In java I got this construction
for (let i = 0; i < x.length-1; I++
And here to avoid outOfBoundsException we are using x.length-1 but how to do the same thing in Kotlin? I got this code so far
x.forEachIndexed { index, _ ->
output.add((x[index+1]-x[index])*10)
}
And it crashes on the last element when we call x[index+1] so I need to handle the last element somehow
Input list
var x = doubleArrayOf(0.0, 0.23, 0.46, 0.69, 0.92, 1.15, 1.38, 1.61)
For a classic Java for loop you got two options in Kotlin.
One would be something like this.
val x = listOf(1,2,3,4)
for (i in 0 .. x.lastIndex){
// ...
}
Using .. you basically go from 0 up to ( and including) the number coresponding to the second item, in this case the last index of the list.( so from 0 <= i <= x.lastIndex)
The second option is using until
val x = listOf(1,2,3,4)
for (i in 0 until x.size){
// ...
}
This is similar to the previous approach, except the fact that until is not inclusive with the last element.(so from 0 <= i < x.size ).
What you probably need is something like this
val x = listOf(1,2,3,4)
for (i in 0 .. x.lastIndex -1){
// ...
}
or alternative, using until, like this
val x = listOf(1,2,3,4)
for (i in 0 until x.size-1){
// ...
}
This should probably avoid the IndexOut of bounds error, since you go just until the second to last item index.
Feel free to ask more if something is not clear.
This is also a great read if you want to learn more about ranges. https://kotlinlang.org/docs/ranges.html#progression
You already have an answer, but this is another option. If you would use a normal list, you would have access to zipWithNext(), and then you don't need to worry about any index, and you can just do:
list.zipWithNext { current, next ->
output.add((next - current)*10)
}
As mentioned by k314159, we can also do asList() to have direct access to zipWithNext and other list methods, without many drawbacks.
array.asList().zipWithNext { current, next ->
output.add(next - current)
}
Related
I have a function with a for-loop:
fun List<Int>.customSum(sumFunction: (Int) -> Boolean): Int {
var sum = 0
for (item in this) {
if (sumFunction(item))
sum += item
}
return sum
}
I want to know how I can write the above in functional style. I know that I have to use this.reduce(), but don't know exactly how to implement it.
return filter(sumFunction).sum()
Should be self-explanatory.
You can’t use reduce because it doesn’t let you reject the first element.
With fold it would be:
return fold(0) { a, b ->
if(sumFunction(b)) a + b else a
}
I can think if two ways to achieve that:
The first one is by using sumOf {...}:
.
fun List<Int>.customSum(sumFunction: (Int) -> Boolean): Int {
return sumOf {
if (sumFunction(it)) it else 0
}
}
The second one is by using filter {...} then sum():
.
fun List<Int>.customSum(sumFunction: (Int) -> Boolean): Int {
return filter(sumFunction).sum()
}
return this.reduce { sum, n -> if (sumFunction(n)) sum + n else 0}
If you really want to use reduce for some reason you can - but you need to add that 0 to the head of the list as your "start state":
fun List<Int>.customSum(sumFunction: (Int) -> Boolean): Int {
val stuff = listOf(0) + this
return stuff.reduce { a, b -> a + if (sumFunction(b)) b else 0 }
}
You have to do that because reduce is really there to combine a bunch of items, which is why for the first iteration you get the first two items in the list. You don't get to handle them separately, which is why you need to throw that 0 in there to get past that first step, and get to a point where you can just do your checking on the second parameter and ignore the first one, treating it as an accumulator instead of another item you also need to check.
That behaviour is what fold is for - with that function you pass in an initial state (which can be a completely different type from your items, since you're not just smushing them together to create a new value like with reduce) and then on each iteration you get that state and an item.
You can handle the item as you like, and then make changes to the accumulator state depending on the result. Which is exactly the behaviour of your for loop! fold is just a functional way to write one. Tenfour04's answer is how you'd do it - it's the right tool for the job here!
Update: I completely overlooked the complexity added by arr.sort() method. So in Kotlin for array of Int, It compiles to use java.util.DualPivotQuicksort see this which in turn has complexity of O(n^2). see this. Other than that, this is also a valid approach.
