I learn scala in university and I cannot understand how to use map, flatmap and Option. Here's couple functions from my lab. I know how to implement first but I have no idea how to deal with second? So, the question: how to implement second function without changing it's signature (using map and flatmap)?
def testCompose[A, B, C, D](f: A => B)
(g: B => C)
(h: C => D): A => D = h compose g compose f
def testMapFlatMap[A, B, C, D](f: A => Option[B])
(g: B => Option[C])
(h: C => D): Option[A] => Option[D] = // help
_.flatMap(f).flatMap(g).map(h)
because:
_ - receive an Option[A]
flatMap(f) - peek inside, return Option[B] (flatMap() won't re-wrap it)
flatMap(g) - peek inside, return Option[C] (flatMap() won't re-wrap it)
map(h) - peek inside, return D (map() will re-wrap it)
Related
The (covariant) functor definition in cats-laws looks like this:
def covariantComposition[A, B, C](fa: F[A], f: A => B, g: B => C): IsEq[F[C]] =
fa.map(f).map(g) <-> fa.map(f.andThen(g))
But if I translate the functor composition rule to Scala, it should be:
def covariantComposition[A, B, C](fa: F[A], f: A => B, g: B => C): IsEq[F[C]] =
fa.map(f).andThen(fa.map(g)) <-> fa.map(f.andThen(g))
Why are they different? Which version is correct?
UPDATE 1 I'm aware of a similar implementation in Haskell, but I haven't had a chance to read it. I wonder if the Haskell version is more by the book.
F(g ∘ f) = F(g) ∘ F(f) is the same as ∀fa, (F(g ∘ f))(fa) = (F(g) ∘ F(f))(fa) (equality of functions is equality of images for all arguments, this is extensionality in HoTT 1 2 3).
The latter is translated as
def covariantComposition[A, B, C](fa: F[A], f: A => B, g: B => C): IsEq[F[C]] =
fa.map(f).map(g) <-> fa.map(f.andThen(g))
(actually, fa.map(f.andThen(g)) <-> fa.map(f).map(g)).
If you'd like to have "point-free" F(g ∘ f) = F(g) ∘ F(f) you could write _.map(f.andThen(g)) <-> _.map(f).map(g) or _.map(f.andThen(g)) <-> (_.map(f)).andThen(_.map(g)) (this is fmap (g . f) = fmap g . fmap f in Haskell, or more precisely, in some "meta-Haskell").
The 2nd code snippet in your question
def covariantComposition[A, B, C](fa: F[A], f: A => B, g: B => C): IsEq[F[C]] =
fa.map(f).andThen(fa.map(g)) <-> fa.map(f.andThen(g))
is incorrect. fa.map(f).andThen... doesn't make sense as it was mentioned in comments. You seem to confuse F and F[A].
In category theory, in general categories, f: A -> B can be just arrows, not necessarily functions (e.g. related pairs in a pre-order if a category is this pre-order), so (F(g ∘ f))(fa) can make no sense. But the category of types in Scala (or Haskell) is a category where objects are types and morphisms are functions.
I think your confusion comes from the different way functor map property can be represented.
trait Functor[F[_]] {
def map1[A, B](f: A => B): F[A] => F[B]
def map2[A, B](f: A => B)(fa: F[A]): F[B]
def map3[A, B](fa: F[A])(f: A => B): F[B]
}
Here... map1 is the haskell aligned definition... and hence the functor law representation used by haskell also works with this one.
So, this haskell
fmap (g . f) = fmap g . fmap f
translates to following Scala
map1( g.compose(f) ) = map1(g).compose( map1(f) )
// or
map1( f.andThen(g) ) <-> map1(f).andThen(map1(g))
But, the thing is that we have few more ways to represent the same map property as given by map2 and map3. The overall essens is still the same, we just switched the representation.
Now, when we add the full object oriented angle to it... the "object-oriented" Functor becomes something like following.
trait List[+A] {
def map(f: A => B): List[B]
}
So... for the "object oriented functor" like List, the same law can be represented as following
listA.map(f).map(g) <-> listA.map(f.andThen(g))
And, you are seeing exactly this.
Option is used for dealing with partiality in Scala, but we can also lift ordinary functions to the context of Options in order to handle errors. When implementing the function map2 I am curious on how to know when to use which functions. Consider the following implementation:
def map2[A,B,C] (ao: Option[A], bo: Option[B]) (f: (A,B) => C): Option[C] =
ao flatMap {aa =>
bo map {bb =>
f(aa, bb)
aa is of type A, and bb is of type B which is then fed to F, giving us a C. However, if we do the following:
def map2_1[A,B,C] (ao: Option[A], bo: Option[B]) (f: (A,B) => C): Option[C] =
ao flatMap {aa =>
bo flatMap {bb =>
f(aa, bb)
aa is still of type A, and bb is still of type B, yet we will have to wrap the last call in Some(f(aa, bb)) in order to get an Option[C] instead of a regular C. Why is this? What does it mean to flatten on BO here?
