maybe it is a stupid question, but I have this doubt and I cannot find a response...
If I have a map operation on a list of complex objects and to make the code more readable I use intermediate variables inside the map the performance can change?
For example this are two versions of the same code:
profilesGroupedWithIds map {
c =>
val blockId = c._2
val entityIds = c._1._2
val entropy = c._1._1
if (separatorID < 0) BlockDirty(blockId, entityIds.swap, entropy)
else BlockClean(blockId, entityIds, entropy)
}
..
profilesGroupedWithIds map {
c =>
if (separatorID < 0) BlockDirty(c._2, c._1._2.swap, c._1._1)
else BlockClean(c._2, c._1._2, c._1._1)
}
As you can see the first version is more readable than the second one.
But the efficiency is the same? Or the three variables that I have created inside the map have to be removed by the garbage collector and this will reduce the perfomance (suppose that 'profilesGroupedWithIds' is a very big list)?
Thanks
Regards
Luca
The vals are just aliases for the tuple elements. So the generated java bytecode will be identical in both cases, and so will be the performance.
More importantly, the first variant is much better code since it clearly conveys the intent.
Here is a third variant that avoids accessing the tuple elements _1 and _2 entirely:
profilesGroupedWithIds map {
case ((entropy,entityIds),blockId) =>
if (separatorID < 0) BlockDirty(blockId, entityIds.swap, entropy)
else BlockClean(blockId, entityIds, entropy)
}
Related
I know that most of the Vector methods are effectively O(1) (constant time) because of the tree they use, but I cannot find any information on the contains method. My first thought is that it would have to be O(n) to check all the elements but I am not sure.
Answering the question in the title, performance characteristics (2.13 docs version) of basic operations head, tail, apply, update, prepend, append, insert are all listed as eC for Vector:
eC The operation takes effectively constant time, but this might depend on some assumptions such as maximum length of a vector or distribution of hash keys.
You are correct contains is O(N), as there is no hashing or nothing else that would avoid the need to compare with all items. Still, if you want to be sure, it is best to check the implementation.
As finding the correct implementation in the library sources can be difficult because of many traits and overrides used to implement the containers, the best way to check this is the debugger. Use a code like:
val v = Vector(0, 1, 2)
v.contains(1)
Use the debugger to step into v.contains and the source you will see is:
def contains[A1 >: A](elem: A1): Boolean = exists (_ == elem)
If you are still not convinced at this point, some more "step into" will lead you to:
def exists(p: A => Boolean): Boolean = {
var res = false
while (!res && hasNext) res = p(next())
res
}
I would like to know if an update operation on a mutable map is better in performance than reassignment.
Lets assume I have the following Map
val m=Map(1 -> Set("apple", "banana"),
2 -> Set("banana", "cabbage"),
3 -> Set("cabbage", "dumplings"))
which I would like to reverse into this map:
Map("apple" -> Set(1),
"banana" -> Set(1, 2),
"cabbage" -> Set(2, 3),
"dumplings" -> Set(3))
The code to do so is:
def reverse(m:Map[Int,Set[String]])={
var rm = Map[String,Set[Int]]()
m.keySet foreach { k=>
m(k) foreach { e =>
rm = rm + (e -> (rm.getOrElse(e, Set()) + k))
}
}
rm
}
Would it be more efficient to use the update operator on a map if it is very large in size?
The code using the update on map is as follows:
def reverse(m:Map[Int,Set[String]])={
var rm = scala.collection.mutable.Map[String,Set[Int]]()
m.keySet foreach { k=>
m(k) foreach { e =>
rm.update(e,(rm.getOrElse(e, Set()) + k))
}
}
rm
}
I ran some tests using Rex Kerr's Thyme utility.