I know It can be solved by keeping multiple arrays or using collections (which is what I ended up submitting), I want to know what I missed in the following approach
fun migratoryBirds(arr: Array<Int>): Int {
var maxCount = 0
var maxType = 0
var count = 0
var type = 0
arr.sort()
println(arr.joinToString(" "))
for (value in arr){
if (type != value){
if (count > maxCount){
maxCount = count
maxType = type
}
// new count values
type = value
count = 1
} else {
count++
}
}
return maxType
}
This code passes every scenario except for Test case 2 which has 73966 items for array. On my local machine, that array of 73k+ elements was causing timeout but I did test for array up-to 20k+ randomly generated value 1..5 and every time it succeeded. But I couldn't manage to pass Test case 2 with this approach. So even though I ended up submitting an answer with collection stream approach, I would really like to know what could I be missing in above logic.
I am running array loop only once Complexity should be O(n), So that could not be reason for failing. I am pre-sorting array in ascending order, and I am checking for > not >=, therefore, If two types end up having same count, It will still return the lower of the two types. And this approach is working correctly even for array of 20k+ elements ( I am getting timeout for anything above 25k elements).
The reason it is failing is this line
arr.sort()
Sorting an array takes O(n logn) time. However using something like a hash map this can be solved in O(n) time.
Here is a quick python solution I made to give you the general idea
# Complete the migratoryBirds function below.
def migratoryBirds(arr):
ans = -1
count = -1
dic = {}
for x in arr:
if x in dic:
dic[x] += 1
else:
dic[x] = 1
if dic[x] > count or dic[x] == count and x < ans:
ans = x
count = dic[x]
return ans
When normally using a for-in-loop, the counter (in this case number) is a constant in each iteration:
for number in 1...10 {
// do something
}
This means I cannot change number in the loop:
for number in 1...10 {
if number == 5 {
++number
}
}
// doesn't compile, since the prefix operator '++' can't be performed on the constant 'number'
Is there a way to declare number as a variable, without declaring it before the loop, or using a normal for-loop (with initialization, condition and increment)?
To understand why i can’t be mutable involves knowing what for…in is shorthand for. for i in 0..<10 is expanded by the compiler to the following:
var g = (0..<10).generate()
while let i = g.next() {
// use i
}
Every time around the loop, i is a freshly declared variable, the value of unwrapping the next result from calling next on the generator.
Now, that while can be written like this:
while var i = g.next() {
// here you _can_ increment i:
if i == 5 { ++i }
}
but of course, it wouldn’t help – g.next() is still going to generate a 5 next time around the loop. The increment in the body was pointless.
Presumably for this reason, for…in doesn’t support the same var syntax for declaring it’s loop counter – it would be very confusing if you didn’t realize how it worked.
(unlike with where, where you can see what is going on – the var functionality is occasionally useful, similarly to how func f(var i) can be).
If what you want is to skip certain iterations of the loop, your better bet (without resorting to C-style for or while) is to use a generator that skips the relevant values:
// iterate over every other integer
for i in 0.stride(to: 10, by: 2) { print(i) }
// skip a specific number
for i in (0..<10).filter({ $0 != 5 }) { print(i) }
let a = ["one","two","three","four"]
// ok so this one’s a bit convoluted...
let everyOther = a.enumerate().filter { $0.0 % 2 == 0 }.map { $0.1 }.lazy
for s in everyOther {
print(s)
}
The answer is "no", and that's a good thing. Otherwise, a grossly confusing behavior like this would be possible:
for number in 1...10 {
if number == 5 {
// This does not work
number = 5000
}
println(number)
}
Imagine the confusion of someone looking at the number 5000 in the output of a loop that is supposedly bound to a range of 1 though 10, inclusive.
Moreover, what would Swift pick as the next value of 5000? Should it stop? Should it continue to the next number in the range before the assignment? Should it throw an exception on out-of-range assignment? All three choices have some validity to them, so there is no clear winner.
To avoid situations like that, Swift designers made loop variables in range loops immutable.
Update Swift 5
for var i in 0...10 {
print(i)
i+=1
}
I want to make a cave explorer game in game maker 8.0.
I've made a block object and an generator But I'm stuck. Here is my code for the generator
var r;
r = random_range(0, 1);
repeat(room_width/16) {
repeat(room_height/16) {
if (r == 1) {
instance_create(x, y, obj_block)
}
y += 16;
}
x += 16;
}
now i always get a blank frame
You need to use irandom(1) so you get an integer. You also should put it inside the loop so it generates a new value each time.
