I had a look at this but still not sure why or when should I use a Guava Immutablemap when I can have a java.util.Map and make it final.
Please can someone explain the advantages.
A final map just means that the reference (pointer) cannot change. You can still mutate the contents of the map (e.g., map.put(k, v), map.remove(k), etc.).
An immutable map is one that you cannot mutate the contents. You cannot add, remove, or clear the map.
For the many advantages of immutable collections, please see here.
Related
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.
I was wondering if there is a good reason to use Collection.empty[T] instead of new Collection[T]() (or the inverse) ? Or is it just a personal preference ?
Thanks.
Calling new Collection[T]() will create a new instance every time. On the other hand, Collection.empty[T] will most likely always return the same singleton object, usually defined somewhere as
object Empty extends Collection[Nothing] ...
which will be much faster. Edit: This is only possible for immutable collections, mutable collections have to return a new instance every time empty is called.
You should always prefer Collection.empty[Type].
In addition to Collection.empty[T] being clearer on the intent, you should favour it for the same reason that you should favour factory methods in general when instantiating a collection: because thoses factories abstract away some implementation details that you might not (or should not) care about.
By example, when you do Seq.empty[String] you actually get an instance of List[String]. You could directly instantiate a List[String] but if all you care about is to have some Seq you would introduce a needless dependency to List (well OK, actually you cannot as it stands, because List is already abstract, but let's pretend we can for the sake of the argument)
The whole point of factories is precisely to have some amount of separation of concern and not bother with unnecessary instantiation details.
As another more elaborate example, let's talk about collection.immutable.HashMap. This one is very much a concrete class so you might think there is no need for a factory here. Except that for optimization purpose the factory in the companion object collection.immutable.HashMap will actually create different sub-classes depending on the number of elements that you initialize the map with (see this question: Scala: how to make a Hash(Trie)Map from a Map (via Anorm in Play)). Obviously, if you directly instantiate collection.immutable.HashMap you will lose this optimization.
Another common optimization for empty is to always return (when it is an immutable collection) the same instance, yet another useful optimization that you would lose by directly instantiating the collection.
So as a rule of thumb, as far as you can you should use the factories that are provided by the various collection companion objects, so as to shield yourself from unneeded dependencies while at the same time benefiting from potential optimizations provided by the collection framework.
empty is just a special case of factory, and so the same logic applies.
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".
Scala has both a mutable and an immutable Map ,
but it has only an immutable List.
If you want a mutable List you need a ListBuffer.
I don't understand why this is so.
Any one knows?.
You can choose between these:
scala.collection.mutable.DoubleLinkedList
scala.collection.mutable.LinkedList
scala.collection.mutable.ListBuffer
scala.collection.mutable.MutableList
So, yes, Scala has mutable lists :-)
I hope that this article may be of some use to you. The diagram at the bottom of the page is particularly useful in providing the mutable and immutable classes.
http://www.scala-lang.org/docu/files/collections-api/collections_1.html
There is a mutable List, but it is called Buffer. The article linked by Graham goes into more depth, but I thought there should be a specific answer to the question as well.
Map is a trait -- like Java's interface --, while List is a class, a concrete implementation of a Seq. There are mutable and immutable Seq, just like for Map.
This may be confusing to Java programmers because, in Java, List is an interface, whose (main) implementations are ArrayList and LinkedList. Alas, Java naming is atrocious. First, ArrayList is not a List by any stretch of imagination. Also, the interface has methods that are not really related to any traditional list.
So, if you want mutable/immutable equivalence, look to concrete subclass implementations of Seq.
Is it lack of time, some technical problem or is there a reason why it should not exist?
It's just a missing case that will presumably eventually be filled in. There is no reason not to do it, and in certain cases it would be considerably faster than the immutable tree (since modifications require log(n) object creations with an immutable tree and only 1 with a mutable tree).
Edit: and in fact it was filled in in 2.12.
Mutable TreeMap.
(There is a corresponding Set also.)
Meanwhile you can use the Java TreeMap, which is exactly what you need.
val m = new java.util.TreeMap[String, Int]()
m.put("aa", 2)
m.put("cc", 3)
I assume that the reason is that having a mutable variant doesn't bring a big benefit. There are some cases mentioned in the other answers when a mutable map could be a bit more efficient, for example when replacing an already existing value: A mutable variant would save creation of new nodes, but the complexity would be still O(log n).
If you want to keep a shared reference to the map, you can use ImmutableMapAdaptor which wraps any immutable map into a mutable structure.
You'll also notice that TreeSet doesn't have a mutable equivalent either. It's because they share the common base class RedBlack, and the underlying data structure that keeps the Trees ordered by elements or keys is a red-black tree. I don't know too much about this data structure, but it's pretty complex (insertion and removal are pretty expensive compared to other Maps), so I assume that had something to do with a mutable variant not being included.
Basically, it's probably because the underlying data structure isn't readily mutable so TreeMap isn't. So, to answer your question, it's a technical problem. It can definitely be done though, there's just not much of a use case for it.
There may be performance reasons for a mutable TreeMap, but usually you can use an immutable map in the same way as you would a mutable one. You just have to assign it to a var rather than a val. It would be the same as for HashMap, which has mutable and immutable variants:
val mh = collection.mutable.HashMap[Int, Int]()
var ih = collection.immutable.HashMap[Int, Int]()
mh += (1 -> 2)
ih += (1 -> 2)
mh // scala.collection.mutable.HashMap[Int,Int] = Map(1 -> 2)
ih // scala.collection.immutable.HashMap[Int,Int] = Map(1 -> 2)