Scala: Mutante vs. Desempenho objeto imutável - OutOfMemoryError

Question

Eu queria comparar as características de immutable.Map e mutable.Map desempenho no Scala para uma operação semelhante (ou seja, fundindo muitos mapas em um único. Ver esta questão ). Eu tenho o que parece ser implementações semelhantes para ambos os mapas mutáveis ??e imutáveis ??(veja abaixo).

Como um teste, eu gerada uma lista contendo 1.000.000 single-item do mapa [Int, Int] e passou esta lista para as funções que eu estava testando. Com memória suficiente, os resultados foram surpreendente: ~ 1200ms para mutable.Map, ~ 1800ms para immutable.Map e ~ 750ms para uma implementação imperativo usando mutable.Map - não sei o que explica a grande diferença lá, mas fique à vontade para comentário sobre isso também.

O que me surpreendeu um pouco, talvez porque eu estou sendo um pouco grossa, é que com a configuração de execução padrão no IntelliJ 8.1, ambas as implementações mutáveis ??atingiu um OutOfMemoryError, mas a coleção imutável não. O teste imutável correu até a conclusão, mas fê-lo muito lentamente - que demora cerca de 28 segundos. Quando eu aumentou a memória JVM max (cerca de 200 MB, não tenho certeza onde o limite é), eu tenho os resultados acima.

De qualquer forma, aqui está o que eu realmente quero saber:

?

Por que as implementações mutáveis ??ficar sem memória, mas a implementação imutável não Eu suspeito que a versão imutável permite que o coletor de lixo para executar e liberar memória antes de as implementações mutáveis ??fazer - e todas essas coletas de lixo explicar a lentidão do imutável prazo de pouca memória - mas eu gostaria de uma explicação mais detalhada do que

.

Implementações abaixo. (Nota: Eu não afirmam que estes são os melhores implementações possível Sinta-se livre para sugerir melhorias..)

  def mergeMaps[A,B](func: (B,B) => B)(listOfMaps: List[Map[A,B]]): Map[A,B] =
    (Map[A,B]() /: (for (m <- listOfMaps; kv <-m) yield kv)) { (acc, kv) =>
      acc + (if (acc.contains(kv._1)) kv._1 -> func(acc(kv._1), kv._2) else kv)
    }

  def mergeMutableMaps[A,B](func: (B,B) => B)(listOfMaps: List[mutable.Map[A,B]]): mutable.Map[A,B] =
    (mutable.Map[A,B]() /: (for (m <- listOfMaps; kv <- m) yield kv)) { (acc, kv) =>
      acc + (if (acc.contains(kv._1)) kv._1 -> func(acc(kv._1), kv._2) else kv)
    }

  def mergeMutableImperative[A,B](func: (B,B) => B)(listOfMaps: List[mutable.Map[A,B]]): mutable.Map[A,B] = {
    val toReturn = mutable.Map[A,B]()
    for (m <- listOfMaps; kv <- m) {
      if (toReturn contains kv._1) {
        toReturn(kv._1) = func(toReturn(kv._1), kv._2)
      } else {
        toReturn(kv._1) = kv._2
      }
    }
    toReturn
  }

Answer 1

Well, it really depends on what the actual type of Map you are using. Probably HashMap. Now, mutable structures like that gain performance by pre-allocating memory it expects to use. You are joining one million maps, so the final map is bound to be somewhat big. Let's see how these key/values get added:

protected def addEntry(e: Entry) { 
  val h = index(elemHashCode(e.key)) 
  e.next = table(h).asInstanceOf[Entry] 
  table(h) = e 
  tableSize = tableSize + 1 
  if (tableSize > threshold) 
    resize(2 * table.length) 
} 

See the 2 * in the resize line? The mutable HashMap grows by doubling each time it runs out of space, while the immutable one is pretty conservative in memory usage (though existing keys will usually occupy twice the space when updated).

