Typescript - multidimensional array initialization - multidimensional-array

I'm playing with Typescript and I wonder, how to properly instantiate and declare multidimensional array. Here's my code:
class Something {
private things: Thing[][];
constructor() {
things = [][]; ??? how instantiate object ???
for(var i: number = 0; i < 10; i++) {
this.things[i] = new Thing[]; ??? how instantiate 1st level ???
for(var j: number = 0; j< 10; j++) {
this.things[i][j] = new Thing(); ??? how instantiate 2nd lvl item ???
}
}
}
}
Can you give me any hint about selected places?

You only need [] to instantiate an array - this is true regardless of its type. The fact that the array is of an array type is immaterial.
The same thing applies at the first level in your loop. It is merely an array and [] is a new empty array - job done.
As for the second level, if Thing is a class then new Thing() will be just fine. Otherwise, depending on the type, you may need a factory function or other expression to create one.
class Something {
private things: Thing[][];
constructor() {
this.things = [];
for(var i: number = 0; i < 10; i++) {
this.things[i] = [];
for(var j: number = 0; j< 10; j++) {
this.things[i][j] = new Thing();
}
}
}
}

Here is an example of initializing a boolean[][]:
const n = 8; // or some dynamic value
const palindrome: boolean[][] = new Array(n)
.fill(false)
.map(() =>
new Array(n).fill(false)
);

If you want to do it typed:
class Something {
areas: Area[][];
constructor() {
this.areas = new Array<Array<Area>>();
for (let y = 0; y <= 100; y++) {
let row:Area[] = new Array<Area>();
for (let x = 0; x <=100; x++){
row.push(new Area(x, y));
}
this.areas.push(row);
}
}
}

You can do the following (which I find trivial, but its actually correct).
For anyone trying to find how to initialize a two-dimensional array in TypeScript (like myself).
Let's assume that you want to initialize a two-dimensional array, of any type. You can do the following
const myArray: any[][] = [];
And later, when you want to populate it, you can do the following:
myArray.push([<your value goes here>]);
A short example of the above can be the following:
const myArray: string[][] = [];
myArray.push(["value1", "value2"]);

Beware of the use of push method, if you don't use indexes, it won't work!
var main2dArray: Things[][] = []
main2dArray.push(someTmp1dArray)
main2dArray.push(someOtherTmp1dArray)
gives only a 1 line array!
use
main2dArray[0] = someTmp1dArray
main2dArray[1] = someOtherTmp1dArray
to get your 2d array working!!!
Other beware! foreach doesn't seem to work with 2d arrays!

Related

ConcurrentModificationException when reinserting a node JavaFX [duplicate]

