Scala immutable map, when to go mutable? - scala

My present use case is pretty trivial, either mutable or immutable Map will do the trick.
Have a method that takes an immutable Map, which then calls a 3rd party API method that takes an immutable Map as well
def doFoo(foo: String = "default", params: Map[String, Any] = Map()) {
val newMap =
if(someCondition) params + ("foo" -> foo) else params
api.doSomething(newMap)
}
The Map in question will generally be quite small, at most there might be an embedded List of case class instances, a few thousand entries max. So, again, assume little impact in going immutable in this case (i.e. having essentially 2 instances of the Map via the newMap val copy).
Still, it nags me a bit, copying the map just to get a new map with a few k->v entries tacked onto it.
I could go mutable and params.put("bar", bar), etc. for the entries I want to tack on, and then params.toMap to convert to immutable for the api call, that is an option. but then I have to import and pass around mutable maps, which is a bit of hassle compared to going with Scala's default immutable Map.
So, what are the general guidelines for when it is justified/good practice to use mutable Map over immutable Maps?
Thanks
EDIT
so, it appears that an add operation on an immutable map takes near constant time, confirming #dhg's and #Nicolas's assertion that a full copy is not made, which solves the problem for the concrete case presented.

Depending on the immutable Map implementation, adding a few entries may not actually copy the entire original Map. This is one of the advantages to the immutable data structure approach: Scala will try to get away with copying as little as possible.
This kind of behavior is easiest to see with a List. If I have a val a = List(1,2,3), then that list is stored in memory. However, if I prepend an additional element like val b = 0 :: a, I do get a new 4-element List back, but Scala did not copy the orignal list a. Instead, we just created one new link, called it b, and gave it a pointer to the existing List a.
You can envision strategies like this for other kinds of collections as well. For example, if I add one element to a Map, the collection could simply wrap the existing map, falling back to it when needed, all while providing an API as if it were a single Map.

Using a mutable object is not bad in itself, it becomes bad in a functional programming environment, where you try to avoid side-effects by keeping functions pure and objects immutable.
However, if you create a mutable object inside a function and modify this object, the function is still pure if you don't release a reference to this object outside the function. It is acceptable to have code like:
def buildVector( x: Double, y: Double, z: Double ): Vector[Double] = {
val ary = Array.ofDim[Double]( 3 )
ary( 0 ) = x
ary( 1 ) = y
ary( 2 ) = z
ary.toVector
}
Now, I think this approach is useful/recommended in two cases: (1) Performance, if creating and modifying an immutable object is a bottleneck of your whole application; (2) Code readability, because sometimes it's easier to modify a complex object in place (rather than resorting to lenses, zippers, etc.)

In addition to dhg's answer, you can take a look to the performance of the scala collections. If an add/remove operation doesn't take a linear time, it must do something else than just simply copying the entire structure. (Note that the converse is not true: it's not beacuase it takes linear time that your copying the whole structure)

I like to use collections.maps as the declared parameter types (input or return values) rather than mutable or immutable maps. The Collections maps are immutable interfaces that work for both types of implementations. A consumer method using a map really doesn't need to know about a map implementation or how it was constructed. (It's really none of its business anyway).
If you go with the approach of hiding a map's particular construction (be it mutable or immutable) from the consumers who use it then you're still getting an essentially immutable map downstream. And by using collection.Map as an immutable interface you completely remove all the ".toMap" inefficiency that you would have with consumers written to use immutable.Map typed objects. Having to convert a completely constructed map into another one simply to comply to an interface not supported by the first one really is absolutely unnecessary overhead when you think about it.
I suspect in a few years from now we'll look back at the three separate sets of interfaces (mutable maps, immutable maps, and collections maps) and realize that 99% of the time only 2 are really needed (mutable and collections) and that using the (unfortunately) default immutable map interface really adds a lot of unnecessary overhead for the "Scalable Language".

Related

Is it possible for a structure made of immutable types to have a cycle?

