\ (x :: f (a->b)) -> x <*> fa) gfab :: g (f b)
We can simplyfy that into:
nestedApply gfab fa = fmap (<*> fa) gfab
Now, this is not the most elegant way to do it, but understanding the process is important.
With
nestedApply :: (Applicative f, Applicative g)
=> g (f (a -> b))
-> f a
-> g (f b )
to get that (a->b) applied to that a in the context f, we need to operate in the context g.
And that's just fmap.
It's clearer with the flipped signature, focusing on its last part
flip nestedApply :: (Applicative f, Applicative g)
=> f a
-> g (f (a -> b)) --- from here
-> g (f b ) --- to here
So what we have here is
nestedApply gffun fx = fmap (bar fx) gffun
with bar fx being applied under the g wraps by fmap for us. Which is
bar fx :: f (a -> b)
-> f b
i.e.
bar :: f a
-> f (a -> b)
-> f b
and this is just <*> isn't it, again flipped. Thus we get the answer,
nestedApply gffun fx = fmap (<*> fx) gffun
As we can see only fmap capabilities of g are used, so we only need
nestedApply :: (Applicative f, Functor g) => ...
in the type signature.
It's easy when writing it on a sheet of paper, in 2D. Which we imitate here with the wild indentation to get that vertical alignment.
Yes we the humans learned to write first, on paper, and to type on a typewriter, much later. The last generation or two were forced into linear typing by the contemporary devices since the young age but now the scribbling and talking (and gesturing and pointing) will hopefully be taking over yet again. Inventive input modes will eventually include 3D workflows and that will be a definite advancement. 1D bad, 2D good, 3D even better. For instance many category theory diagrams are much easier to follow (and at least imagine) when drawn in 3D. The rule of thumb is, it should be easy, not hard. If it's too busy, it probably needs another dimension.
Just playing connect the wires under the wraps. A few self-evident diagrams, and it's done.
Here's some type mandalas for you (again, flipped):
-- <$> -- <*> -- =<<
f a f a f a
(a -> b) f (a -> b) (a -> f b)
f b f b f ( f b) -- fmapped, and
f b -- joined
and of course the mother of all applications,
-- $
a
a -> b
b
a.k.a. Modus Ponens (yes, also flipped).
I have the following two type signatures in Haskell:
foo :: (a -> (a,b)) -> a -> [b]
bar :: (a -> b) -> (a -> b -> c) -> a -> c
I want to write a concrete implementation of these two functions but I'm really struggling to understand where to start.
I understand that foo takes a function (a -> (a,b)) and returns a and a list containing b.
And bar takes a function (b -> c) which returns a function (a -> b -> c) which finally returns a and c.
Can anyone show me an example of a concrete implementation?
How do I know where to start with something like this and what goes on the left side of the definition?
You have some misunderstandings there:
I understand that foo takes a function (a -> (a,b)) and returns a and a list containing b.
No, it doesn't return a. It expects it as another argument, in addition to that function.
And bar takes a function (b -> c) which returns a function (a -> b -> c) which finally returns a and c.
Same here. Given g :: a -> b, bar returns a function bar g :: (a -> b -> c) -> a -> c. This function, in turn, given a function h :: (a -> b -> c), returns a function of type a -> c. And so it goes.
It's just like playing with pieces of a puzzle:
foo :: (a -> (a,b)) -> a -> [b]
-- g :: a -> (a,b)
-- x :: a
-- g x :: (a,b)
foo g x = [b] where
(a,b) = g x
bar :: (a -> b) -> (a -> b -> c) -> a -> c
-- g :: a -> b
-- x :: a
-- g x :: b
-- h :: a -> b -> c
-- h x :: b -> c
-- h x (g x) :: c
bar g h x = c where
c = ....
There's not much free choice for us here. Although, there are more ways to get more values of type b, for foo. Instead of ignoring that a in (a,b) = g x, we can use it in more applications of g, so there actually are many more possibilities there, like
foo2 :: (a -> (a,b)) -> a -> [b]
foo2 g x = [b1,b2] where
(a1,b1) = g x
(a2,b2) = g a1
and many more. Still, the types guide the possible implementations. foo can even make use of foo in its implementation, according to the types:
foo3 :: (a -> (a,b)) -> a -> [b]
foo3 g x = b : bs where
(a,b) = g x
bs = ...
So now, with this implementation, the previous two become its special cases: foo g x === take 1 (foo3 g x) and foo2 g x === take 2 (foo3 g x). Having the most general definition is probably best.
In addition to #will-nes's answer, it will be useful to treat (->) as a right-associative infix operator. So something like f: a -> b -> c is the same as f: a -> (b -> c). So this is saying f is a function that takes a value of type a and returns you a value of type b -> c, which is, another function, one that takes a value of type b and returns you a value of type c.