We all know you can't do the following because of ConcurrentModificationException:
for (Object i : l) {
if (condition(i)) {
l.remove(i);
}
}
But this apparently works sometimes, but not always. Here's some specific code:
public static void main(String[] args) {
Collection<Integer> l = new ArrayList<>();
for (int i = 0; i < 10; ++i) {
l.add(4);
l.add(5);
l.add(6);
}
for (int i : l) {
if (i == 5) {
l.remove(i);
}
}
System.out.println(l);
}
This, of course, results in:
Exception in thread "main" java.util.ConcurrentModificationException
Even though multiple threads aren't doing it. Anyway.
What's the best solution to this problem? How can I remove an item from the collection in a loop without throwing this exception?
I'm also using an arbitrary Collection here, not necessarily an ArrayList, so you can't rely on get.
Iterator.remove() is safe, you can use it like this:
List<String> list = new ArrayList<>();
// This is a clever way to create the iterator and call iterator.hasNext() like
// you would do in a while-loop. It would be the same as doing:
// Iterator<String> iterator = list.iterator();
// while (iterator.hasNext()) {
for (Iterator<String> iterator = list.iterator(); iterator.hasNext();) {
String string = iterator.next();
if (string.isEmpty()) {
// Remove the current element from the iterator and the list.
iterator.remove();
}
}
Note that Iterator.remove() is the only safe way to modify a collection during iteration; the behavior is unspecified if the underlying collection is modified in any other way while the iteration is in progress.
Source: docs.oracle > The Collection Interface
And similarly, if you have a ListIterator and want to add items, you can use ListIterator#add, for the same reason you can use Iterator#remove — it's designed to allow it.
In your case you tried to remove from a list, but the same restriction applies if trying to put into a Map while iterating its content.
This works:
Iterator<Integer> iter = l.iterator();
while (iter.hasNext()) {
if (iter.next() == 5) {
iter.remove();
}
}
I assumed that since a foreach loop is syntactic sugar for iterating, using an iterator wouldn't help... but it gives you this .remove() functionality.
With Java 8 you can use the new removeIf method. Applied to your example:
Collection<Integer> coll = new ArrayList<>();
//populate
coll.removeIf(i -> i == 5);
Since the question has been already answered i.e. the best way is to use the remove method of the iterator object, I would go into the specifics of the place where the error "java.util.ConcurrentModificationException" is thrown.
Every collection class has a private class which implements the Iterator interface and provides methods like next(), remove() and hasNext().
The code for next looks something like this...
public E next() {
checkForComodification();
try {
E next = get(cursor);
lastRet = cursor++;
return next;
} catch(IndexOutOfBoundsException e) {
checkForComodification();
throw new NoSuchElementException();
}
}
Here the method checkForComodification is implemented as
final void checkForComodification() {
if (modCount != expectedModCount)
throw new ConcurrentModificationException();
}
So, as you can see, if you explicitly try to remove an element from the collection. It results in modCount getting different from expectedModCount, resulting in the exception ConcurrentModificationException.
You can either use the iterator directly like you mentioned, or else keep a second collection and add each item you want to remove to the new collection, then removeAll at the end. This allows you to keep using the type-safety of the for-each loop at the cost of increased memory use and cpu time (shouldn't be a huge problem unless you have really, really big lists or a really old computer)
public static void main(String[] args)
{
Collection<Integer> l = new ArrayList<Integer>();
Collection<Integer> itemsToRemove = new ArrayList<>();
for (int i=0; i < 10; i++) {
l.add(Integer.of(4));
l.add(Integer.of(5));
l.add(Integer.of(6));
}
for (Integer i : l)
{
if (i.intValue() == 5) {
itemsToRemove.add(i);
}
}
l.removeAll(itemsToRemove);
System.out.println(l);
}
In such cases a common trick is (was?) to go backwards:
for(int i = l.size() - 1; i >= 0; i --) {
if (l.get(i) == 5) {
l.remove(i);
}
}
That said, I'm more than happy that you have better ways in Java 8, e.g. removeIf or filter on streams.
Same answer as Claudius with a for loop:
for (Iterator<Object> it = objects.iterator(); it.hasNext();) {
Object object = it.next();
if (test) {
it.remove();
}
}
With Eclipse Collections, the method removeIf defined on MutableCollection will work:
MutableList<Integer> list = Lists.mutable.of(1, 2, 3, 4, 5);
list.removeIf(Predicates.lessThan(3));
Assert.assertEquals(Lists.mutable.of(3, 4, 5), list);
With Java 8 Lambda syntax this can be written as follows:
MutableList<Integer> list = Lists.mutable.of(1, 2, 3, 4, 5);
list.removeIf(Predicates.cast(integer -> integer < 3));
Assert.assertEquals(Lists.mutable.of(3, 4, 5), list);
The call to Predicates.cast() is necessary here because a default removeIf method was added on the java.util.Collection interface in Java 8.
Note: I am a committer for Eclipse Collections.
