I want to have a loop that counts from 5 to 1. I know I can do the opposite easily with the .. operator:
for 1..5 { .print }
12345
But when using the .. operator and reversing the sides, it doesn't seem to generate a valid Range that for can work with:
for 5..1 { .print }
Nil
I know I can use the reverse method on the Range object:
for (1..5).reverse { .print }
54321
However, I would expect that the .. operator has a certain way to generate the list of numbers in 1 call. My question is, how to create a Range that counts down using the .. operator?
Just use 5...1 (note the 3 ... ):
for 5...1 { .print }
54321
The iterator that Range.reverse is basically the same iterator as for forwards, but iterating from the end to the beginning. Apart from the extra initial call to .reverse, there should be no difference in execution.
Except for the fact that for 1..5 { } is actually statically optimized to not use a Range internally at all. I guess this is an opportunity for a further statical optimization.
EDIT: https://github.com/rakudo/rakudo/commit/2dd02751da ensures that for (1..5).reverse is optimized in the same way as for 1..5. This makes it about 6x faster.
EDIT: As of https://github.com/rakudo/rakudo/commit/dfd6450d74 , the best way to count down a Range is for 5...1: this uses the same optimization as (1..5).reverse, but is much more readable.
Related
I start with a list of integers from 1 to 1000 in listOfRandoms.
I would like to left join on random from the createDatabase list.
I am currently using a find{} statement within a loop to do this but feel like this is too heavy. Is there not a better (quicker) way to achieve same result?
Psuedo Code
data class DatabaseRow(
val refKey: Int,
val random: Int
)
fun main() {
val createDatabase = (1..1000).map { i -> DatabaseRow(i, Random()) }
val listOfRandoms = (1..1000).map { j ->
val lookup = createDatabase.find { it.refKey == j }
lookup.random
}
}
As mentioned in comments, the question seems to be mixing up database and programming ideas, which isn't helping.
And it's not entirely clear which parts of the code are needed, and which can be replaced. I'm assuming that you already have the createDatabase list, but that listOfRandoms is open to improvement.
The ‘pseudo’ code compiles fine except that:
You don't give an import for Random(), but none of the likely ones return an Int. I'm going to assume that should be kotlin.random.Random.nextInt().
And because lookup is nullable, you can't simply call lookup.random; a quick fix is lookup!!.random, but it would be safer to handle the null case properly with e.g. lookup?.random ?: -1. (That's irrelevant, though, given the assumption above.)
I think the general solution is to create a Map. This can be done very easily from createDatabase, by calling associate():
val map = createDatabase.associate{ it.refKey to it.random }
That should take time roughly proportional to the size of the list. Looking up values in the map is then very efficient (approx. constant time):
map[someKey]
In this case, that takes rather more memory than needed, because both keys and values are integers and will be boxed (stored as separate objects on the heap). Also, most maps use a hash table, which takes some memory.
Since the key is (according to comments) “an ascending list starting from a random number, like 18123..19123”, in this particular case it can instead be stored in an IntArray without any boxing. As you say, array indexes start from 0, so using the key directly would need a huge array and use only the last few cells — but if you know the start key, you could simply subtract that from the array index each time.
Creating such an array would be a bit more complex, for example:
val minKey = createDatabase.minOf{ it.refKey }
val maxKey = createDatabase.maxOf{ it.refKey }
val array = IntArray(maxKey - minKey + 1)
for (row in createDatabase)
array[row.refKey - minKey] = row.random
You'd then access values with:
array[someKey - minKey]
…which is also constant-time.
Some caveats with this approach:
If createDatabase is empty, then minOf() will throw a NoSuchElementException.
If it has ‘holes’, omitting some keys inside that range, then the array will hold its default value of 0 — you can change that by using the alternative IntArray constructor which also takes a lambda giving the initial value.)
Trying to look up a value outside that range will give an ArrayIndexOutOfBoundsException.
Whether it's worth the extra complexity to save a bit of memory will depend on things like the size of the ‘database’, and how long it's in memory for; I wouldn't add that complexity unless you have good reason to think memory usage will be an issue.
How do I break out of a Kotlin repeat loop?
(I see plenty of answers about forEach, but I want to see a repeat-specific answer.)
You can't use a naked return, because that will return from what contains the repeat.
You can't use break, because:
if the repeat is inside a loop, you'll break that loop
if the repeat is not in a loop, you'll get 'break' and 'continue' are only allowed inside a loop
These don't work (they are functionally identical):
repeat(5) { idx ->
println(">> $idx")
if(idx >= 2)
return#repeat // use implicit label
}
repeat(5) #foo{ idx ->
println(">> $idx")
if(idx >= 2)
return#foo // use explicit label
}
In both those cases, you get:
>> 0
>> 1
>> 2
>> 3
>> 4
(The return# in both those blocks actually acts like a continue, which you can see yourself if you add a println after the if-block.)
