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);
for(int i = 0; i < bag.length; i++)
{
if(bag[i].equals(a))
{
tmp = bag[i];
bag[i] = bag[bag.length-1];
bag[bag.length-1] = tmp;
numElements--;
break;
}
}
The goal of this is to find an object in the array and then remove it? is it possible??
Changing the length of an array is not possible. Recall that array is a static data structure whose size is determined before hand. Increasing or decreasing is not supported in this data structure. The fact that one has to increase or decrease the size depending on the usecase means that they have picked up the wrong data structure. They should perhaps go with an ArrayList.
Anyway, coming back to your question, you can simulate the 'size decrease' by maintaining a variable which you let track the array index and decrease the size of this variable. This lets you give the impression of shrinking the array.
The code you have provided does the same. Note however, that you should be using this modified index to track the contents of your array.
for(int i = 0; i < bag.length; i++)
{
if(bag[i].equals(a))
{
tmp = bag[i];
bag[i] = bag[bag.length-1];
bag[bag.length-1] = tmp;
numElements--;
break;
}
}
Whenever a particular bag at a given index equals to the item under question i.e., 'a', we swap elements so that the current bag element to be removed moves to the last and also we reduce the size of our new index - numElements by 1 to simulate this.
If you have the full code with you, please consider adding the following snippet at the end of that program to understand this more:
// Simulation of the array shrinking.
for(int i = 0; i < numElements; i++)
{
System.out.println( bag[i] );
}
// Movement of uninteresting elements to the end of the array.
for(int i = 0; i < bag.length; i++)
{
System.out.println( bag[i] );
}
It's not possible to change the length of an array. You can overwrite the element you wish to remove with the last element of the array and then copy the first bag.length - 1 elements of your array to a new array whose length is bag.length - 1.
for(int i = 0; i < bag.length; i++) {
if(bag[i].equals(a)) {
bag[i] = bag[bag.length-1];
bag = Arrays.copyOf (bag, bag.length - 1);
break;
}
}
public static String[] removeElements(String[] input) {
List<String> result = new ArrayList<String>();
String deleteValue = "somevalue";
for(String item : input)
if(!deleteValue .equals(item))
result.add(item);
return result.toArray(input);
}
This is one method you can fit this into your program.
You cannot decrease the size of an array. okay no problem! you can create your own data structure which supports that right?
Now, create a class named say MyArray with functions like increaseLenght(int) and decreseLength(int). Try it if you want to, will be fun for sure..
You cannot reduce the size of an array. Arrays are fixed length. What you can do is have a variable that indicates how many entries of the array you are using. This is what you are doing with numElements. The standard class ArrayList is implemented like this. The data is kept in an array and a private field size is used. With an ArrayList, when you remove an element, all the elements to the right are shifted left. However I also like your idea.
I would suggest 2 changes.
Make the last element null instead. If you are removing the element, why does it still need to be in the array?
Use numElements - 1 rather than bag.length-1 as the array could be bigger.
With these changes it becomes:
for(int i = 0; i < bag.length; i++)
{
if(bag[i].equals(a))
{
bag[i] = bag[numElements-1];
bag[numElements-1] = null;
numElements--;
break;
}
}
I have an ArrayList of String, and I would like to retrieve the first and last result of the names after calculating the order of alphabets. Below is my code snippet:
ArrayList<String> list = new ArrayList<String>(20);
list.add("Charles Darwin");
list.add("Albert Einstein");
list.add("Issac Newton");
list.add("Tony Hoare");
list.add("Grace Hopper");
list.add("Edgar Dijkstra");
list.add("Ada Lovelace");
list.add("Charles Babbage");
list.add("Stephen Hawking");
String biggest = "";
String smallest = "";
for (int i = 0; i < list.size(); i++) {
String first = list.get(i);
for (int j = 0; j < list.size(); j++) {
String second = list.get(j);
if (!first.equalsIgnoreCase(second)) {
if (first.compareToIgnoreCase(second)>0){
biggest=first;
}
if (first.compareToIgnoreCase(second)<0){
smallest=first;
}
}
}
}
System.out.println(biggest);
System.out.println(smallest);
I am able to retrieve every value for comparison, however, the results are always showing Stephen Hawking as the biggest and smallest.
