I know you shouldn't, I kind of know why. But I mean I don't understand my own code once I am trying really to think what's going on.
So I have an array with bunch of objects. I am iterating over it and once I find an object with specific type, I remove it from the array, and add another object into the array. So something like this:
var arr = parent.allchildren() //getting all the children in array
for ele in arr{
if(ele==somethingHere){
parent.remove(ele)
parent.add(new ele) //add new child into child array
}
}
If I have an array of 1,2,3,4,5, and I remove 3 and add a 6 while iterating, the actual array would be 1,2,4,5,6 but the array I am iterating would still be 1,2,3,4,5.
Which I think it would be fine, because at the end I still get what I want, which removed the element and added the element I need. However modifying the list while iterating it is bad and you shouldn't do that, but for my case I think it does what I need. What could be the potential issue in my case that I can't see?
One thing you may want to think about doing is making all of the changes at the end of the iteration. Instead of making the changes one by one, record the changes you want to make while iterating, and then actually make those changes once your loop is finished.
For example, you could make an array of elements to remove, and an array of elements to add.
//Our array where we record what we want to add
var elementsToAdd = [Any]()
//Our array of what elements we want to remove. We record the index at
//which we want to remove the element from the array
var indexesToRemoveAt = [Int]()
//Getting all the children in array
var arr = parent.allchildren()
//Enumerating an array allows us to access the index at which that
//element occurs. For example, the first element's index would be 0,
//the second element's index would be 1, the third would be 2, and so
//on
for (index,ele) in arr.enumerated() {
if(ele == somethingHere) {
indexesToRemoveAt.append(index)
elementsToAdd.append(newEle)
}
}
//Now that we have recorded the changes we want to make, we could make
//all of the changes at once
arr.remove(at: indexesToRemoveAt)
arr.append(contentsOf: elementsToAdd)
Note that removing array elements at multiple indexes would require the following extension to Array. If you wanted to avoid creating this extension, you could always just loop through the array of indexes and tell the array to remove at each individual index. All this extension function is really doing is looping through the indexes, and removing the array element at said index.
Array extension to remove elements at multiple indexes:
extension Array {
//Allows us to remove at multiple indexes instead of just one
mutating func remove(at indexes: [Int]) {
for index in indexes.sorted(by: >) {
if index <= count-1 {
remove(at: index)
}
}
}
}
I just tested in a playground with the following code:
var arr = ["hi", "bye", "guy", "fry", "sky"]
for a in arr {
if arr.count >= 3 {
arr.remove(at: 2)
}
print(a)
}
print(arr)
This prints:
hi
bye
guy
fry
sky
["hi", "bye"]
So it looks like when you use a for-in loop in Swift, the array is copied and changes you make to it will not affect the array you are iterating over. To answer your question, as long as you understand that this is the behavior, there's nothing wrong with doing this.
Related
I can't seem to find much online, but how can I get the first (and then second and third, later on) entry of an array?
My array is being saved to UserDefaults elsewhere and then pulled for use here.
Thanks!
Did you mean to loop through the elements? You can use for in or forEach for that.
var array: [MyType]
for element in array {
print(element)
}
array.forEach { element in
print(element)
}
Maybe you are trying to iterate value from continually with index also
Here is two way you can get first index value than second index value third index value with index also . hope you will get your result .
First way :
var stringArray = ["a","b","C","D","a","i","x","D"]
for (index, element) in stringArray.enumerated() {
print("Item \(index): \(element)")
}
Another way :
for index in 0 ..< stringArray.count {
print(stringArray[index])
}
let me know if its help you.
EDIT: EVERY ANSWER BELOW ARE WORKING, THANKS FOR HELPING ME!
I'm currently learn about splicing an array in as3.
So here's my code:
//import classes
import flash.utils.Timer;
import flash.events.*;
//variables
var Arr:Array=new Array();
var num:Number=0;
//set a timer and set timer limit of 10 times
var timer:Timer=new Timer(1000,10);
//add a listener to our timer object
timer.addEventListener(TimerEvent.TIMER, tick);
timer.addEventListener(TimerEvent.TIMER_COMPLETE,tock);
//tick function
function tick(e:TimerEvent):void{
//i add an element each time the timer 'ticks'
Arr.push(['index'+num]);
num++;
}
//tock function
function tock(e:TimerEvent):void{
trace('array elements :'+Arr);//traces Arr elemnts
for(var i:int=0;i<Arr.length;i++){
Arr.splice(i,1);// i've tried Arr.splice(0,1), but neither working
trace('elemnts left : '+Arr);
}
I dont really understand the problem, but here's the result:
1.not every elements in Arr array have been removed
2.the maximum Arr's length is ten before spliced.BUT in the loop, its only splicing less than ten times, which it causes the problem above
anybody have an idea for this?