First I created some test data.
val rndm = new util.Random
val dna = Seq('A','C','G','T')
val m = (1 to 4000).map(_ -> Set(rndm.shuffle(dna).mkString
,rndm.shuffle(dna).mkString)).toMap
Then I timed some runs with both the immutable.Map and mutable.Map versions. Here's an example result:
Time: 2.417 ms 95% CI 2.337 ms - 2.498 ms (n=19) // immutable
Time: 1.618 ms 95% CI 1.579 ms - 1.657 ms (n=19) // mutable
Time 2.278 ms 95% CI 2.238 ms - 2.319 ms (n=19) // functional version
As you can see, using a mutable Map with update() has a significant performance advantage.
Just for fun I also compared these results with a more functional version of a Map reverse (or what I call a Map inverter). No var or any mutable type involved.
m.flatten{case(k, vs) => vs.map((_, k))}
.groupBy(_._1)
.mapValues(_.map(_._2).toSet)
This version consistently beat your immutable version but still doesn't come close to the mutable timings.
The trade-of between mutable and immutable collections usually narrows down to this:
immutable collections are safer to share and allows to use structural sharing
mutable collections have better performance
Some time ago I did comparison of performance between mutable and immutable Maps in Scala and the difference was about 2 to 3 times in favor of mutable ones.
So, when performance is not critical I usually go with immutable collections for safety and readability.
For example, in your case functional "scala way" of performing this transformation would be something like this:
m.view
.flatMap(x => x._2.map(_ -> x._1)) // flatten map to lazy view of String->Int pairs
.groupBy(_._1) // group pairs by String part
.mapValues(_.map(_._2).toSet) // extract all Int parts into Set
Although I used lazy view to avoid creating intermediate collections, groupBy still internally creates mutable map (you may want to check it's sources, the logic is pretty similar to what you have wrote), which in turn gets converted to immutable Map which then gets discarded by mapValues.
Now, if you want to squeeze every bit of performance you want to use mutable collections and do as little updates of immutable collections as possible.
For your case is means having Map of mutable Sets as you intermediate buffer:
def transform(m:Map[Int, Set[String]]):Map[String, Set[Int]] = {
val accum:Map[String, mutable.Set[Int]] =
m.valuesIterator.flatten.map(_ -> mutable.Set[Int]()).toMap
for ((k, vals) <- m; v <- vals) {
accum(v) += k
}
accum.mapValues(_.toSet)
}
Note, I'm not updating accum once it's created: I'm doing exactly one map lookup and one set update for each value, while in both your examples there was additional map update.
I believe this code is reasonably optimal performance wise. I didn't perform any tests myself, but I highly encourage you to do that on your real data and post results here.
Also, if you want to go even further, you might want to try mutable BitSet instead of Set[Int]. If ints in your data are fairly small it might yield some minor performance increase.
Just using #Aivean method in a functional way:
def transform(mp :Map[Int, Set[String]]) = {
val accum = mp.values.flatten
.toSet.map( (_-> scala.collection.mutable.Set[Int]())).toMap
mp.map {case(k,vals) => vals.map( v => accum(v)+=k)}
accum.mapValues(_.toSet)
}
I need to iterate on all the enum values, check if they were used to construct an int (called input) and if so, add them to a Set (called usefulEnums). I can either use streams API or iterate over all the enums to do this task. Is there any benefit of using Arrays.stream() over the traditional approach of iterating over the values() array?
enum TestEnum { VALUE1, VALUE2, VALUE3 };
Set<TestEnum> usefulEnums = new HashSet<>();
Arrays.stream(TestEnum.values())
.filter(t -> (input & t.getValue()) != 0)
.forEach(usefulEnums::add);
for (TestEnum t : TestEnum.values()) {
if ((input & t.getValue()) != 0) {
usefulEnums.add(t);
}
}
If you care for efficiency, you should consider:
Set<TestEnum> usefulEnums = EnumSet.allOf(TestEnum.class);
usefulEnums.removeIf(t -> (input & t.getValue()) == 0);
Note that when you have to iterate over all enum constants of a type, using EnumSet.allOf(EnumType.class).stream() avoids the array creation of EnumType.values() entirely, however, most enum types don’t have enough constants for this to make a difference. Further, the JVM’s optimizer may remove the temporary array creation anyway.