So how can I break out of a repeat?
It turns out that repeat (as well as forEach) are not actually loops. They're higher-order functions, that is, they are functions that take functions as parameters.
(I find this frustrating: They look and act like loops, and feature prominently in the Kotlin docs. Why not just go all the way and promote them to proper loops in the language?)
To break out of a repeat loop, this is the best answer I can devise:
run repeatBlock# { // need to make a label outside of the repeat!
repeat(20) { idx ->
println(">> $idx")
if(idx >= 2)
return#repeatBlock
}
}
This is the result of that:
>> 0
>> 1
>> 2
I wish I could do it without introducing a new indent-level, but I don't think it's possible.
That seems like quite an abuse of the repeat function. My understanding is that if you write repeat, you intend it to repeat for that number of times. Any fewer is surprising.
For your example, I would use a for loop:
for (idx in 0 until 20) {
println(">> $idx")
if (idx >= 2)
break
}
I think using the right tool is better than trying to coerce the wrong one (repeat) to do what you want.
repeat itself is implemented as a for loop:
for (index in 0 until times) {
action(index)
}
Trying to use it for anything more than that and you may as well write your own version, rather than wrap up extra behaviour on top of it.
Currently, I am looking into Kotlin and have a question about Sequences vs. Collections.
I read a blog post about this topic and there you can find this code snippets:
List implementation:
val list = generateSequence(1) { it + 1 }
.take(50_000_000)
.toList()
measure {
list
.filter { it % 3 == 0 }
.average()
}
// 8644 ms
Sequence implementation:
val sequence = generateSequence(1) { it + 1 }
.take(50_000_000)
measure {
sequence
.filter { it % 3 == 0 }
.average()
}
// 822 ms
The point here is that the Sequence implementation is about 10x faster.
However, I do not really understand WHY that is. I know that with a Sequence, you do "lazy evaluation", but I cannot find any reason why that helps reducing the processing in this example.
However, here I know why a Sequence is generally faster:
val result = sequenceOf("a", "b", "c")
.map {
println("map: $it")
it.toUpperCase()
}
.any {
println("any: $it")
it.startsWith("B")
}
Because with a Sequence you process the data "vertically", when the first element starts with "B", you don't have to map for the rest of the elements. It makes sense here.
So, why is it also faster in the first example?
Let's look at what those two implementations are actually doing:
The List implementation first creates a List in memory with 50 million elements. This will take a bare minimum of 200MB, since an integer takes 4 bytes.
(In fact, it's probably far more than that. As Alexey Romanov pointed out, since it's a generic List implementation and not an IntList, it won't be storing the integers directly, but will be ‘boxing’ them — storing references to Int objects. On the JVM, each reference could be 8 or 16 bytes, and each Int could take 16, giving 1–2GB. Also, depending how the List gets created, it might start with a small array and keep creating larger and larger ones as the list grows, copying all the values across each time, using more memory still.)
Then it has to read all the values back from the list, filter them, and create another list in memory.
Finally, it has to read all those values back in again, to calculate the average.
The Sequence implementation, on the other hand, doesn't have to store anything! It simply generates the values in order, and as it does each one it checks whether it's divisible by 3 and if so includes it in the average.
(That's pretty much how you'd do it if you were implementing it ‘by hand’.)
You can see that in addition to the divisibility checking and average calculation, the List implementation is doing a massive amount of memory access, which will take a lot of time. That's the main reason it's far slower than the Sequence version, which doesn't!
Seeing this, you might ask why we don't use Sequences everywhere… But this is a fairly extreme example. Setting up and then iterating the Sequence has some overhead of its own, and for smallish lists that can outweigh the memory overhead. So Sequences only have a clear advantage in cases when the lists are very large, are processed strictly in order, there are several intermediate steps, and/or many items are filtered out along the way (especially if the Sequence is infinite!).
In my experience, those conditions don't occur very often. But this question shows how important it is to recognise them when they do!
Leveraging lazy-evaluation allows avoiding the creation of intermediate objects that are irrelevant from the point of the end goal.
Also, the benchmarking method used in the mentioned article is not super accurate. Try to repeat the experiment with JMH.