Make a copy of existing list and iterate over new copy.
for (String str : new ArrayList<String>(listOfStr))
{
listOfStr.remove(/* object reference or index */);
}
People are asserting one can't remove from a Collection being iterated by a foreach loop. I just wanted to point out that is technically incorrect and describe exactly (I know the OP's question is so advanced as to obviate knowing this) the code behind that assumption:
for (TouchableObj obj : untouchedSet) { // <--- This is where ConcurrentModificationException strikes
if (obj.isTouched()) {
untouchedSet.remove(obj);
touchedSt.add(obj);
break; // this is key to avoiding returning to the foreach
}
}
It isn't that you can't remove from the iterated Colletion rather that you can't then continue iteration once you do. Hence the break in the code above.
Apologies if this answer is a somewhat specialist use-case and more suited to the original thread I arrived here from, that one is marked as a duplicate (despite this thread appearing more nuanced) of this and locked.
With a traditional for loop
ArrayList<String> myArray = new ArrayList<>();
for (int i = 0; i < myArray.size(); ) {
String text = myArray.get(i);
if (someCondition(text))
myArray.remove(i);
else
i++;
}
ConcurrentHashMap or ConcurrentLinkedQueue or ConcurrentSkipListMap may be another option, because they will never throw any ConcurrentModificationException, even if you remove or add item.
Another way is to use a copy of your arrayList just for iteration:
List<Object> l = ...
List<Object> iterationList = ImmutableList.copyOf(l);
for (Object curr : iterationList) {
if (condition(curr)) {
l.remove(curr);
}
}
A ListIterator allows you to add or remove items in the list. Suppose you have a list of Car objects:
List<Car> cars = ArrayList<>();
// add cars here...
for (ListIterator<Car> carIterator = cars.listIterator(); carIterator.hasNext(); )
{
if (<some-condition>)
{
carIterator().remove()
}
else if (<some-other-condition>)
{
carIterator().add(aNewCar);
}
}
Now, You can remove with the following code
l.removeIf(current -> current == 5);
I know this question is too old to be about Java 8, but for those using Java 8 you can easily use removeIf():
Collection<Integer> l = new ArrayList<Integer>();
for (int i=0; i < 10; ++i) {
l.add(new Integer(4));
l.add(new Integer(5));
l.add(new Integer(6));
}
l.removeIf(i -> i.intValue() == 5);
Java Concurrent Modification Exception
Single thread
Iterator<String> iterator = list.iterator();
while (iterator.hasNext()) {
String value = iter.next()
if (value == "A") {
list.remove(it.next()); //throws ConcurrentModificationException
}
}
Solution: iterator remove() method
Iterator<String> iterator = list.iterator();
while (iterator.hasNext()) {
String value = iter.next()
if (value == "A") {
it.remove()
}
}
Multi thread
copy/convert and iterate over another one collection. For small collections
synchronize[About]
thread safe collection[About]
I have a suggestion for the problem above. No need of secondary list or any extra time. Please find an example which would do the same stuff but in a different way.
//"list" is ArrayList<Object>
//"state" is some boolean variable, which when set to true, Object will be removed from the list
int index = 0;
while(index < list.size()) {
Object r = list.get(index);
if( state ) {
list.remove(index);
index = 0;
continue;
}
index += 1;
}
This would avoid the Concurrency Exception.
for (Integer i : l)
{
if (i.intValue() == 5){
itemsToRemove.add(i);
break;
}
}
The catch is the after removing the element from the list if you skip the internal iterator.next() call. it still works! Though I dont propose to write code like this it helps to understand the concept behind it :-)
Cheers!
Example of thread safe collection modification:
public class Example {
private final List<String> queue = Collections.synchronizedList(new ArrayList<String>());
public void removeFromQueue() {
synchronized (queue) {
Iterator<String> iterator = queue.iterator();
String string = iterator.next();
if (string.isEmpty()) {
iterator.remove();
}
}
}
}
I know this question assumes just a Collection, and not more specifically any List. But for those reading this question who are indeed working with a List reference, you can avoid ConcurrentModificationException with a while-loop (while modifying within it) instead if you want to avoid Iterator (either if you want to avoid it in general, or avoid it specifically to achieve a looping order different from start-to-end stopping at each element [which I believe is the only order Iterator itself can do]):
*Update: See comments below that clarify the analogous is also achievable with the traditional-for-loop.
final List<Integer> list = new ArrayList<>();
for(int i = 0; i < 10; ++i){
list.add(i);
}
int i = 1;
while(i < list.size()){
if(list.get(i) % 2 == 0){
list.remove(i++);
} else {
i += 2;
}
}
No ConcurrentModificationException from that code.
There we see looping not start at the beginning, and not stop at every element (which I believe Iterator itself can't do).