My desired results are Ada Lovelace as biggest and Tony Hoare as smallest.
You can use Collections.min / max
Your conditional statements seem to wrong.
if (first.compareToIgnoreCase(second)>0){
biggest=first;
}
if (first.compareToIgnoreCase(second)<0){
smallest=first;
}
You're comparing the element in the outer loop to the element in the inner loop. You never make a comparison against the biggest and smallest.
This should help you find the biggest and smallest String in your list.
String biggest = list.get(0);
String smallest = list.get(0);
for (int i = 1; i < list.size(); i++) {
if(list.get(i).compareToIgnoreCase(biggest) > 0)
biggest = list.get(i);
if(list.get(i).compareToIgnoreCase(smallest) < 0)
smallest = list.get(i);
}
Alternatively, you can use Collections.min() and max() as stated in one of the other answers.
are your required to use List ? You might want to see http://docs.oracle.com/javase/7/docs/api/java/util/SortedSet.html. then you can use first() and last() method
I'm new to Java, but I can still recognize multiple issues with this code:
Why are you hard coding initial capacity to 20?
Why are you using indexed loops instead of for each?
Why are you using a nested loop to find min/max?
Use else instead of running the same comparison twice
If first always equals second biggest and smallest will remain uninitiated
And last, and directly dressing your question, all your code does is finding if the last item is greater & smaller than any of the other items in the list, not all of them, since you keep ignoring previous findings and not using biggest/smallest as a condition in any of your comparisons.
I have a Collection of objects I need to iterate over. The collection is of variable size. If the collection has more than 1 object in it, I need to perform special processing on objects 2 .. infinity.
What's the preferred method to do this? For example:
int count = 1;
for (CustomObject co : CustomObjectCollection) {
methodAll(co);
if(count > 1) {
methodSpecial(co);
}
count = count++;
}
What you have will work except for one bug: count = count++ does absolutely nothing. count = count + 1 would work, or count++, but count = count++ is a no-op.
You could also just use a boolean flag if you don't specifically need to track the count.
boolean first = true;
for (CustomObject co : CustomObjectCollection) {
methodAll(co);
if (!first) {
methodSpecial(co);
}
first = false;
}
Which to use
Which to use depends on your specific use case. Assuming that you are not looking to optimise every last op-per-second of performance then go with the option that declares your intent:
count++ is fine if you want to track or use the count outside the loop (as per Louis' answer, it is count++, not count = count++)
for(int i = 0; i < collection.size(); i++) is good as well if the collection supports a get(i) operation. This also lets you skip the first item by initialising int i to a different index. It might be used if you don't want the extra count variable hanging around outside the loop.
the above boolean first = true; (or the inverse boolean notFirst = false;) highlight that you want to treat the first and subsequent elements differently
If you have the List interface on your collection and want to skip a set number of elements, then subList is a good option
Performance
If performance is a concern, then measure it for your platform and implementation, but from general experience, from slowest to fastest, with <=20x difference between the first and last:
for each loop with iterator: slowest
for(int i = 0; i < list.size() ; i++)
declare int size = list.size() then a for(int i = 0; i < size; i++) loop
However these speed results depend on so many things that unless performance is a design goal or an identified issue go with the iterator until you have a reason to use one of the other two - the iterator is generally fast enough.
I think a better way to do this would be to work directly with the iterator like so:
Iterator<CustomObject> it = customObjectCollection.iterator();
if(it.hasNext()) { //first pass don't call methodSpecial
methodAll(it.next());
}
CustomObject customObj;
while(it.hasNext()) { //all the rest 2..infinity
customObj = it.next();
methodAll(customObj);
methodSpecial(customObj);
}
customObj = null; // for garbage collection
This way you don't have to check each iteration if this is the first run or not.
This will work with any Iterable (which is already needed for a foreach loop anyway).
Note: If this is not an ordered collection you might get different elements in the first iteration then you might expect.
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!!!