Initial code produces a list containing 50_000_000 objects:
val list = generateSequence(1) { it + 1 }
.take(50_000_000)
.toList()
then iterates through it and creates another list containing a subset of its elements:
.filter { it % 3 == 0 }
... and then proceeds with calculating the average:
.average()
Using sequences allows you to avoid doing all those intermediate steps. The below code doesn't produce 50_000_000 elements, it's just a representation of that 1...50_000_000 sequence:
val sequence = generateSequence(1) { it + 1 }
.take(50_000_000)
adding a filtering to it doesn't trigger the calculation itself as well but derives a new sequence from the existing one (3, 6, 9...):
.filter { it % 3 == 0 }
and eventually, a terminal operation is called that triggers the evaluation of the sequence and the actual calculation:
.average()
Some relevant reading:
Kotlin: Beware of Java Stream API Habits
Kotlin Collections API Performance Antipatterns
I wrote the following code:
val src = (0 until 1000000).toList()
val dest = ArrayList<Double>(src.size / 2 + 1)
for (i in src)
{
if (i % 2 == 0) dest.add(Math.sqrt(i.toDouble()))
}
IntellJ (in my case AndroidStudio) is asking me if I want to replace the for loop with operations from stdlib. This results in the following code:
val src = (0 until 1000000).toList()
val dest = ArrayList<Double>(src.size / 2 + 1)
src.filter { it % 2 == 0 }
.mapTo(dest) { Math.sqrt(it.toDouble()) }
Now I must say, I like the changed code. I find it easier to write than for loops when I come up with similar situations. However upon reading what filter function does, I realized that this is a lot slower code compared to the for loop. filter function creates a new list containing only the elements from src that match the predicate. So there is one more list created and one more loop in the stdlib version of the code. Ofc for small lists it might not be important, but in general this does not sound like a good alternative. Especially if one should chain more methods like this, you can get a lot of additional loops that could be avoided by writing a for loop.
My question is what is considered good practice in Kotlin. Should I stick to for loops or am I missing something and it does not work as I think it works.
If you are concerned about performance, what you need is Sequence. For example, your above code will be
val src = (0 until 1000000).toList()
val dest = ArrayList<Double>(src.size / 2 + 1)
src.asSequence()
.filter { it % 2 == 0 }
.mapTo(dest) { Math.sqrt(it.toDouble()) }
In the above code, filter returns another Sequence, which represents an intermediate step. Nothing is really created yet, no object or array creation (except a new Sequence wrapper). Only when mapTo, a terminal operator, is called does the resulting collection is created.
If you have learned java 8 stream, you may found the above explaination somewhat familiar. Actually, Sequence is roughly the kotlin equivalent of java 8 Stream. They share similiar purpose and performance characteristic. The only difference is Sequence isn't designed to work with ForkJoinPool, thus a lot easier to implement.
When there is multiple steps involved or the collection may be large, it's suggested to use Sequence instead of plain .filter {...}.mapTo{...}. I also suggest you to use the Sequence form instead of your imperative form because it's easier to understand. Imperative form may become complex, thus hard to understand, when there are 5 or more steps involved in the data processing. If there is just one step, you don't need a Sequence, because it just creates garbage and gives you nothing useful.
You're missing something. :-)
In this particular case, you can use an IntProgression:
val progression = 0 until 1_000_000 step 2
You can then create your desired list of squares in various ways:
// may make the list larger than necessary
// its internal array is copied each time the list grows beyond its capacity
// code is very straight forward
progression.map { Math.sqrt(it.toDouble()) }
// will make the list the exact size needed
// no copies are made
// code is more complicated
progression.mapTo(ArrayList(progression.last / 2 + 1)) { Math.sqrt(it.toDouble()) }
// will make the list the exact size needed
// a single intermediate list is made
// code is minimal and makes sense
progression.toList().map { Math.sqrt(it.toDouble()) }
My advice would be to choose whichever coding style you prefer. Kotlin is both object-oriented and functional language, meaning both of your propositions are correct.
Usually, functional constructs favor readability over performance; however, in some cases, procedural code will also be more readable. You should try to stick with one style as much as possible, but don't be afraid to switch some code if you feel like it's better suited to your constraints, either readability, performance, or both.
The converted code does not need the manual creation of the destination list, and can be simplified to:
val src = (0 until 1000000).toList()
val dest = src.filter { it % 2 == 0 }
.map { Math.sqrt(it.toDouble()) }
And as mentioned in the excellent answer by #glee8e you can use a sequence to do a lazy evaluation. The simplified code for using a sequence:
val src = (0 until 1000000).toList()
val dest = src.asSequence() // change to lazy
.filter { it % 2 == 0 }
.map { Math.sqrt(it.toDouble()) }
.toList() // create the final list
Note the addition of the toList() at the end is to change from a sequence back to a final list which is the one copy made during the processing. You can omit that step to remain as a sequence.