FWIW we also see get being called on list, which could not be done if its reference was just Collection (instead of the more specific List-type of Collection) - List interface includes get, but Collection interface does not. If not for that difference, then the list reference could instead be a Collection [and therefore technically this Answer would then be a direct Answer, instead of a tangential Answer].
FWIWW same code still works after modified to start at beginning at stop at every element (just like Iterator order):
final List<Integer> list = new ArrayList<>();
for(int i = 0; i < 10; ++i){
list.add(i);
}
int i = 0;
while(i < list.size()){
if(list.get(i) % 2 == 0){
list.remove(i);
} else {
++i;
}
}
One solution could be to rotate the list and remove the first element to avoid the ConcurrentModificationException or IndexOutOfBoundsException
int n = list.size();
for(int j=0;j<n;j++){
//you can also put a condition before remove
list.remove(0);
Collections.rotate(list, 1);
}
Collections.rotate(list, -1);
Try this one (removes all elements in the list that equal i):
for (Object i : l) {
if (condition(i)) {
l = (l.stream().filter((a) -> a != i)).collect(Collectors.toList());
}
}
You can use a while loop.
Iterator<Map.Entry<String, String>> iterator = map.entrySet().iterator();
while(iterator.hasNext()){
Map.Entry<String, String> entry = iterator.next();
if(entry.getKey().equals("test")) {
iterator.remove();
}
}
I ended up with this ConcurrentModificationException, while iterating the list using stream().map() method. However the for(:) did not throw the exception while iterating and modifying the the list.
Here is code snippet , if its of help to anyone:
here I'm iterating on a ArrayList<BuildEntity> , and modifying it using the list.remove(obj)
for(BuildEntity build : uniqueBuildEntities){
if(build!=null){
if(isBuildCrashedWithErrors(build)){
log.info("The following build crashed with errors , will not be persisted -> \n{}"
,build.getBuildUrl());
uniqueBuildEntities.remove(build);
if (uniqueBuildEntities.isEmpty()) return EMPTY_LIST;
}
}
}
if(uniqueBuildEntities.size()>0) {
dbEntries.addAll(uniqueBuildEntities);
}
If using HashMap, in newer versions of Java (8+) you can select each of 3 options:
public class UserProfileEntity {
private String Code;
private String mobileNumber;
private LocalDateTime inputDT;
// getters and setters here
}
HashMap<String, UserProfileEntity> upMap = new HashMap<>();
// remove by value
upMap.values().removeIf(value -> !value.getCode().contains("0005"));
// remove by key
upMap.keySet().removeIf(key -> key.contentEquals("testUser"));
// remove by entry / key + value
upMap.entrySet().removeIf(entry -> (entry.getKey().endsWith("admin") || entry.getValue().getInputDT().isBefore(LocalDateTime.now().minusMinutes(3)));
The best way (recommended) is use of java.util.concurrent package. By
using this package you can easily avoid this exception. Refer
Modified Code:
public static void main(String[] args) {
Collection<Integer> l = new CopyOnWriteArrayList<Integer>();
for (int i=0; i < 10; ++i) {
l.add(new Integer(4));
l.add(new Integer(5));
l.add(new Integer(6));
}
for (Integer i : l) {
if (i.intValue() == 5) {
l.remove(i);
}
}
System.out.println(l);
}
Iterators are not always helpful when another thread also modifies the collection. I had tried many ways but then realized traversing the collection manually is much safer (backward for removal):
for (i in myList.size-1 downTo 0) {
myList.getOrNull(i)?.also {
if (it == 5)
myList.remove(it)
}
}
In case ArrayList:remove(int index)- if(index is last element's position) it avoids without System.arraycopy() and takes not time for this.
arraycopy time increases if(index decreases), by the way elements of list also decreases!
the best effective remove way is- removing its elements in descending order:
while(list.size()>0)list.remove(list.size()-1);//takes O(1)
while(list.size()>0)list.remove(0);//takes O(factorial(n))
//region prepare data
ArrayList<Integer> ints = new ArrayList<Integer>();
ArrayList<Integer> toRemove = new ArrayList<Integer>();
Random rdm = new Random();
long millis;
for (int i = 0; i < 100000; i++) {
Integer integer = rdm.nextInt();
ints.add(integer);
}
ArrayList<Integer> intsForIndex = new ArrayList<Integer>(ints);
ArrayList<Integer> intsDescIndex = new ArrayList<Integer>(ints);
ArrayList<Integer> intsIterator = new ArrayList<Integer>(ints);
//endregion
// region for index
millis = System.currentTimeMillis();
for (int i = 0; i < intsForIndex.size(); i++)
if (intsForIndex.get(i) % 2 == 0) intsForIndex.remove(i--);
System.out.println(System.currentTimeMillis() - millis);
// endregion
// region for index desc
millis = System.currentTimeMillis();
for (int i = intsDescIndex.size() - 1; i >= 0; i--)
if (intsDescIndex.get(i) % 2 == 0) intsDescIndex.remove(i);
System.out.println(System.currentTimeMillis() - millis);
//endregion
// region iterator
millis = System.currentTimeMillis();
for (Iterator<Integer> iterator = intsIterator.iterator(); iterator.hasNext(); )
if (iterator.next() % 2 == 0) iterator.remove();
System.out.println(System.currentTimeMillis() - millis);
//endregion
for index loop: 1090 msec
for desc index: 519 msec---the best
for iterator: 1043 msec
you can also use Recursion
Recursion in java is a process in which a method calls itself continuously. A method in java that calls itself is called recursive method.

