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
Related
I have a function like this:
fun randomWalk(numSteps: Int): Int {
var n = 0
repeat(numSteps) { n += (-1 + 2 * Random.nextInt(2)) }
return n.absoluteValue
}
This works fine, except that it uses a mutable variable, and I would like to make everything immutable when possible, for better safety and readability. So I came up with an equivalent version that doesn't use any mutable variables:
fun randomWalk_seq(numSteps: Int): Int =
generateSequence(0) { it + (-1 + 2 * Random.nextInt(2)) }
.elementAt(numSteps)
.absoluteValue
This also works fine and produces the same results, but it takes 3 times longer.
I used the following way to measure it:
#OptIn(ExperimentalTime::class)
fun main() {
val numSamples = 100000
val numSteps = 15708
repeat(5) {
val randomWalkSamples: IntArray
val duration = measureTime {
randomWalkSamples = IntArray(numSamples) { randomWalk(numSteps) }
}
println(duration)
}
}
I know it's a bit hacky (I could have used JMH but this is just a quick test - at least I know that measureTime uses a monotonic clock). The results for the iterative (mutable) version:
2.965358406s
2.560777033s
2.554363661s
2.564279403s
2.608323586s
As expected, the first line shows it took a bit longer on the first run due to the warming up of the JIT, but the next 4 lines have fairly small variation.
After replacing randomWalk with randomWalk_seq:
6.636866719s
6.980840906s
6.993998111s
6.994038706s
7.018054467s
Somewhat surprisingly, I don't see any warmup time - the first line is always lesser duration than the following 4 lines, every time I run this. And also, every time I run it, the duration keeps increasing, with line 5 always being the greatest duration.
Can someone explain the findings, and also is there any way of making this function not use any mutable variables but still have performance that is close to the mutable version?
Your solution is slower for two main reasons: boxing and the complexity of the iterator used by generateSequence()'s Sequence implementation.
Boxing happens because a Sequence uses its types generically, so it cannot use primitive 32-bit Ints directly, but must wrap them in classes and unwrap them when retrieving the items.
You can see the complexity of the iterator by Ctrl+clicking the generateSequence function to view the source code.
#Михаил Нафталь's suggestion is faster because it avoids the complex iterator of the sequence, but it still has boxing.
I tried writing an overload of sumOf that uses IntProgression directly instead of Iterable<T>, so it won't use boxing, and that resulted in equivalent performance to your imperative code with the var. As you can see, it's inline and when put together with the { -1 + 2 * Random.nextInt(2) } lambda suggested by #Михаил Нафталь, then the resulting compiled code will be equivalent to your imperative code.
inline fun IntProgression.sumOf(selector: (Int) -> Int): Int {
var sum: Int = 0.toInt()
for (element in this) {
sum += selector(element)
}
return sum
}
Ultimately, I don't think you're buying yourself much in the way of code clarity by removing a single var in such a small function. I would say the sequence code is arguably harder to read. vars may add to code complexity in complex algorithms, but I don't think they do in such simple algorithms, especially when there's only one of them and it's local to the function.
Equivalent immutable one-liner is:
fun randomWalk2(numSteps: Int) =
(1..numSteps).sumOf { -1 + 2 * Random.nextInt(2) }.absoluteValue
Probably, even more performant would be to replace
with
so that you'll have one multiplication and n additions instead of n multiplications and (2*n-1) additions:
fun randomWalk3(numSteps: Int) =
(-numSteps + 2 * (1..numSteps).sumOf { Random.nextInt(2) }).absoluteValue
Update
As #Tenfour04 noted, there is no specific stdlib implementation for IntProgression.sumOf, so it's resolved to Iterable<T>.sumOf, which will add unnecessary overhead for int boxing.
So, it's better to use IntArray here instead of IntProgression:
fun randomWalk4(numSteps: Int) =
(-numSteps + 2 * IntArray(numSteps).sumOf { Random.nextInt(2) }).absoluteValue
Still encourage you to check this all with JMH
I think:"Removing mutability without losing speed" is wrong title .because
mutability thing comes to deal with the flow that program want to achieve .
you are using var inside function.... and 100% this var will not ever change from outside this function and that is mutability concept.
if we git rid off from var everywhere why we need it in programming ?
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'm implementing a simple Bloom filter as an exercise.
Bloom filters require multiple hash functions, which for practical purposes I don't have.
Assuming I want to have 3 hash functions, isn't it enough to just take the hash of the object I'm checking membership for, hashing it (with murmur3) and then add +1, +2, +3 (for the 3 different hashes) before hashing them again?
