I have an issue removing the 1st and 2nd element of my list even by using the iterator.
I have read the following threads but can't fix my issue (those were the most relevant but I checked other material as well):
ConcurrentModificationException when trying remove element from list
Iterating through a Collection, avoiding ConcurrentModificationException when removing objects in a loop
So my code looks like this:
List<List<String>> list = cnf.read();
List<List<String>> nlist = new ArrayList<>();
for (List<String> l : list) {
if (l.size() <= 3) {
nlist.add(l);
} else {
int size = l.size();
while (size > 3) {
List<String> three = l.subList(0, 2);
three.add("Y" + (count++));
//Iterator itr = l.iterator();
ListIterator itr = l.listIterator();
int v = 0;
while (itr.hasNext()) {
itr.next();
if (v == 0 || v == 1) {
itr.remove();
v++;
}
}
l.add(0, "Y" + (count++));
size--;
nlist.add(three);
}
nlist.add(l);
}
}
for (List<String> l : nlist) {
System.out.println(l.toString());
System.out.println(l.size());
}
I get a ConcurrentModificationException at the print statement here :
System.out.println(l.toString());
I tried using iterators for my 2 for loops as well but It doesn't seem to make a difference!
I am new to posting questions so let me know If I am doing it right!
Thank you.
After A long debugging, here is the solution.
The sublist function passes by reference and not by value, a sublist created by ArrayList.subList call keeps a reference to the original list and accesses its elementData array directly.
For this reason, when adding an element to the "three" list, we alter the state of the original list. this happens here:
three.add("Y" + (count++));
A way of fixing it for this specific case is to create and initialize the "three" list the following way:
String one = l.get(0);
String two = l.get(1);
List<String> three = new ArrayList<>();
three.add(one);
three.add(two);
three.add("Y" + (count));
This allows us to manipulate our lists without getting Concurrency Exceptions (ConcurrentModificationException). However, if you are manipulating big lists, I would suggest you use another less hardcoded method for list creation.
I will mark this thread as answered and hope it helps people.
Good afternoon everyone, I am currently studying for my Java Final and I have a review exercise that asks the reader to create a program that asks the user to input 10 integers and then to use a method to remove duplicates and display the distinct list. The method is provided for you as well.
I've gotten the majority of the code written, in fact I thought I was done until I realized that the for loop is removing more than just duplicates..
Here is my code:
public class lab25 {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner input = new Scanner(System.in);
int i;
//Create array list
ArrayList<Integer> numbers = new ArrayList<>();
System.out.println("Please enter 10 numbers!");
//Populate
for(i=0; i<10; i++) {
numbers.add(input.nextInt());
}
System.out.println("Your numbers are: " + numbers.toString());
removeDuplicate(numbers);
System.out.println("The distinct numbers are: " +numbers.toString());
input.close();
}
public static void removeDuplicate(ArrayList<Integer> list) {
int i;
for(i=0; i<list.size(); i++) {
if(list.contains(list.get(i))) {
list.remove(i);
}
}
}
}
Just curious what I have done wrong here? I think my issue might lie in my for loop.. Thanks to all who answer.
list.contains(list.get(i)) always returns true, since the i'th element of the List is contained in the List.
Therefore removeDuplicate is trying to remove all the elements (but you only remove half of them, since after removing the i'th element you skip the new i'th element).
There are many ways to remove duplicates. The most efficient involve using a HashSet. If you want to find duplicates using only List methods, you can check if list.lastIndexOf(list.get(i)) > i.
The expression list.contains(list.get(i)) is always true, since you're asking if the list contains some element from the list. You need to check if list.get(i) is contained in the first i-1 items in the list, which I recommend doing with a loop.
Be aware that a loop with list.remove will run slowly, since removing item i from an ArrayList is done by replacing item i with i+1, then replacing item i+1 with i+2 and so on. This means it takes around length^2 time to make a loop that calls remove in every iteration. The function list.contains has the same problem, as it has to go through the entire list. This may not matter if you have 10 items, but if you had a list with a million items, it would take a long time to run.