But for this specific task, where the result is supposed to be a Set<TestEnum>, using an EnumSet instead of a HashSet may even improve subsequent operations working with the Set. Creating an EnumSet holding all constants and removing unintented constants like in the solution above, means just initializing a long with 0b111, followed by clearing the bits of nonmatching elements.
For this short operation the for loop is going to be faster (nano-seconds faster), but to me the stream operation is more verbose, it tells exactly what is being done here. It's like reading diagonally.
Also you could collect directly to a HashSet:
Arrays.stream(TestEnum.values())
.filter(t -> (input & t.getValue()) != 0)
.collect(Collectors.toCollection(HashSet::new));
Valuable input from Holger as usual makes this even nicer:
EnumSet<TestEnum> filtered = EnumSet.allOf(TestEnum.class).stream()
.filter(t -> (input & t.getValue()) != 0)
.collect(Collectors.toCollection(() -> EnumSet.noneOf(TestEnum.class)));
I have a class with few Int and Double fields. What is the fastes way to copy all data from one object to another?
class IntFields {
private val data : Array[Int] = Array(0,0)
def first : Int = data(0)
def first_= (value: Int) = data(0) = value
def second : Int = data(1)
def second_= (value : Int) = data(1) = value
def copyFrom(another : IntFields) =
Array.copy(another.data,0,data,0,2)
}
This is the way I may suggest. But I doubt it is really effective, since I have no clear understanding scala's internals
update1:
In fact I'm searching for scala's equivalent of c++ memcpy. I need just take one simple object and copy it contents byte by byte.
Array copying is just a hack, I've googled for normal scala supported method and find none.
update2:
I've tried to microbenchmark two holders: simple case class with 12 variables and one backed up with array. In all benchmarks (simple copying and complex calculations over collection) array-based solution works slower for about 7%.
So, I need other means for simulating memcpy.
Since both arrays used for Array.copy are arrays of primitive integers (i.e. it is not the case that one of the holds boxed integers, in which case a while loop with boxing/unboxing would have been used to copy the elements), it is equally effective as the Java System.arraycopy is. Which is to say - if this were a huge array, you would probably see the difference in performance compared to a while loop in which you copy the elements. Since the array only has 2 elements, it is probably more efficient to just do:
def copyFrom(another: IntFields) {
data(0) = another.data(0)
data(1) = another.data(1)
}
EDIT:
I'd say that the fastest thing is to just copy the fields one-by-one. If performance is really important, you should consider using Unsafe.getInt - some report it should be faster than using System.arraycopy for small blocks: https://stackoverflow.com/questions/5574241/interesting-uses-of-sun-misc-unsafe
I wanted to compare the performance characteristics of immutable.Map and mutable.Map in Scala for a similar operation (namely, merging many maps into a single one. See this question). I have what appear to be similar implementations for both mutable and immutable maps (see below).
As a test, I generated a List containing 1,000,000 single-item Map[Int, Int] and passed this list into the functions I was testing. With sufficient memory, the results were unsurprising: ~1200ms for mutable.Map, ~1800ms for immutable.Map, and ~750ms for an imperative implementation using mutable.Map -- not sure what accounts for the huge difference there, but feel free to comment on that, too.
What did surprise me a bit, perhaps because I'm being a bit thick, is that with the default run configuration in IntelliJ 8.1, both mutable implementations hit an OutOfMemoryError, but the immutable collection did not. The immutable test did run to completion, but it did so very slowly -- it takes about 28 seconds. When I increased the max JVM memory (to about 200MB, not sure where the threshold is), I got the results above.
Anyway, here's what I really want to know:
Why do the mutable implementations run out of memory, but the immutable implementation does not? I suspect that the immutable version allows the garbage collector to run and free up memory before the mutable implementations do -- and all of those garbage collections explain the slowness of the immutable low-memory run -- but I'd like a more detailed explanation than that.
Implementations below. (Note: I don't claim that these are the best implementations possible. Feel free to suggest improvements.)
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
}
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.