Consider the following:
case class Node(var left: Option[Node], var right: Option[Node])
It's easy to see how you could traverse this, search it, whatever. But now imagine you did this:
val root = Node(None, None)
root.left = root
Now, this is bad, catastrophic. In fact, you type it into a REPL, you'll get a StackOverflow (hey, that would be a good name for a band!) and a stack trace a thousand lines long. If you want to try it, do this:
{ root.left = root }: Unit
to suppress the REPL well-intentioned attempt to print out the results.
But to construct that, I had to specifically give the case-class mutable members, something I would never do in real life. If I use ordinary mutable members, I get a problem with construction. The closest I can come is
case class Node(left: Option[Node], right: Option[Node])
val root: Node = Node(Some(loop), None)
Then root has the rather ugly value Node(Some(null),None), but it's still not cyclic.
So my question is, if a data-structure is transitively immutable (that is, all of its members are either immutable values or references to other data-structures that are themselves transitively immutable), is it guaranteed to be acyclic?
It would be cool if it were.
Yes, it is possible to create cyclic data structures even with purely immutable data structures in a pure, referentially transparent, effect-free language.
The "obvious" solution is to pull out the potentially cyclic references into a separate data structure. For example, if you represent a graph as an adjacency matrix, then you don't need cycles in your data structure to represent cycles in your graph. But that's cheating: every problem can be solved by adding a layer of indirection (except the problem of having too many layers of indirection).
Another cheat would be to circumvent Scala's immutability guarantees from the outside, e.g. on the default Scala-JVM implementation by using Java reflection methods.
It is possible to create actual cyclic references. The technique is called Tying the Knot, and it relies on laziness: you can actually set the reference to an object that you haven't created yet because the reference will be evaluated lazily, by which time the object will have been created. Scala has support for laziness in various forms: lazy vals, by-name parameters, and the now deprecated DelayedInit. Also, you can "fake" laziness using functions or method: wrap the thing you want to make lazy in a function or method which produces the thing, and it won't be created until you call the function or method.
So, the same techniques should be possible in Scala as well.
How about using lazy with call by name ?
scala> class Node(l: => Node, r: => Node, v: Int)
// defined class Node
scala> lazy val root: Node = new Node(root, root, 5)
// root: Node = <lazy>

Immutable DataStructures In Scala

We know that Scala supports immutable data structures..i.e each time u update the list it will create a new object and reference in the heap.
Example
val xs:List[Int] = List.apply(22)
val newList = xs ++ (33)
So when i append the second element to a list it will create a new list which will contain both 22 and 33.This exactly works like how immutable String works in Java.
So the question is each time I append a element in the list a new object will be created each time..This ldoes not look efficient to me.
is there some special data structures like persistent data structures are used when dealing with this..Does anyone know about this?
Appending to a list has O(n) complexity and is inefficient. A general approach is to prepend to a list while building it, and finally reverse it.
Now, your question on creating new object still applies to the prepend. Note that since xs is immutable, newList just points to xs for the rest of the data after the prepend.
While #manojlds is correct in his analysis, the original post asked about the efficiency of duplicating list nodes whenever you do an operation.
As #manojlds said, constructing lists often require thinking backwards, i.e., building a list and then reversing it. There are a number of other situations where list building requires "needless" copying.
To that end, there is a mutable data structure available in Scala called ListBuffer which you can use to build up your list and then extract the result as an immutable list:
val xsa = ListBuffer[Int](22)
xsa += 33
val newList = xsa.toList
However, the fact that the list data structure is, in general, immutable means that you have some very useful tools to analyze, de-compose and re-compose the list. Many builtin operations take advantage of the immutability. By extension, your own programs can also take advantage of this immutability.

