If have a workflow that removes elements of a List by a certain criteria. However certain items are skipped? Why is this happening?
List<Integer> listWithAge = new ArrayList<>();
int randomNumber = 100;
for (int i = 0; i < randomNumber; i++) {
listWithAge.add(i);
}
// this is my loop
for (int i = 0; i < listWithAge.size(); i++) {
System.out.println(i);
if ((listWithAge.get(i) % 3) == 2) listWithAge.remove(i);
}
Above code is my loop. I replaced my condition with something simpler. If I run this code my second loop only runs for 67 turns instead of 100.
It is problematic to iterate over a list and remove elements while iterating over it.
If you think about how the computer has to reconcile it, it makes sense...
Here's a thought experiment for you to go through.
If you have a list that is size 10 and you want to remove elements 1, 5, and 9 then you would think maybe the following would work:
List<String> listOfThings = ...some list with 10 things in it...;
list.remove(0);
list.remove(4);
list.remove(8);
However, after the first remove command, the list is only size 9.. Then after the second command, it's size has become 8. At this point, it hardly even makes sense to do list.remove(8) anymore because you're looking at an 8-element list and the largest index is 7.
You can also see now that the 2nd command didn't even remove the element now that you wanted.
If you want to keep this style of "remove as I go" syntax, the more appropriate way is to use Iterators. Here's an SO that talks about it and shows you the syntax you would need (see the question). It's easy to read up on elsewhere too.
How Iterator's remove method actually remove an object
Skipping a value would be the result of your list getting out of sync with your loop index because the list is reduced in size. This causes you to hop over some locations since the reduction in size affects future locations that have not been reached.
So the first thing you could do is simply correct the synchronization by decrementing i when you remove a value from the list. This will keep index at the same spot as the list shifts "left" caused by the removal.
for (int i = 0; i < listWithAge.size(); i++) {
if ((listWithAge.get(i) % 3) == 2) listWithAge.remove(i--);
}
The other option is to loop thru the list backwards.
for (int i = listWithAge.size()-1; i >= 0; i--) {
if ((listWithAge.get(i) % 3) == 2) {
listWithAge.remove(i);
}
}
This way, no values should be skipped since the removing of the element does affect the loop index's future positions relative to the changing size of the list.
But the best way would be to use an iterator as has already been mentioned by
Atmas
As a side note, I recommend you always use blocks {} even for single statements as I did above in the if block. It will save you some serious debugging time in the future when you decide you need to add additional statements and then wonder why things are no longer working.
And deleting like this from a list is very expensive, especially for large lists. I would suggest that if you don't have duplicate values, you use a Set. Otherwise, instead of deleting matching values, add the non-matching to a second list.
List<Integer> listWithAge = new ArrayList<>();
int randomNumber = 100;
for (int i = 0; i < randomNumber; i++) {
listWithAge.add(i);
}
// this is my loop
List<Integer> itemsToBeDeleted = new ArrayList<>();
for (int i = 0; i < listWithAge.size(); i++) {
System.out.println(i);
if ((listWithAge.get(i) % 3) == 2) {
itemsToBeDeleted.add(i);
}
//delete all outside the loop
//deleting inside the loop messes the indexing of the array
listWithAge.removeAll(itemsToBeDeleted);
I'm trying to iterate through an ArrayList and print each element in it but it only outputs the first element in the ArrayList. Then there's an infinite loop and it keeps printing out the first element.
ArrayList<String> startTime = new ArrayList<String>();
for (int i = 0; i < startTime.size(); i++) {
String getTime = startTime.get(i);
getTime = convertTime(getTime);
startTime.add(i, getTime);
System.out.println(startTime.get(i));
}
When you do startTime.add(i, getTime) you are adding an element in the i position. That means in your next iteration, you are going to have an extra element in your list. When you increment the counter and check startTime.size(), it's always going to be true. Hence your infinite loop.
As a suggestion, if you want to add your getTime element, you might want to use some sort of auxiliary structure, like another List.
Definitly use advanced for-loops:
ArrayList<String> startTime = new ArrayList<String>();
for(String aStartTime: startTime){
// do something
}
Try using this for-loop:
for(int i=0,n=startTime.size();i<n;i++){
//your for-loop code here
}
When you were adding elements to startTime, it was increasing its size. Therefore, it upped the boundary that i had to meet infinitely. In my example, n will be set to startTime's size at the beginning of the loop and won't change as the loop executes.
You can try doing the loop backward (from the last element to the first element ) such as:
// assume that I have ArrayList variable named *list*
for(int i = list.size() - 1 ; i >= 0 ; i--){
list.add(//something);
}
this code won't give you an infinite loop since the loop control variable never going to change ( i >= 0 ) even though your list size keeps changing. The lower bound will always be 0 and the i will decrease towards 0
I am trying to sort a LinkedList in Java. I need to go through mylist from back to front. The elements in the list are objects from my class CustomElement. If they match a certain pattern I want to put them up front.
