I am programming a game and almost have the save-file system complete. I have two Vectors (one holds the name of the savegame, one holds the sessionID).
At launch, the program will read in data from a file and add that information to the Vectors. Then another method is called to check if the files shown in the Vector acctualy exist. If not, they will be removed from the Vectors. At the end, the Vectors are printed to and rewrite the file.
The problem I'm having is the for loop isn't checking every item in the Vector, because Vector.size() is decreasing when items are removed.Is there a better way to form the for loop, or is there a workaround I can use?
private static void slistCleanup() throws IOException {
private static Vector<String> saveNames = new Vector<String>();
private static Vector<Integer> sessionIDs = new Vector<Integer>();
Scanner slistReader = new Scanner(new FileReader(sessionList));
File tempSave;
String path;
int run = 1;
String tempName = " ";
int tempID = 0;
for (int x = 0; x < saveNames.size(); x++) {
path = currentDir + "\\saves\\" + sessionIDs.elementAt(x) + ".sav";
tempSave = new File(path);
System.out.println("-----------------------"); //debug
System.out.println("current pass: " + run);
System.out.println("tempSave Path: " + tempSave.getAbsolutePath()); //debug
System.out.println("tempSave exists: " + tempSave.exists()); //debug
System.out.println("-----------------------"); //debug
run++; //debug
if (!tempSave.exists()) {
saveNames.remove(x);
sessionIDs.remove(x);
}
}
for (int x = 0; x < saveNames.size(); x++) {
System.out.println(saveNames.elementAt(x));
System.out.println(sessionIDs.elementAt(x));
}
slistReader.close();
}
If you need more code, let me know.
Loop backwards:
for (int x = saveNames.size()-1; x >= 0; x--)
One way that would require few changes to your existing code would be to traverse the vector in the reverse direction.
for (int x = saveNames.size() - 1; x >= 0; x--) {
...
}
As Fildor noted in the comments, you can do this with iterators
Iterator namesItr = saveNames.iterator();
Iterator sessionItr = sessionIDs.iterator();
while(namesItr.hasNext() && sessionItr.hasNext()) {
Object currentName = namesItr.next();
Object currentSession = sessionItr.next();
if (!tempSave.exists()) {
namesItr.remove();
sessionItr.remove();
}
}
Always loop backwards through an object when you are removing items from it, so:
for (int x = saveNames.size()-1; x >=0; x--) {
That way the removed items don't cause you a problem.
The reason for the problem is that you are starting at x=0; you delete x=0 (so x=1 is the new x=0, x=2 is the new x=1 etc) but you move on to x=1, skipping one.
On the other hand if you start at saveNames.size()-1:
You start at (for example) 9, delete it, 9 is now empty but we move on to 8 anyway. 8 is unaffected because its before 9
If you are having trouble because you're deleting items from the array as your looping over it, you can make a new array before the loop and put items you're keeping into it. After the loop is done, the new array will only contain items you're keeping, and you can set the original array to the new array you created.
You can just add the indexes you want to remove, and the going from last to first remove them. Or you can also use
x--;
inside your loop when you remove a element.
Best and simple way to remove items via loops in my opinion is:
scan the items using for loop and add the itemsToDelete list the items you want to delete.
list.removeAll(itemsToDelete)
The simplest thing to do is just take out x++.
Explanation: When you remove saveNames.remove(x) and x=0 then what was in index 1 moves to index 0. X is still 0 and it'll now remove the 2nd item which is now at index 0. Once all the items are removed saveNames.size() will be 0, x is no longer less than saveNames.size() so the loop will break.
Related
I have a small problem, I want to go through a list and compare two objects of the array. Each object has 3 elements, I use a StringTokenizer to be able to remove the separator, so each object has 3 elements. I would like to know how to make a method that gets the third element of each object and compare them. And if that element is less than another delete that element and the 2 before it.
I tried to make them with an iterator but I wouldn't know very well that it started from the 3 element and increased the position by 3.
