Deleting while iterating in java! but deleting other then the current object - java

iterator.remove(). is good however my question: if I want while to iterate, for a kind of condition, to delete another object from the array List.
Example (al being the arraylist)
for (Iterator i = al.iterator(); i.hasNext(); ){
IEvent event =(IEvent) i.next();
if (nbSendingWTS > 0 || nbSendingCTS > 0){
i.remove();
al.remove(swtsee);
al.remove(sdctsee);
System.out.println("dropping evtg");
}
This is giving me an error: Exception in thread "main" java.util.ConcurrentModificationException
Also the normal iteration:
for(IEVEnt event:al){}
is giving an error
To be more clear the swtsee a d sdctsee are taken from previous iterations on the arraylist and saved so i can delete if I have the new condition. So is there a way when I detect them to shift them to higher indexes and then I use a reverse iteration?
What to do?

You can't remove element as discussed by you.
Do not delete while iterating.
Keep a Hash for all the objects you want to delete.
Do a second iteration which delete using .remove() if the object is in Hash.

You can't delete if you are using for each style or iterator.
Use normal for loop like following
for(int i=0; i<al.size ; i++){
if(something){
al.remove(i)
i--;
}
}
This will work.

To remove with an Iterator, you collect your stuff in a new Collection, and remove in a final step, for example:
// list := List (1, 2, 4, 3, 4, 9, 6, 5, 7, 8);
List <Integer> toRemove = new ArrayList <Integer> ();
for (int i : list)
if (i % 2 == 0)
toRemove.add (i);
list.removeAll (toRemove);
I can't see how a1 is connected to your i. As long as it isn't iterated over, it should be secure to call those 2 a1.remove (...)-ings while iterating.

For your reference Java Collections
your code should work fine commenting the following lines
for (Iterator i = al.iterator(); i.hasNext(); )
{
IEvent event =(IEvent) i.next();
if (nbSendingWTS > 0 || nbSendingCTS > 0)
{
// You have got the iterator for the underlying array list(al)
**only remove the elements through iterator.**
i.remove();
// after remove thru iterator
// you are structurally modifiying arraylist directly(al.remove())
// which gives u concurrent modification
// al.remove(swtsee);
// al.remove(sdctsee);
System.out.println("dropping evtg");
}
}
and the best way to do is
List<Integer> l = new ArrayList<Integer>();
List<Integer> itemsToRemove = new ArrayList<Integer>();
for (int i=0; i < 10; ++i) {
l.add(new Integer(1));
l.add(new Integer(2));
}
for (Integer i : l)
{
if (i.intValue() == 2)
itemsToRemove.add(i);
}
l.removeAll(itemsToRemove);
System.out.println(l);

Related

ConcurrentModificationException When removing element using list iterator java

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.

Concerns with listiterator remove java

I am trying to work the Josephus problem in an arrayList data structure and the task is to use ListIterator. I am getting stuck at the removing part.
Lets say I have following 10 soldiers - 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
And I have to step every 3rd soldier - which means remove 3, then 6, then 9, and so on.
I am able to successfully remove 3, but next time it removes 7 and not 6. Can you please review my code and tell me what I am doing wrong:
private static int suicide (List<Integer> list, int step) {
ListIterator<Integer> itr = list.listIterator();
while(itr.hasNext()&& list.size() != 1){
if((itr.nextIndex()+1) % step == 0) {
System.out.println(itr.previousIndex()+1);
itr.next();
itr.remove();
itr.next();
}
else {
itr.next();
}
if(!itr.hasNext() ){
itr=list.listIterator();
}
}
return 0;
}
One possibility is, that you make a copy of the list, and remove the items in that copy, still iterating the "original" (and skipping on iterator step in the copy).
ListIterator itr = list.listIterator();
ListIterator itr2 = copy.listIterator();
while(itr.hasNext()&& list.size() != 1){
if((itr.nextIndex()+1) % step == 0) {
System.out.println(itr.previousIndex()+1);
itr.next();
itr2.next();
itr2.remove();
itr.next();
}
else {
itr.next();
itr2.next();
}
Try this;
int i = 3;
int step = 1;
while (list.size() > i) {
list.remove(i);
step++;
i = 3 * step - step + 1;
}
Revert you design logic !
Building a new List with accepted elements only is a really best approach.
Removing is always harder to understand and to maintain.
For example you methode suicide should return a new List<> with the elements you want.
Your code will be really simpler :
List<Integer> cleanedList = new ArrayList<Integer>();
for(Integer soldier : list) {
if(solider%3!=0){
cleanedList.add(soldier);
}
}
return cleanedList;
The most of the time I prefer API that creates new objets. If you apply this rule you code will always get simpler code ! And updating parameters is not a good behavior : if you call the method twice, you obtains two differents results.