Please help me out
There are simpler and faster options:
You can just set the Array's length to 0. That will effectively remove all its elements at once.
Arr.length = 0;
You can create a new empty instance of the Array class. That will not destroy the original object immediately, but if there are no references to it, it will be consumed by the Garbage Collector eventually, so you won't need to think of it.
// You can omit () with the "new" operator if there are no mandatory arguments.
Arr = new Array;
for(var i:int=0;i<Arr.length;i++)
This is why it doesn't splice everything. Every time this loop runs, Arr.length is decreased by 1 since you spliced it, so once it gets to i==5(sixth loop), the conditions become fulfilled as 'i'(5) is no longer less than Arr.length(5 left in array), thus the loop stops.
You conditions should be to splice as long as the array has more than 0 items. Try this instead:
for(var i:int=0;Arr.length>0;i++)
Also, splice works like this. Arr.splice(INDEX, AMOUNT TO REMOVE). In this case you can splice at index0 to remove them one by one from the bottom. thus the right way to write this is:
Arr.splice(0,1)
If your goal is just to empty the array, simply do
Arr.length = 0;
On a side note, you dont need to put those square brackets when pushing a new array,
Arr.push('index '+num);
works just as well.
you can write in this way. It will remove the entire elements
function tock(e:TimerEvent):void{
var i = Arr.length
while (i--) {
...
if (...) {
Arr.splice(i, 1);
}
}
}
otherwise you just reinitialize that array(Arr)
var word = "morning"
var arr = Array(word)
for s in 0...word.count {
print(arr[s])
}
This will not print. Of course, if I substitute a number for s, the code works fine.
Why will it not accept a variable in the array access braces? Is this peculiar to Swift?
I've spent a long time trying to figure this out, and it's nothing to do with s being optional.
Anyone understand this?
you are using inclusive range ... instead of ..<, so s goes from 0 to 7, not 0 to 6.
However, in arr the index goes from 0 to 6 because there are 7 characters.
Thus, when the program tries to access arr[7], it throws an index out of range error.
If you were coding on Xcode, the debugger would have told you that there is no arr[7].
As for the code, here is a better way to print every item in arr than using an index counter:
var word = "morning"
var arr = Array(word)
for s in arr {
print(s)
}
This is called a "foreach loop", for each item in arr, it assigns it to s, performs the code in the loop, and moves on to the next item, assigns it to s, and so on.
When you have to access every element in an array or a collection, foreach loop is generally considered to be a more elegant way to do so, unless you need to store the index of a certain item during the loop, in which case the only option is the range-based for loop (which you are using).
Happy coding!
When I run it, it prints the array then throws the error Fatal error: Index out of range. To fix this, change the for loop to:
for s in 0..<word.count {
print(arr[s])
}
try this
when you use a word to recognize
size of Array your Array index start as 0 so array last index must be equal with your (word.count - 1)
var word = "morning"
var arr = Array(word)
for s in 0...(word.count-1) {
print(arr[s])
}
Basically avoid index based for loops as much as possible.
To print each character of a string simply use
var word = "morning"
for s in word { // in Swift 3 word.characters
print(s)
}
To solve the index issue you have to use the half-open range operator ..< since indexes are zero-based.
I have this foreach loop to check for collision and i want platform(movieclip) to be removed in case of collision. So far i've come up with this:
if (mcContent.mcPlayer.y + mcContent.mcPlayer.height > platformCloud.y)
{
mcContent.mcPlayer.y = platformCloud.y - mcContent.mcPlayer.height - 1;
jump();
mcContent.removeChild(platformCloud);
//platformsCloud.splice(platformCloud);
}
What this is doing is, removing the movieclip (ok so far so good) but without the splice, when the loop runs again through the array it is still there. So with the splice that is commented out there's 1 little problem, it removes all the movieclips from the array, apprently.