Scala - encapsulating data in objects

Motivations
This question is about working with Lists of data in Scala, and about resorting to either tuples or class objects for holding data. Perhaps some of my assumptions are wrong, so there it goes.
My current approach
As I understand, tuples do not afford the possibility of elegantly addressing their elements beyond the provided ._1, ._2, etc. I can use them, but code will be a bit unpleasant wherever data is extracted far from the lines of code that had defined it.
Also, as I understand, a Scala Map can only use a single type declaration for its values, so it can't diversify the value type of its values except for the case of type inheritance. (to the later point, considering the use of a type hierarchy for Map values "diversity" - may seem to be very artificial unless a class hierarchy fits any "model" intuition to begin with).
So, when I need to have lists where each element contains two or more named data entities, e.g. as below one of type String and one of type List, each accessible through an intelligible name, I resort to:
case class Foo (name1: String, name2: List[String])
val foos: List[Foo] = ...
Then I can later access instances of the list using .name1 and .name2.
Shortcomings and problems I see here
When the list is very large, should I assume this is less performant or more memory consuming than using a tuple as the List's type? alternatively, is there a different elegant way of accomplishing struct semantics in Scala?
In terms of performance, I don't think there is going to be any distinction between a tuple and an instance of a cases class. In fact, a tuple is an instance of a case class.
Secondly, if you're looking for another, more readable way to get the data out of the tuple, I suggest you consider pattern matching:
val (name1, name2) = ("first", List("second", "third"))

Should Scala immutable case classes be defined to hold Seq[T], immutable.Seq[T], List[T] or Vector[T]?

If we want to define a case class that holds a single object, say a tuple, we can do it easily:
sealed case class A(x: (Int, Int))
In this case, retrieving the "x" value will take a small constant amount of time, and this class will only take a small constant amount of space, regardless of how it was created.
Now, let's assume we want to hold a sequence of values instead; we could it like this:
sealed final case class A(x: Seq[Int])
This might seem to work as before, except that now storage and time to read all of x is proportional to x.length.
However, this is not actually the case, because someone could do something like this:
val hugeList = (1 to 1000000000).toList
val a = A(hugeList.view.filter(_ == 500000000))
In this case, the a object looks like an innocent case class holding a single int in a sequence, but in fact it requires gigabytes of memory, and it will take on the order of seconds to access that single element every time.
This could be fixed by specifying something like List[T] as the type instead of Seq[T]; however, this seems ugly since it adds a reference to a specific implementation, while in fact other well behaved implementations, like Vector[T], would also do.
Another worrying issue is that one could pass a mutable Seq[T], so it seems that one should at least use immutable.Seq instead of scala.collection.Seq (although the compiler can't actually enforce the immutability at the moment).
Looking at most libraries it seems that the common pattern is to use scala.collection.Seq[T], but is this really a good idea?
Or perhaps Seq is being used just because it's the shortest to type, and in fact it would be best to use immutable.Seq[T], List[T], Vector[T] or something else?
New text added in edit
Looking at the class library, some of the most core functionality like scala.reflect.api.Trees does in fact use List[T], and in general using a concrete class seems a good idea.
But then, why use List and not Vector?
Vector has O(1)/O(log(n)) length, prepend, append and random access, is asymptotically smaller (List is ~3-4 times bigger due to vtable and next pointers), and supports cache efficient and parallelized computation, while List has none of those properties except O(1) prepend.
So, personally I'm leaning towards Vector[T] being the correct choice for something exposed in a library data structure, where one doesn't know what operations the library user will need, despite the fact that it seems less popular.
First of all, you talk both about space and time requirements. In terms of space, your object will always be as large as the collection. It doesn't matter whether you wrap a mutable or immutable collection, that collection for obvious reasons needs to be in memory, and the case class wrapping it doesn't take any additional space (except its own small object reference). So if your collection takes "gigabytes of memory", that's a problem of your collection, not whether you wrap it in a case class or not.
You then go on to argue that a problem arises when using views instead of eager collections. But again the question is what the problem actually is? You use the example of lazily filtering a collection. In general running a filter will be an O(n) operation just as if you were iterating over the original list. In that example it would be O(1) for successive calls if that collection was made strict. But that's a problem of the calling site of your case class, not the definition of your case class.
The only valid point I see is with respect to mutable collections. Given the defining semantics of case classes, you should really only use effectively immutable objects as arguments, so either pure immutable collections or collections to which no instance has any more write access.
There is a design error in Scala in that scala.Seq is not aliased to collection.immutable.Seq but a general seq which can be either mutable or immutable. I advise against any use of unqualified Seq. It is really wrong and should be rectified in the Scala standard library. Use collection.immutable.Seq instead, or if the collection doesn't need to be ordered, collection.immutable.Traversable.
So I agree with your suspicion:
Looking at most libraries it seems that the common pattern is to use scala.collection.Seq[T], but is this really a good idea?
No! Not good. It might be convenient, because you can pass in an Array for example without explicit conversion, but I think a cleaner design is to require immutability.