The easiest ways is to use Stream.distinct():
public static List<Integer> removeDuplicate(List<Integer> list) {
return list.stream().distinct().collect(Collectors.toList());
}
In case you are free to choose collection, you should use LinkedHashSet instead. It holds ordered unique numbers.
A solution could be this one. I startet at the end of the list that I don't delete indexes the loop has to visit in the future.
public static void removeDuplicate(ArrayList<Integer> list) {
int i = list.size() - 1;
while (i > -1) {
// check for duplicate
for (int j = 0; j < i; j++) {
if (list.get(i) == list.get(j)) {
// is duplicate: remove
list.remove(i);
break;
}
}
i--;
}
}
You are taking the list.get(i) which of course is present in the list, and you will delete all of the values in the end.
You could remove them by using a set:
Set<String> hs = new HashSet<>();
hs.addAll(numbers);
numbers.clear();
numbers.addAll(hs);
If you want to keep the current order and do not want to use set.
List<String> notduplicatedList =
new ArrayList<>(new LinkedHashSet<>(String));
I have a homework question that asks me to make a list iterator that starts at the end of the list. I have no idea how to do this. Here is the code
public class LinkedListTester7
{
public static void main(String [] args)
{
LinkedList<Integer> list = new LinkedList<Integer>();
for (int i = 0 ; i < 100 ; i = i + 2)
list.add(i) ;
//-----------Start below here. To do: approximate lines of code = 3
// 1. make a list iterator that starts at the end of the list ;
ListIterator<Integer> litr = list.listIterator();
//2. while hasPrevious ;
while (litr.hasPrevious()) {
//3. print what is returned by previous() followed by a blank without a newline
System.out.println(litr.previous()+ " ");
}
//-----------------End here. Please do not remove this comment. Reminder: no changes outside the todo regions.
System.out.println() ;
}
}
Read the java.util.List API documentation...
ListIterator<Integer> litr = list.listIterator(list.size());
Aactually a A ListIterator has no current element; its cursor position always lies between the element that would be returned by a call to previous() and the element that would be returned by a call to next().
If you want the first call to next()/previous() to return the last element you could use listIterator(int index) and pass it list.size() -1 if you want to iterate forward or list.size() if you want to iterate backward.
I think this might be the code you're looking for:
for (int i = litr.size(); i >= 0; i--) {
System.out.println(litr.get(i));
}
Basically it starts at the end of the array and prints the results from last to first.
This is what's starting at the END of the ArrayList:
i = litr.size();
i-- counts backwards through the array until it reaches 0.
This is a follow up to my previous question :
Collection - Iterator.remove() vs Collection.remove()
The below two pieces of code , which apparently differs only by a single line , but one throws exception and other don't . Can you please explain the difference ?
List<String> list = new ArrayList<String>
(Arrays.asList("noob1","noob2","noob3"));
System.out.println(list);
for (String str : list) {
if (str.equals("noob2")) {
list.remove(str);
}
}
runs fine , but if i change the condition to
if (!str.equals("noob2"))
the code throws exception !
What happens in this situation is you are removing the second list element.
List<String> list = new ArrayList<String>
(Arrays.asList("noob1", "noob2", "noob3", "noob4"));
System.out.println(list);
for (Iterator<String> iterator = list.iterator(); iterator.hasNext(); ) {
String str = iterator.next();
if (str.equals("noob3")) {
System.out.println("Checking "+str);
list.remove(str);
}
}
System.out.println(list);
prints
[noob1, noob2, noob3, noob4]
Checking noob1
Checking noob2
Checking noob3
[noob1, noob2, noob4]
By removing the second last element you have reduced the size to the number of elements which you have iterated over.
// from ArrayList.Itr
public boolean hasNext() {
return cursor != size;
}
This causes the loop to exit early before the concurrent modifcation check is performed in next(). If you remove any other element next() is called and you get a CME.