How can i splice only the current index that is being checked?
.splice() accepts a start index and an amount of items to remove, not the object you want to remove from the array.
Parameters
startIndex:int — An integer that specifies the index of the element in the array where the insertion or deletion begins. You can use a negative integer to specify a position relative to the end of the array (for example, -1 is the last element of the array).
deleteCount:uint — An integer that specifies the number of elements to be deleted. This number includes the element specified in the startIndex parameter. If you do not specify a value for the deleteCount parameter, the method deletes all of the values from the startIndex element to the last element in the array. If the value is 0, no elements are deleted.
You want to do this:
var index:int = platformsCloud.indexOf(platformCloud);
platformsCloud.splice(index, 1);
Why not just create a new array of the items to keep? Use Array.push to add new items. This may actually be more efficient than modifying the existing array. It also doesn't require keeping track of indices (which are required to use Array.splice).
Example code:
var keptPlatforms = [];
// do stuff
if (mcContent.mcPlayer.y + mcContent.mcPlayer.height > platformCloud.y)
{
mcContent.mcPlayer.y = platformCloud.y - mcContent.mcPlayer.height - 1;
jump();
mcContent.removeChild(platformCloud);
} else {
keptPlatforms.push(platformCloud);
}
// later, after this cycle, use the new Array
platformClouds = keptPlatforms;
Now, the reason platformsCloud.splice(platformCloud) removes all items is because the first argument is coerced to an integer so it is equivalent to platformsCloud.splice(0) which says "remove the 0th-indexed item to the end of the array". And, this does indeed clear the array.
To use Array.splice, you'd have to do something like:
// inside a loop this approach may lead to O(n^2) performance
var i = platformClouds.indexOf(platformCloud);
if (i >= 0) {
platformClouds.splice(i, 1); // remove 1 item at the i'th index
}
Concatenating an element x to an array items is easy in D, it's as if it were an array list:
arr ~= x;
but how do I remove an element at index i from items?
(Caveat: If I remove an element and then add a new element, the array must not be reallocated. So a simple slice won't work.)
Update:
Based on CyberShadow's answer about using assumeSafeAppend, I wrote this code:
static void removeAt(T)(ref T[] arr, size_t index)
{
foreach (i, ref item; arr[index .. $ - 1])
item = arr[i + 1];
arr = arr[0 .. $ - 1];
arr.assumeSafeAppend();
}
However, the problem happens when you have something like:
auto superArr = [0, 1, 2, 3, 4]; //Must not be modified
auto arr = superArr[0 .. $ - 1];
writeln(superArr);
arr.removeAt(0); //Should copy the slice and modify the copy
writeln(superArr); //but obviously doesn't
The base array of slice should not be modified if an element is removed from the slice; instead, the slice needs to be copied.
But I have no way of knowing if an array is a slice of a bigger array... so that doesn't work.
Any suggestions?
Copying my answer on digitalmars.D (thanks for forwarding):
As has been mentioned, std.algorithm.remove can be of help. You may want to look at three of its capabilities in particular: (a) remove multiple offsets in one pass, e.g. remove(a, 0, 4) removes the first and fifth element, (b) you can remove subranges, e.g. remove(a, tuple(1, 3)) removes the second through fourth element, and (c) if you don't care about the order in which elements are left after removal you may want to look into unstable remove, which does considerably less work.
Andrei
(Caveat: If I remove an element and then add a new element, the array must not be reallocated. So a simple slice won't work.)
The assumeSafeAppend function will tell the runtime not to reallocate the array when appending to it (i.e. it is an affirmation from the user that there aren't other slices which might be stomped by an append).
remove from std.algorithm does an in-place remove. If you're using std.container, there's also Array.linearRemove.
Well if order is of no importance you can copy the last element to the location of removal then reduce the array length by one.
If you just want to remove the first or last elements use slices:
array = array [1..$]
array = array [0..$-1]
Or a general way which works for a middle one as well:
array = array [0..unlucky] ~ array [unlucky+1..$]
If the elements aren't basic elements such as structs, floats, ints then arrays are implicitly arrays of pointers and this is an efficient operation.
There's no automated way of doing this, you'll have to shuffle the array items along, reset .length and then catenate.