So the types in your example can be re-written as follows
foo :: (a -> (a,b)) -> (a -> [b])
bar :: (a -> b) -> ((a -> (b -> c)) -> (a -> c))
Similarly, you can think of arguments to a function in pieces as well, as being left-associative (like + and -), though there's no explicit operator in this case. foo a b c d e is the same as ((((foo a) b) c) d) e. For example, let's say we have a function f: Int -> Int -> Int (which is the same as f: Int -> (Int -> Int)). You don't have to provide both arguments at once. So you can write g = f 1, which has the type (Int -> Int). And then you can provide an argument to g, like g 2, which has the type Int. f 1 2 and let g = f 1 in g 2 are more or less the same. Here's a more concrete example of how this works:
Prelude> f = (+)
Prelude> g = f 1
Prelude> g 2
3
Prelude> :t f
f :: Num a => a -> a -> a
Prelude> :t g
g :: Num a => a -> a
Prelude> :t g 2
g 2 :: Num a => a
In #will-nes's sample implementation examples, he defines the functions with all of the arguments up front, but you don't have to think of them that way. Just think of f: a -> b -> c as taking a value of type a and returning another function. While most of the methods you encounter will use all of their arguments up-front, there might be cases in which you don't want to do that. Here's an example:
veryExpensive :: A -> B
unstagedFun :: A -> (B -> C) -> C
unstagedFun a f = f (veryExpensive a)
stagedFun :: A -> (B -> C) -> C
stagedFun a = let b = veryExpensive a in \f -> f b
(You can also rewrite the latter as let b = veryExpensive a in ($ b))
Of course, with compiler optimizations, I wouldn't be surprised if the unstaged version staged automatically, but hopefully this offers some motivation for thinking of functions as not having multiple arguments, but rather, as a single argument, but they may return other functions that may themselves return functions (but also only take a single argument).
I was wondering how to write my own sortOn function.
I made a sortBy function and an on function as shown bellow but can't figure out how to combine them and what additional code to add. sortOn is like sortBy but the given function (in here named comp) is applied only once for every element of the list
sortBy :: (a -> a -> Ordering) -> [a] -> [a]
sortBy comp [] = []
sortBy comp [x] = [x]
sortBy comp (x:xs) = insert x (sortBy comp xs)
where
insert x [] = [x]
insert x (y:ys)
| (comp x y == LT) || (comp x y == EQ) = x:y:ys
| otherwise = y:(insert x ys)
on :: (b -> b -> c) -> (a -> b) -> a -> a -> c
on b f x y = b (f x) (f y)
Here's a hint.
If you have a list [a] and you just sort it, the sort function will implicitly make use of the Ord instance for a and specifically the function:
compare :: a -> a -> Ordering
to figure out the relative ordering of pairs of a elements.
Now, if you have a list [a] and a transformation function b, and you want to use sortOn to sort the list of the transformed values, you'll need to figure out the relative ordering of pairs of b elements. How will you do this? Well, you'll implicitly use the Ord instance for b and specifically the function:
compare :: b -> b -> Ordering
In other words, when you try to define:
sortOn :: (Ord b) => (a -> b) -> [a] -> [a]
sortOn f lst = ...
you'll have arguments of type:
f :: a -> b
lst :: [a]
and additional objects of type:
sortBy :: (a -> a -> Ordering) -> [a] -> [a]
on :: (b -> b -> c) -> (a -> b) -> a -> a -> c
compare :: b -> b -> Ordering
Now, can you see how to put them together to define sortOn?
SPOILERS
Further hint: What's the type of compare `on` f?
Further further hint: It's a -> a -> Ordering.
# let rec map1 f l = match l with
[]->[]
|h::t -> f h::map1 f t;;
val map1 : ('a -> 'b) -> 'a list -> 'b list = <fun>
I am new to OCaml , I have two questions:
In the third line, why there is a f before h :: map1 f t? f should be a argument in the map1 function. Why the book's example puts it seperately?
In the first example ('a -> 'b) -> 'a list -> 'b list = <fun> why there is a b list?
the book explains that b list is the result of the function f and a list is the argument of the function f. However, why there is no a, b list in the following example? It also has a function f and it also puts f separately in the third line.
# let rec apply f n x=
if n = 0 then x
else f ( apply f (n-1) x);;
val apply : ('a -> 'a) -> int -> 'a -> 'a = <fun>
|h::t -> f h::map1 f t
the precedence rules of Ocaml's syntax means that the above match clause is parsed as
|h::t -> (f h)::(map1 f t)
and of course f h is the application of function f to argument h
In words, when the list l is matching the pattern h::t (so l is a proper list of head h and tail t), a pair is made :: (or built, or constructed) of head f h and tail map1 f t
A typical use would be first to have a function from integers to strings:
let nextasstr n = Printf.sprintf "(%d)" (n+1);;
So nextasstr 2 is the string "(3)" without the quotes. Of course [2;3] is a list of integers, i.e. a int list
Then map1 nextasstr [2;3] is evaluated to [ "(3)"; "(4)" ], a list of strings, i.e. a string list; you see that the second argument has a type different of the result. (this should give an insight on the 'a list vs 'b list difference and the typing map1 : ('a -> 'b) -> 'a list -> 'b list) with the first argument being a arbitrary function of type 'a -> 'b
You should see the Ocaml MOOC, follow the Ocaml tutorial, read its documentations. This may take weeks of work.