BTW Something which also bypasses the check is
for (Iterator<String> iterator = list.iterator(); iterator.hasNext(); ) {
String str = iterator.next();
System.out.println("Checking "+str);
if (str.equals("noob2")) {
list.remove("noob1");
list.remove("noob3");
}
}
as long as the size of the collection is the same as the index it is up to, the check is not performed.
The for loop is just a simplified syntax for an iterator scan of the list. The iterator may throw an exception if the list is modified under it, but it is not guaranteed. Because of hasNext, iterators are often working one element ahead, making the first case less likely to be affected by list modification. By the time "noob2" is removed, the iterator already knows about "noob3".
Actually you should never remove collections' elements during "casual" iterating. When you have to modify your collection in some loop you have to use iterator to make these operations.
public class Test {
public static void main(String... args) {
List<String> list = new ArrayList<String>(Arrays.asList("noob1", "noob2", "noob3"));
System.out.println(list);
for (Iterator<String> iterator = list.iterator(); iterator.hasNext();) {
String str = iterator.next();
if (!str.equals("noob2")) {
iterator.remove();
}
}
System.out.println(list);
}
}
I suppose the exception is thown because you are trying to change a collection you are looping on... and not because the if condition.
I suggest you to create a new list only containing the items that verify the condition. Add them to the new list and avoid to change the original collection.
It's because you are trying to remove from a Collection you are currently iterating through. Making a minor alteration you can do what you want to do:
String[] strValues = {"noob1","noob2","noob3"}; // <<< Array
List<String> list = new ArrayList<String>(Arrays.asList(strValues));
System.out.println(list);
for (String str : strValues) { // << List is duplicate of array so can iterate through array
if (!str.equals("noob2")) {
list.remove(str);
}
}
That should work. Hopefully
Well, your first case doesn't throw the Exception because, the iterator returns false for Iterator.hasNext() at index 2 as you remove the element at index 1.
Iterator<String> itr = list.iterator();
while(itr.hasNext()){
String s= itr.next();
if(s.equals("noob2")){
list.remove(s); // size of the list is 2 here
System.out.println(itr.hasNext());// this returns false as it doesn't have anything at index 2 now.(on 2nd iteration )
}
}
You can test it clearly using a simple for-loop:
for (int i=0; i<list.size(); i++) {
if (list.get(i).equals("noob2")) {
System.out.println(list.get(i));
System.out.println(list.size());
list.remove(list.get(i));
System.out.println(list.size());
}
}
Output:
[noob1, noob2, noob3]
noob2
3
2
Notice the size of the list after you remove the element, which fails after incrementing. 2<2 which is false
I have an ArrayList that I want to iterate over. While iterating over it I have to remove elements at the same time. Obviously this throws a java.util.ConcurrentModificationException.
What is the best practice to handle this problem? Should I clone the list first?
I remove the elements not in the loop itself but another part of the code.
My code looks like this:
public class Test() {
private ArrayList<A> abc = new ArrayList<A>();
public void doStuff() {
for (A a : abc)
a.doSomething();
}
public void removeA(A a) {
abc.remove(a);
}
}
a.doSomething might call Test.removeA();
Two options:
Create a list of values you wish to remove, adding to that list within the loop, then call originalList.removeAll(valuesToRemove) at the end
Use the remove() method on the iterator itself. Note that this means you can't use the enhanced for loop.
As an example of the second option, removing any strings with a length greater than 5 from a list:
List<String> list = new ArrayList<String>();
...
for (Iterator<String> iterator = list.iterator(); iterator.hasNext(); ) {
String value = iterator.next();
if (value.length() > 5) {
iterator.remove();
}
}
From the JavaDocs of the ArrayList
The iterators returned by this class's iterator and listIterator
methods are fail-fast: if the list is structurally modified at any
time after the iterator is created, in any way except through the
iterator's own remove or add methods, the iterator will throw a
ConcurrentModificationException.
You are trying to remove value from list in advanced "for loop", which is not possible, even if you apply any trick (which you did in your code).