Iterating over basic “for” loop using Handlebars.js

I’m new to Handlebars.js and just started using it. Most of the examples are based on iterating over an object. I wanted to know how to use handlebars in basic for loop.
Example.
for(i=0 ; i<100 ; i++) {
create li's with i as the value
}
How can this be achieved?
There's nothing in Handlebars for this but you can add your own helpers easily enough.
If you just wanted to do something n times then:
Handlebars.registerHelper('times', function(n, block) {
var accum = '';
for(var i = 0; i < n; ++i)
accum += block.fn(i);
return accum;
});
and
{{#times 10}}
<span>{{this}}</span>
{{/times}}
If you wanted a whole for(;;) loop, then something like this:
Handlebars.registerHelper('for', function(from, to, incr, block) {
var accum = '';
for(var i = from; i < to; i += incr)
accum += block.fn(i);
return accum;
});
and
{{#for 0 10 2}}
<span>{{this}}</span>
{{/for}}
Demo: http://jsfiddle.net/ambiguous/WNbrL/
Top answer here is good, if you want to use last / first / index though you could use the following
Handlebars.registerHelper('times', function(n, block) {
var accum = '';
for(var i = 0; i < n; ++i) {
block.data.index = i;
block.data.first = i === 0;
block.data.last = i === (n - 1);
accum += block.fn(this);
}
return accum;
});
and
{{#times 10}}
<span> {{#first}} {{#index}} {{#last}}</span>
{{/times}}
If you like CoffeeScript
Handlebars.registerHelper "times", (n, block) ->
(block.fn(i) for i in [0...n]).join("")
and
{{#times 10}}
<span>{{this}}</span>
{{/times}}
This snippet will take care of else block in case n comes as dynamic value, and provide #index optional context variable, it will keep the outer context of the execution as well.
/*
* Repeat given markup with given times
* provides #index for the repeated iteraction
*/
Handlebars.registerHelper("repeat", function (times, opts) {
var out = "";
var i;
var data = {};
if ( times ) {
for ( i = 0; i < times; i += 1 ) {
data.index = i;
out += opts.fn(this, {
data: data
});
}
} else {
out = opts.inverse(this);
}
return out;
});
Couple of years late, but there's now each available in Handlebars which allows you to iterate pretty easily over an array of items.
https://handlebarsjs.com/guide/builtin-helpers.html#each

Multi-dimensional Arrays in AS3

I am currently playing around with flex, I have C++ background, so I am not used to AS3.
The problem is in the main *.mxml file I have fx:script block and I try to define a multidimensional array like that:
public var Board:Array = new Array(25);
I use a function to initialize the 2d-array:
public function initBoard():void {
var i:int;
var j:int;
for (i = 0; i < 25; i++) {
Board[i] = new Array(40);
for (j = 0; i < 40; j++) {
Board[i][j] = 0;
}
}
}
This function gets called later on in the main loop to init and reset the "board" why doesn't it work. The only difference to the AS3 documentation is that it gets done in a function. Is there a scope problem?
Thanking you in anticipation,
Niklas Voss
P.S. I hope someone can tell me why it doesn't work and how to do it right...
You have i where there should be j.
for (j = 0; i < 40; j++) {
This should solve your problems.
for (j = 0; j < 40; j++) {
You don't need to define an array length in AS3 - I just use the [] operator for creating a new array. Also you used i where j was needed in the innermost for loop.
function initBoard():Array
{
var board:Array = [];
var i:int = 0;
var j:int;
for(i; i<25; i++)
{
board[i] = [];
j = 0;
for(j; j<40; j++)
{
board[i][j] = 0;
}
}
return board;
}
trace(initBoard());