Last and not least, one could do the simpler:
def map2_2[A,B,C] (ao: Option[A], bo: Option[B]) (f: (A,B) => C): Option[C] = for {
as <- ao
bs <- bo
} yield(f(as,bs))
I know that for-comprehensions are syntactic sugar for ForEach'es, maps and flatmaps etc, but how do I, as a developer, know that the compiler will choose MAP with bs <- bo, and not flatMap?
I think I am on the verge of understanding the difference, yet nested flatmaps confuse me.
Taking the last question first, the developer knows what the compiler will do with for because the behaviour is defined and predictable: All <- turn into flatMap except the last one which will be either map or foreach depending on whether or not there is a yield.
The broader question seems to be about the difference between map and flatMap. The difference should be clear from the signatures e.g. for List these are the (simplified) signatures:
def map[B] (f: A => B) : List[B]
def flatMap[B](f: A => List[B]): List[B]
So map just replaces the values in a List with new values by applying f to each element of type A to generate a B.
flatMap generates a new list by concatenating the results of calling f on each element of the original List. It is equivalent to map followed by flatten (hence the name).
Intuitively, map is a one-for-one replacement whereas flatMap allows each element in the original List to generate 0 or more new elements.
I tried to do research but still not yet figure out what is the terminology of Scala, related to lower case a,b as per the code below
def curry[A, B, C](f: (A, B) => C): A => (B => C) = a => b => f(a, b)
Why is that a,b appears on the right hand side?
I know that it is a part of Algebraic Data Type but still could not find a match definition for this.
Update based on Tim's answer, "Scala knows the type of a and b from the return type A => (B => C). a is type A, b is type B."
I want to ask about how Scala knows the type of a and b, i.e: the mechanism behind? What is the terminology of this?
I guess this is a language feature. Please suggest a foundation guideline to fully understand and practice to gain intuition when reading these complex code.
Update from Mario Galic's comment: ... Scala compiler can perform type inference based on the signature of curry ... Please clarify: if the left hand side (i.e the signature) is too obvious, why we need to have the right hand side definition? I mean, there is only 1 way to infer the logic of the left hand side, then, what is the need of creating the right hand side content?
P/S: I wish that I could mark each feedback as the answer because each provides different aspect which helps me to fully grasp the meaning.
It might help to add full type annotations
def curry[A, B, C](f: (A, B) => C): A => (B => C) =
(a: A) => ((b: B) => f(a, b): C)
Note how curry is a method that takes a function as input and also returns a function as output. You might be wondering where do a and b come from in the output function
(a: A) => ((b: B) => f(a, b): C)
but note that they are just the means of declaring the parameters of the output function. You are free to give them any name, for example the following would also work
(x: A) => ((y: B) => f(x, y): C)
The key is to understand that functions are first class values in Scala, so you can pass them in as arguments to other functions and return them as return values from other functions, in the same way you would do with familiar values like say integer 42. Writing value 42 is straightforward, but writing down function value is more verbose since you have to specify the parameters like a and b but nevertheless conceptually it is still just a value. Hence we could say curry is a method that takes a value and returns a value, but these values happen to be function values.
As we all know, it's pretty easy to create a tuple: (1,'a'). And the type of said tuple is pretty simple: (Int,Char). That type designation, however, is a convenient alternative for the more verbose type designation Tuple2[Int,Char]. In fact, the tuple creation itself is a convenient alternate syntax to the more direct new Tuple2(1,'a').
It's a similar story with functions.
The type designation Char => Int is a convenient alternative to the more verbose Function1[Char,Int]. And, after studying the ScalaDocs page, we learn that a simple function like...
val ctoi = (c:Char) => c.toInt
...is the equivalent of...
val ctoi = new Function1[Char, Int] {
def apply(c: Char): Int = c.toInt
}
So, armed with this information, we can now translate...
def curry[A,B,C](f: (A, B) => C): A => (B => C) =
a => b => f(a, b)
...into its equivalent...
def curry[A,B,C](f: Function2[A,B,C]): Function1[A,Function1[B,C]] =
new Function1[A,Function1[B,C]] {
def apply(a:A) = new Function1[B,C] {def apply(b:B) = f(a,b)}
}
With this it's a little easier to see how a => b => ..., while a bit confusing at first, is actually a very convenient way to designate the names of the arguments being passed in to the hidden apply() methods.