Better way is to code iterator level as other advised here.
I wonder how people have not suggested traditional for loop approach.
for( int i = 0; i < lStringList.size(); i++ )
{
String lValue = lStringList.get( i );
if(lValue.equals("_Not_Required"))
{
lStringList.remove(lValue);
i--;
}
}
This works as well.
In Java 8 you can use the Collection Interface and do this by calling the removeIf method:
yourList.removeIf((A a) -> a.value == 2);
More information can be found here
You should really just iterate back the array in the traditional way
Every time you remove an element from the list, the elements after will be push forward. As long as you don't change elements other than the iterating one, the following code should work.
public class Test(){
private ArrayList<A> abc = new ArrayList<A>();
public void doStuff(){
for(int i = (abc.size() - 1); i >= 0; i--)
abc.get(i).doSomething();
}
public void removeA(A a){
abc.remove(a);
}
}
While iterating the list, if you want to remove the element is possible. Let see below my examples,
ArrayList<String> names = new ArrayList<String>();
names.add("abc");
names.add("def");
names.add("ghi");
names.add("xyz");
I have the above names of Array list. And i want to remove the "def" name from the above list,
for(String name : names){
if(name.equals("def")){
names.remove("def");
}
}
The above code throws the ConcurrentModificationException exception because you are modifying the list while iterating.
So, to remove the "def" name from Arraylist by doing this way,
Iterator<String> itr = names.iterator();
while(itr.hasNext()){
String name = itr.next();
if(name.equals("def")){
itr.remove();
}
}
The above code, through iterator we can remove the "def" name from the Arraylist and try to print the array, you would be see the below output.
Output : [abc, ghi, xyz]
Do the loop in the normal way, the java.util.ConcurrentModificationException is an error related to the elements that are accessed.
So try:
for(int i = 0; i < list.size(); i++){
lista.get(i).action();
}
Here is an example where I use a different list to add the objects for removal, then afterwards I use stream.foreach to remove elements from original list :
private ObservableList<CustomerTableEntry> customersTableViewItems = FXCollections.observableArrayList();
...
private void removeOutdatedRowsElementsFromCustomerView()
{
ObjectProperty<TimeStamp> currentTimestamp = new SimpleObjectProperty<>(TimeStamp.getCurrentTime());
long diff;
long diffSeconds;
List<Object> objectsToRemove = new ArrayList<>();
for(CustomerTableEntry item: customersTableViewItems) {
diff = currentTimestamp.getValue().getTime() - item.timestamp.getValue().getTime();
diffSeconds = diff / 1000 % 60;
if(diffSeconds > 10) {
// Element has been idle for too long, meaning no communication, hence remove it
System.out.printf("- Idle element [%s] - will be removed\n", item.getUserName());
objectsToRemove.add(item);
}
}
objectsToRemove.stream().forEach(o -> customersTableViewItems.remove(o));
}
One option is to modify the removeA method to this -
public void removeA(A a,Iterator<A> iterator) {
iterator.remove(a);
}
But this would mean your doSomething() should be able to pass the iterator to the remove method. Not a very good idea.
Can you do this in two step approach :
In the first loop when you iterate over the list , instead of removing the selected elements , mark them as to be deleted. For this , you may simply copy these elements ( shallow copy ) into another List.
Then , once your iteration is done , simply do a removeAll from the first list all elements in the second list.
In my case, the accepted answer is not working, It stops Exception but it causes some inconsistency in my List. The following solution is perfectly working for me.
List<String> list = new ArrayList<>();
List<String> itemsToRemove = new ArrayList<>();
for (String value: list) {
if (value.length() > 5) { // your condition
itemsToRemove.add(value);
}
}
list.removeAll(itemsToRemove);
In this code, I have added the items to remove, in another list and then used list.removeAll method to remove all required items.