In the second statement, you are generating a random real value and storing it in r. What you actually require is choosing one of the two values. I recommend that you use the function choose(...) for this. Here goes the corrected statement:
r = choose(0,1); //Choose either 0 or 1 and store it in r
Also, move the above statement to the inner loop. (Because you want to decide whether you want to place a block at the said (x,y) location at every spot, right?)
Also, I recommend that you substitute sprite_width and sprite_height instead of using the value 16 directly, so that any changes you make to the sprite will adjust the resulting layout of the blocks accordingly.
Here is the code with corrections:
var r;
repeat(room_width/sprite_width) {
repeat(room_height/sprite_height) {
r = choose(0, 1);
if (r == 1)
instance_create(x, y, obj_block);
y += sprite_height;
}
x += sprite_width;
}
That should work. I hope that helps!
Looks like you are only creating a instance if r==1. Shouldn't you create a instance every time?
Variable assignment r = random_range(0, 1); is outside the loop. Therefore performed only once before starting the loop.
random_range(0, 1) returns a random real number between 0 and 1 (not integer!). But you have if (r == 1) - the probability of getting 1 is a very small.
as example:
repeat(room_width/16) {
repeat(room_height/16) {
if (irandom(1)) {
instance_create(x, y, obj_block)
}
y += 16;
}
x += 16;
}
Here's a possible, maybe even better solution:
length = room_width/16;
height = room_height/16;
for(xx = 0; xx < length; xx+=1)
{
for(yy = 0; yy < height; yy+=1)
{
if choose(0, 1) = 1 {
instance_create(xx*16, yy*16, obj_block); }
}
}
if you want random caves, you should probably delete random sections of those blocks,
not just single ones.
For bonus points, you could use a seed value for the random cave generation. You can also have a pathway random generation that will have a guaranteed path to the finish with random openings and fake paths that generate randomly from that path. Then you can fill in the extra spaces with other random pieces.
But in regards to your code, you must redefine the random number each time you are placing a block, which is why all of them are the same. It should be called inside of the loops, and should be an integer instead of a decimal value.
Problem is on the first line, you need to put r = something in the for cycle
I've got a particle "engine" whom I've implementing a Pool system to and I've tested two different ways of rendering every Particle in a list. Please note that the Pooling really doesn't have anything with the problem to do. I just followed a tutorial and tried to use the second method when I noticed that they behaved differently.
The first way:
for (int i = 0; i < particleList.size(); i++) {
Iterator<Particle> it = particleList.iterator();
while (it.hasNext()) {
Particle p = it.next();
if (p.isDead()){
it.remove();
}
p.render(batch, delta);
}
}
Which works just fine. My particles are sharp and they move with the correct speed.
The second way:
Particle p;
for (int i = 0; i < particleList.size(); i++) {
p = particleList.get(i);
p.render(batch, delta);
if (p.isDead()) {
particleList.remove(i);
bulletPool.free(p);
}
}
Which makes all my particles blurry and moving really slow!
The render method for my particles look like this:
public void render(SpriteBatch batch, float delta) {
sprite.setX(sprite.getX() + (dx * speed) * delta * Assets.FPS);
sprite.setY(sprite.getY() + (dy * speed) * delta * Assets.FPS);
ttl--;
sprite.setScale(sprite.getScaleX() - 0.002f);
if (ttl <= 0 || sprite.getScaleX() <= 0)
isDead = true;
sprite.draw(batch);
}
Why do the different rendering methods provide different results?
Thanks in advance
You are mutating (removing elements from) a list while iterating over it. This is a classic way to make a mess.
The Iterator must have code to handle the delete case correctly. But your index-based for loop does not. Specifically when you call particleList.remove(i) the i is now "out of sync" with the content of the list. Consider what happens when you remove the element at index 3: 'i' will increment to 4, but the old element 4 got shuffled down into index 3, so it will get skipped.
I assume you're avoiding the Iterator to avoid memory allocations. So, one way to side-step this issue is to reverse the loop (go from particleList.size() down to 0). Alternatively, you can only increment i for non-dead particles.