It is important to highlight the comments by #hotkey saying that you should not always assume that another iteration or a copy of a list causes worse performance than lazy evaluation. #hotkey says:
Sometimes several loops. even if they copy the whole collection, show good performance because of good locality of reference. See: Kotlin's Iterable and Sequence look exactly same. Why are two types required?
And excerpted from that link:
... in most cases it has good locality of reference thus taking advantage of CPU cache, prediction, prefetching etc. so that even multiple copying of a collection still works good enough and performs better in simple cases with small collections.
#glee8e says that there are similarities between Kotlin sequences and Java 8 streams, for detailed comparisons see: What Java 8 Stream.collect equivalents are available in the standard Kotlin library?
Let's say I'd like to iterate through a generic iterator in reverse, without knowing about the internals of the iterator and essentially not cheating via untyped magic and assuming this could be any type of iterable, which serves a iterator; can we optimise the reverse of a iterator at runtime or even via macros?
Forwards
var a = [1, 2, 3, 4].iterator();
// Actual iteration bellow
for(i in a) {
trace(i);
}
Backwards
var a = [1, 2, 3, 4].iterator();
// Actual reverse iteration bellow
var s = [];
for(i in a) {
s.push(i);
}
s.reverse();
for(i in s) {
trace(i);
}
I would assume that there has to be a simpler way, or at least fast way of doing this. We can't know a size because the Iterator class doesn't carry one, so we can't invert the push on to the temp array. But we can remove the reverse because we do know the size of the temp array.
var a = [1,2,3,4].iterator();
// Actual reverse iteration bellow
var s = [];
for(i in a) {
s.push(i);
}
var total = s.length;
var totalMinusOne = total - 1;
for(i in 0...total) {
trace(s[totalMinusOne - i]);
}
Is there any more optimisations that could be used to remove the possibility of the array?
It bugs me that you have to duplicate the list, though... that's nasty. I mean, the data structure would ALREADY be an array, if that was the right data format for it. A better thing (less memory fragmentation and reallocation) than an Array (the "[]") to copy it into might be a linked List or a Hash.
But if we're using arrays, then Array Comprehensions (http://haxe.org/manual/comprehension) are what we should be using, at least in Haxe 3 or better:
var s = array(for (i in a) i);
Ideally, at least for large iterators that are accessed multiple times, s should be cached.
To read the data back out, you could instead do something a little less wordy, but quite nasty, like:
for (i in 1-s.length ... 1) {
trace(s[-i]);
}
But that's not very readable and if you're after speed, then creating a whole new iterator just to loop over an array is clunky anyhow. Instead I'd prefer the slightly longer, but cleaner, probably-faster, and probably-less-memory:
var i = s.length;
while (--i >= 0) {
trace(s[i]);
}
First of all I agree with Dewi Morgan duplicating the output generated by an iterator to reverse it, somewhat defeats its purpose (or at least some of its benefits). Sometimes it's okay though.
Now, about a technical answer:
By definition a basic iterator in Haxe can only compute the next iteration.
On the why iterators are one-sided by default, here's what we can notice:
if all if iterators could run backwards and forwards, the Iterator classes would take more time to write.
not all iterators run on collections or series of numbers.
E.g. 1: an iterator running on the standard input.
E.g. 2: an iterator running on a parabolic or more complicated trajectory for a ball.
E.g. 3: slightly different but think about the performance problems running an iterator on a very large single-linked list (eg the class List). Some iterators can be interrupted in the middle of the iteration (Lambda.has() and Lambda.indexOf() for instance return as soon as there is a match, so you normally don't want to think of what's iterated as a collection but more as an interruptible series or process iterated step by step).
While this doesn't mean you shouldn't define two-ways iterators if you need them (I've never done it in Haxe but it doesn't seem impossible), in the absolute having two-ways iterators isn't that natural, and enforcing Iterators to be like that would complicate coding one.
An intermediate and more flexible solution is to simply have ReverseXxIter where you need, for instance ReverseIntIter, or Array.reverseIter() (with using a custom ArrayExt class). So it's left for every programmer to write their own answers, I think it's a good balance; while it takes more time and frustration in the beginning (everybody probably had the same kind of questions), you end up knowing the language better and in the end there are just benefits for you.
Complementing the post of Dewi Morgan, you can use for(let i = a.length; --i >= 0;) i; if you wish to simplify the while() method. if you really need the index values, I think for(let i=a.length, k=keys(a); --i in k;) a[k[i]]; is the best that give to do keeping the performance. There is also for(let i of keys(a).reverse()) a[i]; which has cleaner writing, but its iteration rate increases 1n using .reduce()