Now, as for other performance problems, you are creating a list of keys and values in the first two versions. That means that, before you join any maps, you already have each Tuple2 (the key/value pairs) in memory twice! Plus the overhead of List, which is small, but we are talking about more than one million elements times the overhead.

You may want to use a projection, which avoids that. Unfortunately, projection is based on Stream, which isn't very reliable for our purposes on Scala 2.7.x. Still, try this instead:

for (m <- listOfMaps.projection; kv <- m) yield kv

A Stream doesn't compute a value until it is needed. The garbage collector ought to collect the unused elements as well, as long as you don't keep a reference to the Stream's head, which seems to be the case in your algorithm.

EDIT

Complementing, a for/yield comprehension takes one or more collections and return a new collection. As often as it makes sense, the returning collection is of the same type as the original collection. So, for example, in the following code, the for-comprehension creates a new list, which is then stored inside l2. It is not val l2 = which creates the new list, but the for-comprehension.

val l = List(1,2,3)
val l2 = for (e <- l) yield e*2

Now, let's look at the code being used in the first two algorithms (minus the mutable keyword):

(Map[A,B]() /: (for (m <- listOfMaps; kv <-m) yield kv)) 

The foldLeft operator, here written with its /: synonymous, will be invoked on the object returned by the for-comprehension. Remember that a : at the end of an operator inverts the order of the object and the parameters.

Now, let's consider what object is this, on which foldLeft is being called. The first generator in this for-comprehension is m <- listOfMaps. We know that listOfMaps is a collection of type List[X], where X isn't really relevant here. The result of a for-comprehension on a List is always another List. The other generators aren't relevant.

So, you take this List, get all the key/values inside each Map which is a component of this List, and make a new List with all of that. That's why you are duplicating everything you have.

(in fact, it's even worse than that, because each generator creates a new collection; the collections created by the second generator are just the size of each element of listOfMaps though, and are immediately discarded after use)

The next question -- actually, the first one, but it was easier to invert the answer -- is how the use of projection helps.

When you call projection on a List, it returns new object, of type Stream (on Scala 2.7.x). At first you may think this will only make things worse, because you'll now have three copies of the List, instead of a single one. But a Stream is not pre-computed. It is lazily computed.

What that means is that the resulting object, the Stream, isn't a copy of the List, but, rather, a function that can be used to compute the Stream when required. Once computed, the result will be kept so that it doesn't need to be computed again.

Also, map, flatMap and filter of a Stream all return a new Stream, which means you can chain them all together without making a single copy of the List which created them. Since for-comprehensions with yield use these very functions, the use of Stream inside the prevent unnecessary copies of data.

Now, suppose you wrote something like this:

val kvs = for (m <- listOfMaps.projection; kv <-m) yield kv
(Map[A,B]() /: kvs) { ... }

In this case you aren't gaining anything. After assigning the Stream to kvs, the data hasn't been copied yet. Once the second line is executed, though, kvs will have computed each of its elements, and, therefore, will hold a complete copy of the data.

Now consider the original form::

(Map[A,B]() /: (for (m <- listOfMaps.projection; kv <-m) yield kv)) 

In this case, the Stream is used at the same time it is computed. Let's briefly look at how foldLeft for a Stream is defined:

override final def foldLeft[B](z: B)(f: (B, A) => B): B = { 
  if (isEmpty) z 
  else tail.foldLeft(f(z, head))(f) 
} 

If the Stream is empty, just return the accumulator. Otherwise, compute a new accumulator (f(z, head)) and then pass it and the function to the tail of the Stream.

Once f(z, head) has executed, though, there will be no remaining reference to the head. Or, in other words, nothing anywhere in the program will be pointing to the head of the Stream, and that means the garbage collector can collect it, thus freeing memory.

The end result is that each element produced by the for-comprehension will exist just briefly, while you use it to compute the accumulator. And this is how you save keeping a copy of your whole data.

Finally, there is the question of why the third algorithm does not benefit from it. Well, the third algorithm does not use yield, so no copy of any data whatsoever is being made. In this case, using projection only adds an indirection layer.