The question doesn't really make sense, but in case this helps here is a breakdown of that line:
def curry[A, B, C](f: (A, B) => C): A => (B => C) = a => b => f(a, b)
This splits into a definition and an implementation with = inbetween. The definition is
def curry[A, B, C](f: (A, B) => C): A => (B => C)
Breaking it down further, A, B, and C are type parameters, meaning that any three types can be used when calling this function.
Next comes the single argument to the function:
f: (A, B) => C
The value of this argument is a function that takes two values (one of type A and one of type B) are returns a single value of type C.
Next comes the type of the result:
A => (B => C)
This is a function that takes a single argument of type A and returns a function that takes a single argument of type B and returns a result of type C.
So curry is a function that takes a function of type (A, B) => C) and returns a function of type A => (B => C). This implements the process known as currying (hence the name).
Now for the implementation (the other side of the =):
a => b => f(a, b)
Adding some brackets might make this clearer:
a => (b => f(a, b))
This is a function that take a and returns b => f(a, b). a is the argument for this function. So that leaves this
b => f(a, b)
This is a simple function with an argument b that returns f(a, b).
Scala knows the type of a and b from the return type A => (B => C). a is type A, b is type B.
I am new to Scala and I just started learning it and now trying some exercises. This one in particular I have a trouble understanding.
I understand up to the (f: (A, B) => C) part, but the rest I dont quite get it. Can someone please explain what's happening after the anonymous function part?
Thanks!
This is the function:
def curry[A, B, C](f: (A, B) => C): A => (B => C) = a => b => f(a, b)
def curry a method named "curry"
[A, B, C] will deal with 3 different types
(f it will receive an argument that we'll name "f"
: (A, B) => C) that argument is type "function that takes A,B and returns C"
: A => (B => C) "curry" returns type "function that takes A and returns function that takes B and returns C"
= here's the "curry" code
a => b => f(a, b) function that takes an argument (we'll call "a") and returns a function that takes an argument (we'll call "b") that returns the value returned after "a" and "b" are passed to "f()"
def map2[A,B,C] (a: Par[A], b: Par[B]) (f: (A,B) => C) : Par[C] =
(es: ExecutorService) => {
val af = a (es)
val bf = b (es)
UnitFuture (f(af.get, bf.get))
}
def map3[A,B,C,D] (pa :Par[A], pb: Par[B], pc: Par[C]) (f: (A,B,C) => D) :Par[D] =
map2(map2(pa,pb)((a,b)=>(c:C)=>f(a,b,c)),pc)(_(_))
I have map2 and need to produce map3 in terms of map2. I found the solution in GitHub but it is hard to understand. Can anyone put a sight on it and explain map3 and also what this does (())?
On a purely abstract level, map2 means you can run two tasks in parallel, and that is a new task in itself. The implementation provided for map3 is: run in parallel (the task that consist in running in parallel the two first ones) and (the third task).
Now down to the code: first, let's give name to all the objects created (I also extended _ notations for clarity):
def map3[A,B,C,D] (pa :Par[A], pb: Par[B], pc: Par[C]) (f: (A,B,C) => D) :Par[D] = {
def partialCurry(a: A, b: B)(c: C): D = f(a, b, c)
val pc2d: Par[C => D] = map2(pa, pb)((a, b) => partialCurry(a, b))
def applyFunc(func: C => D, c: C): D = func(c)
map2(pc2d, pc)((c2d, c) => applyFunc(c2d, c)
}
Now remember that map2 takes two Par[_], and a function to combine the eventual values, to get a Par[_] of the result.
The first time you use map2 (the inside one), you parallelize the first two tasks, and combine them into a function. Indeed, using f, if you have a value of type A and a value of type B, you just need a value of type C to build one of type D, so this exactly means that partialCurry(a, b) is a function of type C => D (partialCurry itself is of type (A, B) => C => D).
Now you have again two values of type Par[_], so you can again map2 on them, and there is only one natural way to combine them to get the final value.
The previous answer is correct but I found it easier to think about like this:
def map3[A, B, C, D](a: Par[A], b: Par[B], c: Par[C])(f: (A, B, C) => D): Par[D] = {
val f1 = (a: A, b: B) => (c: C) => f(a, b, c)
val f2: Par[C => D] = map2(a, b)(f1)
map2(f2, c)((f3: C => D, c: C) => f3(c))
}
Create a function f1 that is a version of f with the first 2 arguments partially applied, then we can map2 that with a and b to give us a function of type C => D in the Par context (f1).
Finally we can use f2 and c as arguments to map2 then apply f3(C => D) to c to give us a D in the Par context.
Hope this helps someone!