val-mutable versus var-immutable in Scala

Are there any guidelines in Scala on when to use val with a mutable collection versus using var with an immutable collection? Or should you really aim for val with an immutable collection?
The fact that there are both types of collection gives me a lot of choice, and often I don't
know how to make that choice.
Pretty common question, this one. The hard thing is finding the duplicates.
You should strive for referential transparency. What that means is that, if I have an expression "e", I could make a val x = e, and replace e with x. This is the property that mutability break. Whenever you need to make a design decision, maximize for referential transparency.
As a practical matter, a method-local var is the safest var that exists, since it doesn't escape the method. If the method is short, even better. If it isn't, try to reduce it by extracting other methods.
On the other hand, a mutable collection has the potential to escape, even if it doesn't. When changing code, you might then want to pass it to other methods, or return it. That's the kind of thing that breaks referential transparency.
On an object (a field), pretty much the same thing happens, but with more dire consequences. Either way the object will have state and, therefore, break referential transparency. But having a mutable collection means even the object itself might lose control of who's changing it.
If you work with immutable collections and you need to "modify" them, for example, add elements to them in a loop, then you have to use vars because you need to store the resulting collection somewhere. If you only read from immutable collections, then use vals.
In general, make sure that you don't confuse references and objects. vals are immutable references (constant pointers in C). That is, when you use val x = new MutableFoo(), you'll be able to change the object that x points to, but you won't be able to change to which object x points. The opposite holds if you use var x = new ImmutableFoo(). Picking up my initial advice: if you don't need to change to which object a reference points, use vals.
The best way to answer this is with an example. Suppose we have some process simply collecting numbers for some reason. We wish to log these numbers, and will send the collection to another process to do this.
Of course, we are still collecting numbers after we send the collection to the logger. And let's say there is some overhead in the logging process that delays the actual logging. Hopefully you can see where this is going.
If we store this collection in a mutable val, (mutable because we are continuously adding to it), this means that the process doing the logging will be looking at the same object that's still being updated by our collection process. That collection may be updated at any time, and so when it's time to log we may not actually be logging the collection we sent.
If we use an immutable var, we send an immutable data structure to the logger. When we add more numbers to our collection, we will be replacing our var with a new immutable data structure. This doesn't mean collection sent to the logger is replaced! It's still referencing the collection it was sent. So our logger will indeed log the collection it received.
I think the examples in this blog post will shed more light, as the question of which combo to use becomes even more important in concurrency scenarios: importance of immutability for concurrency. And while we're at it, note the preferred use of synchronised vs #volatile vs something like AtomicReference: three tools
var immutable vs. val mutable
In addition to many excellent answers to this question. Here is a simple example, that illustrates potential dangers of val mutable:
Mutable objects can be modified inside methods, that take them as parameters, while reassignment is not allowed.
import scala.collection.mutable.ArrayBuffer
object MyObject {
def main(args: Array[String]) {
val a = ArrayBuffer(1,2,3,4)
silly(a)
println(a) // a has been modified here
}
def silly(a: ArrayBuffer[Int]): Unit = {
a += 10
println(s"length: ${a.length}")
}
}
Result:
length: 5
ArrayBuffer(1, 2, 3, 4, 10)
Something like this cannot happen with var immutable, because reassignment is not allowed:
object MyObject {
def main(args: Array[String]) {
var v = Vector(1,2,3,4)
silly(v)
println(v)
}
def silly(v: Vector[Int]): Unit = {
v = v :+ 10 // This line is not valid
println(s"length of v: ${v.length}")
}
}
Results in:
error: reassignment to val
Since function parameters are treated as val this reassignment is not allowed.