Instead of using For each loop, use normal for loop. for example,the below code removes all the element in the array list without giving java.util.ConcurrentModificationException. You can modify the condition in the loop according to your use case.
for(int i=0; i<abc.size(); i++) {
e.remove(i);
}
Sometimes old school is best. Just go for a simple for loop but make sure you start at the end of the list otherwise as you remove items you will get out of sync with your index.
List<String> list = new ArrayList<>();
for (int i = list.size() - 1; i >= 0; i--) {
if ("removeMe".equals(list.get(i))) {
list.remove(i);
}
}
You can also use CopyOnWriteArrayList instead of an ArrayList. This is the latest recommended approach by from JDK 1.5 onwards.
Do somehting simple like this:
for (Object object: (ArrayList<String>) list.clone()) {
list.remove(object);
}
An alternative Java 8 solution using stream:
theList = theList.stream()
.filter(element -> !shouldBeRemoved(element))
.collect(Collectors.toList());
In Java 7 you can use Guava instead:
theList = FluentIterable.from(theList)
.filter(new Predicate<String>() {
#Override
public boolean apply(String element) {
return !shouldBeRemoved(element);
}
})
.toImmutableList();
Note, that the Guava example results in an immutable list which may or may not be what you want.
for (A a : new ArrayList<>(abc)) {
a.doSomething();
abc.remove(a);
}
"Should I clone the list first?"
That will be the easiest solution, remove from the clone, and copy the clone back after removal.
An example from my rummikub game:
SuppressWarnings("unchecked")
public void removeStones() {
ArrayList<Stone> clone = (ArrayList<Stone>) stones.clone();
// remove the stones moved to the table
for (Stone stone : stones) {
if (stone.isOnTable()) {
clone.remove(stone);
}
}
stones = (ArrayList<Stone>) clone.clone();
sortStones();
}
I arrive late I know but I answer this because I think this solution is simple and elegant:
List<String> listFixed = new ArrayList<String>();
List<String> dynamicList = new ArrayList<String>();
public void fillingList() {
listFixed.add("Andrea");
listFixed.add("Susana");
listFixed.add("Oscar");
listFixed.add("Valeria");
listFixed.add("Kathy");
listFixed.add("Laura");
listFixed.add("Ana");
listFixed.add("Becker");
listFixed.add("Abraham");
dynamicList.addAll(listFixed);
}
public void updatingListFixed() {
for (String newList : dynamicList) {
if (!listFixed.contains(newList)) {
listFixed.add(newList);
}
}
//this is for add elements if you want eraser also
String removeRegister="";
for (String fixedList : listFixed) {
if (!dynamicList.contains(fixedList)) {
removeResgister = fixedList;
}
}
fixedList.remove(removeRegister);
}
All this is for updating from one list to other and you can make all from just one list
and in method updating you check both list and can eraser or add elements betwen list.
This means both list always it same size
Use Iterator instead of Array List
Have a set be converted to iterator with type match
And move to the next element and remove
Iterator<Insured> itr = insuredSet.iterator();
while (itr.hasNext()) {
itr.next();
itr.remove();
}
Moving to the next is important here as it should take the index to remove element.
List<String> list1 = new ArrayList<>();
list1.addAll(OriginalList);
List<String> list2 = new ArrayList<>();
list2.addAll(OriginalList);
This is also an option.
If your goal is to remove all elements from the list, you can iterate over each item, and then call:
list.clear()
What about of
import java.util.Collections;
List<A> abc = Collections.synchronizedList(new ArrayList<>());
ERROR
There was a mistake when I added to the same list from where I took elements:
fun <T> MutableList<T>.mathList(_fun: (T) -> T): MutableList<T> {
for (i in this) {
this.add(_fun(i)) <--- ERROR
}
return this <--- ERROR
}
DECISION
Works great when adding to a new list:
fun <T> MutableList<T>.mathList(_fun: (T) -> T): MutableList<T> {
val newList = mutableListOf<T>() <--- DECISION
for (i in this) {
newList.add(_fun(i)) <--- DECISION
}
return newList <--- DECISION
}
Just add a break after your ArrayList.remove(A) statement