flatMap signature:
/* applies a transformation of the monad "content" by composing
* this monad with an operation resulting in another monad instance
* of the same type
*/
def flatMap(f: A => M[B]): M[B]
is there anyway to understand by its signature (input to output) (except for the name flat) that its flattening the structure? or must I read its implementation to understand that? isn't good coding practice means I can understand by the signature of the function (input to output even without function name) exactly what it does? if so how does flatMap follows this basic programming practice? or does it violates it?
It helps to understand that the type of the thing that flatMap is defined on is also M and would be something like Traversable[A] or a subclass such as a List[A], or another Monad such as Option or Future. Therefore the left hand side of the function definition that flatMap requires (f: A => M[B]) hints to us that this function processes the individual items contained within the Monad, and produces M[B] for each one. The fact that it then returns M[B] rather than M[M[B]] is a hint that some flattening is taking place.
However I disagree that it should be possible to know exactly what a function does by its signature. For example:
Trait Number {
def plus(that: Number): Number
def minus(that: Number): Number
}
Without the function names and possible also some documentation, I don't think it is reasonable to expect that another person can know what plus and minus do.
Does flatMap functional signature (input -> output) proposes its doing any flattening?
Yes, it does.
In fact, join (or flatten), map and unit form a minimal set of primitive functions needed to implement a monad.
flatMap can be implemented in terms of these other functions.
//minimal set
def join[A](mma: M[M[A]]): M[A] = ???
def map[A](ma: M[A], f: A => B): M[B] = ???
def unit[A](a: A): M[A] = ???
def flatMap[A](ma: M[A], f: A => M[B]): M[B] = join(map(ma))
It should now be clear that flatMap flattens the structure of the mapped monad.
For comparison, here's another mimimal set of primitives, where join/flatten is implemented in terms of flatMap.
// minimal set
def unit[A](a: A): M[A] = ???
def flatMap[A](ma: M[A], f: A => M[B]): M[B] = ???
def map[A](ma: M[A], f: A => B): M[B] = flatMap(ma, a => unit(f(a)))
def join[A](mma: M[M[A]]): M[A] = flatMap(mma, ma => ma)
Related
I took the scala odersky course and thought that the function that Flatmap takes as arguments , takes an element of Monad and returns a monad of different type.
trait M[T] {
def flatMap[U](f: T => M[U]): M[U]
}
On Monad M[T] , the return type of function is also the same Monad , the type parameter U might be different.
However I have seen examples on internet , where the function returns a completely different Monad. I was under impression that return type of function should the same Monad. Can someone simplify the below to explain how flapmap results in the actual value instead of Option in the list.
Is the List not a Monad in Scala.
val l= List(1,2,3,4,5)
def f(x:int) = if (x>2) Some(x) else None
l.map(x=>f(x))
//Result List[Option[Int]] = List( None , None , Some(3) , Some(4) , Some(5))
l.flatMap(x=>f(x))
//Result: List(3,4,5)
Let's start from the fact that M[T]is not a monad by itself. It's a type constructor. It becomes a monad when it's associated with two operators: bind and return (or unit). There are also monad laws these operators must satisfy, but let's omit them for brevity. In Haskell the type of bind is:
class Monad m where
...
(>>=) :: m a -> (a -> m b) -> m b
where m is a type constructor. Since Scala is OO language bind will look like (first argument is self):
trait M[T] {
def bind[U](f: T => M[U]): M[U]
}
Here M === m, T === a, U === b. bind is often called flatMap. In a pure spherical world in a vacuum that would be a signature of flatMap in OO language. Scala is a very practical language, so the real signature of flatMap for List is:
final def flatMap[B, That](f: (A) ⇒ GenTraversableOnce[B])(implicit bf: CanBuildFrom[List[A], B, That]): That
It's not bind, but will work as a monadic bind if you provide f in the form of (A) => List[B] and also make sure that That is List[B]. On the other hand Scala is not going to watch your back if you provide something different, but will try to find some meaningful conversion (e.g. CanBuildFrom or something else) if it exists.
UPDATE
You can play with scalac flags (-Xlog-implicits, -Xlog-implicit-conversions) to see what's happening:
scala> List(1).flatMap { x => Some(x) }
<console>:1: inferred view from Some[Int] to scala.collection.GenTraversableOnce[?] via scala.this.Option.option2Iterable[Int]: (xo: Option[Int])Iterable[Int]
List(1).flatMap { x => Some(x) }
^
res1: List[Int] = List(1)
Hmm, perhaps confusingly, the signature you gave is not actually correct, since it's really (in simplified form):
def flatMap[B](f: (A) ⇒ GenTraversableOnce[B]): Traversable[B]
Since the compiler is open-source, you can actually see what it's doing (with its full signature):
def flatMap[B, That](f: A => GenTraversableOnce[B])(implicit bf: CanBuildFrom[Repr, B, That]): That = {
def builder = bf(repr) // extracted to keep method size under 35 bytes, so that it can be JIT-inlined
val b = builder
for (x <- this) b ++= f(x).seq
b.result
}
So you can see that there is actually no requirement that the return type of f be the same as the return type of flatMap.