My problem is that if I detect that the element in my list with index 5 for example matches my pattern and I move it to index 0, the previous element with index 4 has index 5 now, right? That is why I want the for loop to check the element with index 5 again: i++. But that's causing an infinite loop, whreas the method's working fine without i++, but not the way that I want it, because it's skipping the element with index 4 (now 5).
Is it gernerally possible to raise the variable i inside the for loop? And if yes, what am I doing wrong.
for (int i = mylist.size() - 1; i >= 0; i--) {
if (mylist.get(i) matches a certain pattern) {
CustomElement helper = mylist.get(i);
mylist.remove(i);
mylist.add(0, helper);
i++;
}
}
Yes, it is possible to modify i inside your for loop, if it weren't possible, you wouldn't be getting this infinite loop.
What must be happening, is that if (mylist.get(i) matches a certain pattern) continues to be true after a certain point, and you never get to a point where i >= 0 is not true.
So, if myList.get(0) matches your pattern, you'll just put it back at index 0, and keep checking it forever.
It is, but in your case using get(i) for a linked list will give quadratic performance.
If you don't mind your "matching" items being reversed in order then you'd be better creating a new list:
final LinkedList<CustomElement> newList = new LinkedList<> ();
for (final CustomElement e: myList)
{
if (e matches your pattern) { newList.addFirst (e); }
else { newList.addLast (e); }
}
myList = newList;
Then all problems with index variables disappear...
(You could also achieve linear performance whilst modifying the existing list, but it's a little more complicated.)
I'm receiving two object's List and I should compare all the objects...
For this I've been using the next loop:
for (int i=0;it2.hasNext();i++ ) {
objetoDb4o=it2.next();
tratdb4o=(Tratam)objetoDb4o;
for (int j=0;it.hasNext();j++ ) {
objetoJson = it.next();
tratJson = (Tratam)objetoJson;
}
}
it and it2 are iterators.. The values of size are correct (4 and 4)
But the loop doesn't work.. the first for iterates only one time, and the second for iterates three times.
I can not find the mistake! Can you help me please??
You need to reset the iterator it before you enter the inner loop, otherwise you'll go through the second list only once during the first iteration of the outer loop; the remaining iterations will stop at it.hasNext(), skipping the inner loop altogether.
I got a weird problem.
I thought this would cost me few minutes, but I am struggling for few hours now...
Here is what I got:
for (int i = 0; i < size; i++){
if (data.get(i).getCaption().contains("_Hardi")){
data.remove(i);
}
}
The data is the ArrayList.
In the ArrayList I got some strings (total 14 or so), and 9 of them, got the name _Hardi in it.
And with the code above I want to remove them.
If I replace data.remove(i); with a System.out.println then it prints out something 9 times, what is good, because _Hardi is in the ArrayList 9 times.
But when I use data.remove(i); then it doesn't remove all 9, but only a few.
I did some tests and I also saw this:
When I rename the Strings to:
Hardi1
Hardi2
Hardi3
Hardi4
Hardi5
Hardi6
Then it removes only the on-even numbers (1, 3, 5 and so on).
He is skipping 1 all the time, but can't figure out why.
How to fix this? Or maybe another way to remove them?
The Problem here is you are iterating from 0 to size and inside the loop you are deleting items. Deleting the items will reduce the size of the list which will fail when you try to access the indexes which are greater than the effective size(the size after the deleted items).
There are two approaches to do this.
Delete using iterator if you do not want to deal with index.
for (Iterator<Object> it = data.iterator(); it.hasNext();) {
if (it.next().getCaption().contains("_Hardi")) {
it.remove();
}
}
Else, delete from the end.
for (int i = size-1; i >= 0; i--){
if (data.get(i).getCaption().contains("_Hardi")){
data.remove(i);
}
}
You shouldn't remove items from a List while you iterate over it. Instead, use Iterator.remove() like:
for (Iterator<Object> it = list.iterator(); it.hasNext();) {
if ( condition is true ) {
it.remove();
}
}
Every time you remove an item, you are changing the index of the one in front of it (so when you delete list[1], list[2] becomes list[1], hence the skip.
Here's a really easy way around it: (count down instead of up)
for(int i = list.size() - 1; i>=0; i--)
{
if(condition...)
list.remove(i);
}
Its because when you remove an element from a list, the list's elements move up. So if you remove first element ie at index 0 the element at index 1 will be shifted to index 0 but your loop counter will keep increasing in every iteration. so instead you of getting the updated 0th index element you get 1st index element. So just decrease the counter by one everytime you remove an element from your list.
You can use the below code to make it work fine :
for (int i = 0; i < data.size(); i++){
if (data.get(i).getCaption().contains("_Hardi")){
data.remove(i);
i--;
}
}
It makes perfect sense if you think it through. Say you have a list [A, B, C]. The first pass through the loop, i == 0. You see element A and then remove it, so the list is now [B, C], with element 0 being B. Now you increment i at the end of the loop, so you're looking at list[1] which is C.