As the murmur3 function has a very good avalanche effect (really spreads out results) wouldn't this for all purposes be reasonable?
Pseudo-code:
function generateHashes(obj) {
long hash = murmur3_hash(obj);
long hash1 = murmur3_hash(hash+1);
long hash2 = murmur3_hash(hash+2);
long hash3 = murmur3_hash(hash+3);
(hash1, hash2, hash3)
}
If not, what would be a simple, useful approach to this? I'd like to have a solution that would allow me to easily scale for more hash functions if needed be.
AFAIK, the usual approach is to not actually use multiple hash functions. Rather, hash once and split the resulting hash into 2, 3, or how many parts you want for your Bloom filter. So for example create a hash of 128 bits and split it into 2 hashes 64 bit each.
https://github.com/Claudenw/BloomFilter/wiki/Bloom-Filters----An-overview
The hashing functions of Bloom filter should be independent and random enough. MurmurHash is great for this purpose. So your approach is correct, and you can generate as many new hashes your way. For the educational purposes it is fine.
But in real world, running hashing function multiple times is slow, so the usual approach is to create ad-hoc hashes without actually calculating the hash.
To correct #memo, this is done not by splitting the hash into multiple parts, as the width of the hash should remain constant (and you can't split 64 bit hash to more than 64 parts ;) ). The approach is to get a two independent hashes and combine them.
function generateHashes(obj) {
// initialization phase
long h1 = murmur3_hash(obj);
long h2 = murmur3_hash(h1);
int k = 3; // number of desired hash functions
long hash[k];
// generation phase
for (int i=0; i<k; i++) {
hash[i] = h1 + (i*h2);
}
return hash;
}
As you see, this way creating a new hash is a simple multiply-add operation.
It would not be a good approach. Let me try and explain. Bloom filter allows you to test if an element most likely belongs to a set, or if it absolutely doesn’t. In others words, false positives may occur, but false negatives won’t.
Reference: https://sc5.io/posts/what-are-bloom-filters-and-why-are-they-useful/
Let us consider an example:
You have an input string 'foo' and we pass it to the multiple hash functions. murmur3 hash gives the output K, and subsequent hashes on this hash value give x, y and z
Now assume you have another string 'bar' and as it happens, its murmur3 hash is also K. The remaining hash values? They will be x, y and z because in your proposed approach the subsequent hash functions are not dependent on the input, but instead on the output of first hash function.
long hash1 = murmur3_hash(hash+1);
long hash2 = murmur3_hash(hash+2);
long hash3 = murmur3_hash(hash+3);
As explained in the link, the purpose is to perform a probabilistic search in a set. If we perform search for 'foo' or for 'bar' we would say that it is 'likely' that both of them are present. So the % of false positives will increase.
In other words this bloom filter will behave like a simple hash-function. The 'bloom' aspect of it will not come into picture because only the first hash function is determining the outcome of search.
Hope I was able to explain sufficiently. Let me know in comments if you have some more follow-up queries. Would be happy to assist.
When there is a collection and you must perform two or more operations on all of its elements, what is faster?:
val f1: String => String = _.reverse
val f2: String => String = _.toUpperCase
val elements: Seq[String] = List("a", "b", "c")
iterate multiple times and perform one operation on one loop
val result = elements.map(f1).map(f2)
This approach does have the advantage, that the result after application of the first function could be reused.
iterate one time and perform all operation on each element together
val result = elements.map(element => f2(f1(element)))
or
val result = elements.map(element => f1.compose(f2)
Is there any difference in performance between these two approaches? And if yes, which is faster?
Here's the thing, transformation of a collection is more or less of runtime O(N) , * runtime cost of all the functions applied. So I doubt the 2nd set of choices you present above would make even the slightest difference in runtime. The first option you list, is a different story. New collection creation can be avoided, because that could result in overhead. That's where "view" collections come in (see this good example I spotted)
In Scala, what does "view" do?
If you had the apply several mapping operations you might do this:
val result = elements.view.map(f1).map(f2).force
(force at the end, causes all functions to evaluate)
The 2nd set of examples above would maybe be a tiny bit faster, but the "view" option could make your code more readable if you had a lot of these or complex anonymous functions used in the mapping.
Composing functions to produce a single pass transformation will probably gain you some performance, but will quickly become unreadable. Consider using views as an alernative. While this will create intermediate collections:
val result = elements.map(f1).map(f2)
This will perform lazy evaluation and will perform functional composition the same way you do:
val result = elements.view.map(f1).map(f2)
Notice that result type will be SeqView so you might want to convert it to list later with toList.
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.