The flatmap found in the standard library is a much more general and flexible method than the monadic bind method like the flatMap from Odersky's example.
For example, the full signature of flatmap on List is
def flatMap[B, That](f: (A) ⇒ GenTraversableOnce[B])(implicit bf: CanBuildFrom[List[A], B, That]): That
Instead of requiring the function passed into flatmap to return a List, it is able to return any GenTraversableOnce object, a very generic type.
flatmap then uses the implicit CanBuildFrom mechanism to determine the appropriate type to return.
So when you use flatmap with a function that returns a List, it is a monadic bind operation, but it lets you use other types as well.
For a coding assignment for school I have to do stuff with flatmap but I have absolutely no idea what it does and I've read a couple pages online and read in my textbook but I still have no true understanding of what it does. I know what map does but for some reason it's hard for me to wrap my head around flatmap. Can anyone help? Thanks.
Just to add some more information- when I look at examples online I DO see how flatmap returns something different from map. But what is flatmap actually doing when its being called on something? How does flatmap actually work? What is it doing before it returns the result?
Here's an analogy.
Imagine you have a big bag filled with shopping vouchers for cartons of eggs. If you have a function which is "use the voucher to buy a carton of eggs" and you called bigBagOfVouchers.map(buyCartonOfEggs), you'd have a bag of cartons of eggs.
However, if you called bigBagOfVouchers.flatMap(buyCartonOfEggs), you'd have a bag of eggs - without any cartons.
flatMap flattens the result by one level. What might have been Bag[Carton[Egg]] is now Bag[Egg].
Functors define map which have type
trait Functor[F[_]] {
def map[A, B](f: A => B)(v: F[A]): F[B]
}
Monads are functors which support two additional operations:
trait Monad[M[_]] extends Functor[M] {
def pure[A](v: A): M[A]
def join[A](m: M[M[A]]): M[A]
}
Join flattens nested values e.g. if m is List then join has type
def joinList[A](l: List[List[A]]): List[A]
If you have a monad m and you map over it, what happens if b is the same monadic type? For example:
def replicate[A](i: Int, value: A): List[A] = ???
val f = new Functor[List] {
def map[A, B](f: A => B)(v: List[A]) = v.map(f)
}
then
f.map(x => replicate(x, x))(List(1,2,3)) == List(List(1), List(2,2), List(3,3,3))
This has type List[List[Int]] while the input is a List[Int]. It's fairly common with a chain of operations to want each step to return the same input type. Since List can also be made into a monad, you can easily create such a list using join:
listMonad.join(List(List(1), List(2,2), List(3,3,3))) == List(1,2,2,3,3,3)
Now you might want to write a function to combine these two operations into one:
trait Monad[M] {
def flatMap[A, B](f: A => M[B])(m: M[A]): M[B] = join(map(f)(m))
}
then you can simply do:
listMonad.flatMap(List(1,2,3), x => replicate(x, x)) == List(1,2,2,3,3,3)
Exactly what flatMap does depends on the monad type constructor M (List in this example) since it depends on map and join.
In reactive programming, you often come into the situation in which you need to use flatMap to convert Future[Future[List]] to Future[List]. For example, you have two functions: get Users from database and process retrieved Users; and both return Future[List[User]]. If you apply map to get and process, the result will be Future[Future[List[User]]] which does not make any sense. Instead, you should use flatMap:
def main(): Future[List[User]] = getUsers flatMap processUsers
def getUsers: Future[List[User]]
def processUsers(users: List[User]): Future[List[User]]
(It seems like I ask too many basic questions, but Scala is very difficult, especially when it comes to the language feature or syntactic sugar, and I come from a Java background..)
I am watching this video: Scalaz State Monad on YouTube. The lecturer wrote these code:
trait State[S, +A] {
def run(initial: S):(S, A)
def map[B](f: A=>B): State[S, B] =
State { s =>
val (s1, a) = run(s)
(s1, f(a))
}
He said the State on the forth line is a "lambda" expression (I may have misheard). I'm a bit confused about this. He did have an object State, which looks like this:
Object State {
def apply[S, A](f: S=>(S, A)): State[S, A] =
new State[S, A] {
def run(i: S) = f(i)
}
}
The lecturer said he was invoking the apply method of Object State by invoking like that, and yes he did pass in a function that returns a correct tuple, but what is that { s=>? Where the hell did he get the s and is this about the implicit mechanism in Scala?