One solution is to decrement i whenever you remove an item, so that it "canceles out" the subsequent increment. A better solution, as matt b points out above, is to use an Iterator<T> which has a built-in remove() function.
Speaking generally, it's a good idea, when facing a problem like this, to bring out a piece of paper and pretend you're the computer -- go through each step of the loop, writing down all of the variables as you go. That would have made the "skipping" clear.
I don't understand why this solution is the best for most of the people.
for (Iterator<Object> it = data.iterator(); it.hasNext();) {
if (it.next().getCaption().contains("_Hardi")) {
it.remove();
}
}
Third argument is empty, because have been moved to next line. Moreover it.next() not only increment loop's variable but also is using to get data. For me use for loop is misleading. Why you don't using while?
Iterator<Object> it = data.iterator();
while (it.hasNext()) {
Object obj = it.next();
if (obj.getCaption().contains("_Hardi")) {
it.remove();
}
}
Because your index isn't good anymore once you delete a value
Moreover you won't be able to go to size since if you remove one element, the size as changed.
You may use an iterator to achieve that.
for (Iterator<Object> it = data.iterator(); it.hasNext();) {
if ( it.getCaption().contains("_Hardi")) {
it.remove(); // performance is low O(n)
}
}
If your remove operation is required much on list. Its better you use LinkedList which gives better performance Big O(1) (roughly).
Where in ArrayList performance is O(n) (roughly) . So impact is very high on remove operation.
It is late but it might work for someone.
Iterator<YourObject> itr = yourList.iterator();
// remove the objects from list
while (itr.hasNext())
{
YourObject object = itr.next();
if (Your Statement) // id == 0
{
itr.remove();
}
}
In addition to the existing answers, you can use a regular while loop with a conditional increment:
int i = 0;
while (i < data.size()) {
if (data.get(i).getCaption().contains("_Hardi"))
data.remove(i);
else i++;
}
Note that data.size() must be called every time in the loop condition, otherwise you'll end up with an IndexOutOfBoundsException, since every item removed alters your list's original size.
This happens because by deleting the elements you modify the index of an ArrayList.
import java.util.ArrayList;
public class IteratorSample {
public static void main(String[] args) {
// TODO Auto-generated method stub
ArrayList<Integer> al = new ArrayList<Integer>();
al.add(1);
al.add(2);
al.add(3);
al.add(4);
System.out.println("before removal!!");
displayList(al);
for(int i = al.size()-1; i >= 0; i--){
if(al.get(i)==4){
al.remove(i);
}
}
System.out.println("after removal!!");
displayList(al);
}
private static void displayList(ArrayList<Integer> al) {
for(int a:al){
System.out.println(a);
}
}
}
output:
before removal!!
1
2
3
4
after removal!!
1
2
3
There is an easier way to solve this problem without creating a new iterator object. Here is the concept. Suppose your arrayList contains a list of names:
names = [James, Marshall, Susie, Audrey, Matt, Carl];
To remove everything from Susie forward, simply get the index of Susie and assign it to a new variable:
int location = names.indexOf(Susie);//index equals 2
Now that you have the index, tell java to count the number of times you want to remove values from the arrayList:
for (int i = 0; i < 3; i++) { //remove Susie through Carl
names.remove(names.get(location));//remove the value at index 2
}
Every time the loop value runs, the arrayList is reduced in length. Since you have set an index value and are counting the number of times to remove values, you're all set. Here is an example of output after each pass through:
[2]
names = [James, Marshall, Susie, Audrey, Matt, Carl];//first pass to get index and i = 0
[2]
names = [James, Marshall, Audrey, Matt, Carl];//after first pass arrayList decreased and Audrey is now at index 2 and i = 1
[2]
names = [James, Marshall, Matt, Carl];//Matt is now at index 2 and i = 2
[2]
names = [James, Marshall, Carl];//Carl is now at index 3 and i = 3
names = [James, Marshall,]; //for loop ends
Here is a snippet of what your final method may look like:
public void remove_user(String name) {
int location = names.indexOf(name); //assign the int value of name to location
if (names.remove(name)==true) {
for (int i = 0; i < 7; i++) {
names.remove(names.get(location));
}//end if
print(name + " is no longer in the Group.");
}//end method
This is a common problem while using Arraylists and it happens due to the fact that the length (size) of an Arraylist can change. While deleting, the size changes too; so after the first iteration, your code goes haywire. Best advice is either to use Iterator or to loop from the back, I'll recommend the backword loop though because I think it's less complex and it still works fine with numerous elements:
//Let's decrement!
for(int i = size-1; i >= 0; i--){
if (data.get(i).getCaption().contains("_Hardi")){
data.remove(i);
}
}
Still your old code, only looped differently!
I hope this helps...
Merry coding!!!