More efficient way to remove an element from an array in Actionscript 3

I have an array of objects. Each object has a property called name. I want to efficiently remove an object with a particular name from the array. Is this the BEST way?
private function RemoveSpoke(Name:String):void {
var Temp:Array=new Array;
for each (var S:Object in Spokes) {
if (S.Name!=Name) {
Temp.push(S);
}
}
Spokes=Temp;
}
If you are willing to spend some memory on a lookup table this will be pretty fast:
private function remove( data:Array, objectTable:Object, name:String):void {
var index:int = data.indexOf( objectTable[name] );
objectTable[name] = null;
data.splice( index, 1 );
}
The test for this looks like this:
private function test():void{
var lookup:Object = {};
var Spokes:Array = [];
for ( var i:int = 0; i < 1000; i++ )
{
var obj:Object = { name: (Math.random()*0xffffff).toString(16), someOtherProperty:"blah" };
if ( lookup[ obj.name ] == null )
{
lookup[ obj.name ] = obj;
Spokes.push( obj );
}
}
var t:int = getTimer();
for ( var i:int = 0; i < 500; i++ )
{
var test:Object = Spokes[int(Math.random()*Spokes.length)];
remove(Spokes,lookup,test.name)
}
trace( getTimer() - t );
}
myArray.splice(myArray.indexOf(myInstance), 1);
The fastest way will be this:
function remove(array: Array, name: String): void {
var n: int = array.length
while(--n > -1) {
if(name == array[n].name) {
array.splice(n, 1)
return
}
}
}
remove([{name: "hi"}], "hi")
You can also remove the return statement if you want to get rid of all alements that match the given predicate.
I don't have data to back it up but my guess is that array.filter might be the fastest.
In general you should prefer the old for-loop over "for each" and "for each in" and use Vector if your elements are of the same type. If performance is really important you should consider using a linked list.
Check out Grant Skinners slides http://gskinner.com/talks/quick/ and Jackson Dunstan's Blog for more infos about optimization.
If you don't mind using the ArrayCollection, which is a wrapper for the Array class, you could do something like this:
private function RemoveSpoke(Name:String, Spokes:Array):Array{
var ac:ArrayCollection = new ArrayCollection(Spokes);
for (var i:int=0, imax:int=ac.length; i<imax; i++) {
if (Spokes[i].hasOwnProperty("Name") && Spokes[i].Name === Name) {
ac.removeItemAt(i);
return ac.source;
}
}
return ac.source;
}
You could also use ArrayCollection with a filterFunction to get a view into the same Array object
Perhaps this technique (optimized splice method by CJ's) will further improve the one proposed by Quasimondo:
http://cjcat.blogspot.com/2010/05/stardust-v11-with-fast-array-splicing_21.html
Here's an efficient function in terms of reusability, allowing you to do more than remove the element. It returns the index, or -1 if not found.
function searchByProp(arr:Array, prop:String, value:Object): int
{
var item:Object;
var n: int = arr.length;
for(var i:int=n;i>0;i--)
{
item = arr[i-1];
if(item.hasOwnProperty(prop))
if( value == item[prop] )
return i-1;
}
return -1;
}

How to create an object from 2 arrays?

So I hava array Links and array Params with same langth N
So what I need is to create an object where for each link from Links I will be able to see
param from Params
And than for example to be abble to call something like
for each( item in object)
if (item.param == "some value") {
// some code
} else...
How to do such thing (Code exaMple, please)
From Array: You could first build a list with elements composed of a item and a param (supposing the length is indeed the same for both lists)
var items:Array = new Array();
for(var i:uint = 0; i < links.length; i++) {
links:Array .push({link:links[i], param:params[i]});
}
You can then filter them easily:
items.forEach(checkValue);
for(var i:uint = 0; i < items.length; i++) {
if (items[i].param == "some value") {
// some code
} else{
...
}
}

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