ConcurrentModification Exception while Iterating [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);
A simple test as example:
#Test
public void testRemoveIfOneList() {
List<String> outer = new ArrayList<>();
outer.add("one");
outer.add("two");
outer.add("three");
outer.removeIf(o -> o.length() == 3);
assertEquals(1, outer.size());
}
It even works when you compare two lists and want to remove from both.
#Test
public void testRemoveIfTwoLists() {
List<String> outer = new ArrayList<>();
outer.add("one");
outer.add("two");
outer.add("three");
List<String> inner = new ArrayList<>();
inner.addAll(outer);
// first, it removes from inner, and if anything is removed, then removeIf() returns true,
// leading to removing from outer
outer.removeIf(o -> inner.removeIf(i -> i.equals(o)));
assertEquals(0, outer.size());
assertEquals(0, inner.size());
}
However, if one of the list has duplicates, make sure it's iterated in the inner loop, because for inner list, it will remove all elements meeting the criteria, but for outer list, when any element is removed, it will return immediately and stops checking.
This test will fail:
#Test
public void testRemoveIfTwoListsInnerHasDuplicates() {
List<String> outer = new ArrayList<>();
outer.add("one");
outer.add("one");
outer.add("two");
outer.add("two");
outer.add("three");
outer.add("three");
List<String> inner = new ArrayList<>();
inner.addAll(outer); // both have duplicates
// remove all elements from inner(executed twice), then remove from outer
// but only once! if anything is removed, it will return immediately!!
outer.removeIf(o -> inner.removeIf(i -> i.equals(o)));
assertEquals(0, inner.size()); // pass, inner all removed
assertEquals(0, outer.size()); // will fail, outer has size = 3
}
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.

How to find sublist (intersection) among items in one Arraylist of integer in java?

I have a Hashtable<List<Integer>, List<Triples>> ResultIPM2 = new HashTable<>(), I want to work on the key part which is aList<Integer>contains three patterns {[2,3,4] , [3,2,4,3] , [2,4,3]}.What I want to do is, I want to remove [2,4,3] because it is a subset of [3,2,4,3]. How can I report just [2,4,3] and [3,2,4,3] .
Note that the intersection must be in the same order as the smaller list.
For example, if instead of [3,2,4,3] we had a pattern [3,2,3,4] we wouldn't remove any pattern. by "remove" I mean not showing the pattern when reporting it.Here is a piece of my code:
Please don't care about the other answers because I had to edit this old question in order to be able to ask a new question.Thank you
Set<List<Integer>> key = resultsIPM2.keySet();
boolean maximal = false;
for(List<Integer> p1 : key){
for(List<Integer> p2:key){
if(!p2.equals(p1)){
if(p1.containsAll(p2)){
System.out.println(p1);
}
}
}
}
You can't use the enhanced for loop to remove from the list. Use an explicit Iterator.
List<List<Pair>> newPattern = new CopyOnWriteArrayList<>();
//Algorithm : end of Line 14
Iterator<List<Pair> iter = ptList.iterator();
while(iter.hasNext()){
List<Pair> list = iter.next();
Iterator<Pair> pairIter = list.iterator();
while(pairIter.hasNext()){
Pair pair = pairIter.next();
if (getFrequency(pair) < minSupp){
pairIter.remove();
newPattern.add(list);
}
}
}
Iterator.remove() is the only safe way to modify a collection during iteration.
Use Iterator instead of the for-each construct when you need to remove the current element.
The for-each construct hides the iterator, so you cannot call remove. Therefore, the for-each construct is not usable for filtering.
Now regarding the code which you said dint work is because
Iterator <Pair> iter = pairList.iterator();
should be something like this
Iterator<List<Pair> iter = ptList.iterator();
then again when you do iter.next();
It will give you List<Pair> list not this Pair pair. So you need to iterate over the list again the compare and remove.
Eran gave you a very nice example follow that
You can not modify a List while you are in a foreach loop of this list. You can use a normal for-loop which just increments an int to get your element:
List<List<Pair>> newPattern = new CopyOnWriteArrayList<>();
//Algorithm : end of Line 14
for(int i = 0; i < ptList.size(); i++){
List<Pair> list = ptList.get(i);
for(int j = 0; j < list.size(); j++){
Pair pair = list.get(j);
if (getFrequency(pair) < minSupp){
list.remove(j);
j--;
newPattern.add(list);
}
}
}

Java - Collection.remove() behaves differently in different conditions

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

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