Also the function that is passed in map function changes A to B already, how can it still pass the generics check (from [S, A] to [S, B] assuming A and B are two different types, or are they the same type?If true, why use B not A in map function?)?
The video link is here and is set to be at 18:17
http://youtu.be/Jg3Uv_YWJqI?t=18m17s
Thank you (Monad is a bit daunting to anyone that's not so familiar with functional programming)
You are on right way here.
He passes the function as parameter, but what is the function in general? Function takes argument and return a value. Returned value here is tuple (s1, f(a)), but also we need to reffer to function argument. s in your example is function argument.
It is same as
def map[B](f: A=>B): State[S, B] = {
def buildState(s: S) = {
val (s1, a) = run(s)
(s1, f(a))
}
State(buildState)
}
But, since we dont need this function anywhere else we just inline it in apply call.
I'm working on this Functional Programming in Scala exercise:
// But what if our list has an element type that doesn't have a Monoid instance?
// Well, we can always map over the list to turn it into a type that does.
As I understand this exercise, it means that, if we have a Monoid of type B, but our input List is of type A, then we need to convert the List[A] to List[B], and then call foldLeft.
def foldMap[A, B](as: List[A], m: Monoid[B])(f: A => B): B = {
val bs = as.map(f)
bs.foldLeft(m.zero)((s, i) => m.op(s, i))
}
Does this understanding and code look right?
First I'd simplify the syntax of the body a bit:
def foldMap[A, B](as: List[A], m: Monoid[B])(f: A => B): B =
as.map(f).foldLeft(m.zero)(m.ops)
Then I'd move the monoid instance into its own implicit parameter list:
def foldMap[A, B](as: List[A])(f: A => B)(implicit m: Monoid[B]): B =
as.map(f).foldLeft(m.zero)(m.ops)
See the original "Type Classes as Objects and Implicits" paper for more detail about how Scala implements type classes using implicit parameter resolution, or this answer by Rex Kerr that I've also linked above.
Next I'd switch the order of the other two parameter lists:
def foldMap[A, B](f: A => B)(as: List[A])(implicit m: Monoid[B]): B =
as.map(f).foldLeft(m.zero)(m.ops)
In general you want to place the parameter lists containing parameters that change the least often first, in order to make partial application more useful. In this case there may only be one possible useful value of A => B for any A and B, but there are lots of values of List[A].
For example, switching the order allows us to write the following (which assumes a monoid instance for Bar):
val fooSum: List[Foo] => Bar = foldMap(fooToBar)
Finally, as a performance optimization (mentioned by stew above), you could avoid creating an intermediate list by moving the application of f into the fold:
def foldMap[A, B](f: A => B)(as: List[A])(implicit m: Monoid[B]): B =
as.foldLeft(m.zero) {
case (acc, a) => m.op(acc, f(a))
}
This is equivalent and more efficient, but to my eye much less clear, so I'd suggest treating it like any optimization—if you need it, use it, but think twice about whether the gains are really worth the loss of clarity.
Is there any quick way to use as a concrete function (of type, say, (A) => B) as a PartialFunction[A, B]? The most concise syntax I know of is:
(a: A) => a match { case obj => func(obj) }
Is there an implicit conversion anywhere, something like:
implicit def funcAsPartial[A, B](func: A => B) = new PartialFunction[A, B] {
def isDefinedAt(a: A) = true
def apply(a: A) = func(a)
}
I guess I just wrote what I was looking for, but does this already exist in the Scala libraries?
Doing this with an implicit conversion is dangerous, for the same reason that (A) => B should not inherit from PartialFunction[A, B]. That is, the contract of PartialFunction guarantees that you can safely* call apply wherever isDefinedAt returns true. Function1's contract provides no such guarantee.
Your implicit conversion will result in a PartialFunction that violates its contract if you apply it to a function that is not defined everywhere. Instead, use a pimp to make the conversion explicit:
implicit def funcAsPartial[A, B](f: A => B) = new {
/** only use if `f` is defined everywhere */
def asPartial(): PartialFunction[A, B] = {
case a => f(a)
}
def asPartial(isDefinedAt: A => Boolean): PartialFunction[A, B] = {
case a if isDefinedAt(a) => f(a)
}
}
// now you can write
val f = (i: Int) => i * i
val p = f.asPartial // defined on all integers
val p2 = f.asPartial(_ > 0) // defined only on positive integers
* As discussed in the comments, it may not be entirely clear what "safety" means here. The way I think about it is that a PartialFunction explicitly declares its domain in the following precise sense: if isDefinedAt returns true for a value x, then apply(x) can be evaluated in a way that is consistent with the intent of the function's author. That does not imply that apply(x) will not throw an exception, but merely that the exception was part of the design of the function (and should be documented).
No, I tried to find one a few months ago and ended up writing my own that's essentially the same as yours.