Iterator<Integer> it = lisM.iterator();
int num;
while (it.hasNext()){
num = it.next();
System.out.println(num);
}
Is --> if, I was wrong to put it in the picture
This only answers part of your question. I could not understand the question completely, please edit it and I can edit my answer.
You should not remove items from a list whilst in a for loop, therefore you can, for example, create another boolean list with the same size divided by 3 and just fill it with true Booleans then set the position divided by 3 to false if you want to delete the three items. Then you can create a new list, iterate over the boolean list and add 3 "Objects" which are actually Strings (thanks #JB Nizet) at a time, every time the boolean list element is true. When it is false you just don't add the elements and by doing so you are practically deleting the two elements before that element together with that element.
You casted a String to an int, that does not work you have to parse the Strings.
I corrected some of your code and added the boolean list here:
ArrayList<String> lisM = new ArrayList<>(); // here I initialise the list as an array list with strings.
ArrayList<Boolean> booleanList = new ArrayList<>();
for (int i = 0; i < lisM.size() / 3; i++) {
booleanList.add(true);
}
for(int i = 3; i < lisM.size();i+=3) {
int m = Integer.parseInt(lisM.get(i)); // here I changed the casting to parsing and moved it out of the for loop, there is no need to initialize it again every single time since you do not change it in the second for loop.
for (int j = 6; j < lisM.size(); j += 6) {
int m1 = Integer.parseInt(lisM.get(j));// here I changed the casting to parsing again.
if (m > m1) { // this makes no sense here because you are going over all of the elements of the list and comparing them to all of them. But I kept it here for the sake of example.
booleanList.set(i/3,false);
}
// if you want to go over the whole list you will have to clear the list and start over again for every element.
}
}
and here is how you could create the new list without the elements you do not want:
ArrayList<String> newLisM = new ArrayList<>();
for (int i = 0; i <booleanList.size(); i++) {
if(booleanList.get(i))
for (int j = 0; j < 3; j++) {
newLisM.add(lisM.get(i+j));
}
}
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;
}
}
In java programming:
I have something along the lines of
ArrayList<String> list = new ArrayList<String>();
int listSize = list.size();
for(int x = 0; x< listSize; x++){
list.remove(x);
listSize = list.size();
}
will this work well enough? or will the for loop keep running. Thanks!
I'm a bit concerned at what you're attempting to do here. Any time someone wants to remove items from a collection in a loop without using an Iterator, alarms usually go off.
So, to answer the question as stated: If you change your loop invariant, then you'll get some funky behavior. This is why it's often called an "invariant"; you don't want to change it. You're certainly welcome to, but you're playing with fire here.
But, let's assume that we have a list of 5 elements in it and we execute your code as is.
List<String> stringList = Arrays.asList("word", "another word", "let's go", "keep it up", "get moving");
int length = stringList.size();
for(int i = 0; i < length; i++) {
System.out.println(stringList.remove(i));
length = stringList.size();
}
System.out.println("done");
If you're expecting it to print out every word in the list, then think again. It will skip every element because we are changing the content of the list while iterating over it. When deleting elements from an ArrayList, the underlying implementation will shift over elements to the left in where it was deleted. So, if you delete from position 0, every element now lives to the left of its original position (so element 1 is now element 0).
A quick run-through:
First iteration, i = 0; "word" is removed. Size is changed to 4, i increases to 1.
Second iteration, i = 1; "let's go" is removed. Size is changed to 3, i increases to 2.
Third iteration, i = 2; "get moving" is removed. SIze is changed to 2, i increases to 3.
No further iteration happens as the condition is not satisfiable.
If you want to cleanly remove all items from the collection without printing them out, then ArrayList#clear() is what you're looking for. If you want to remove elements one at a time, then use an Iterator instead:
List<String> stringList = new ArrayList<>(Arrays.asList("word", "another word", "let's go", "keep it up", "get moving"));
for(Iterator<String> stringIterator = stringList.iterator(); stringIterator.hasNext(); ) {
System.out.println(stringIterator.next());
stringIterator.remove();
}
System.out.println("done");
The loop won't keep running and it will end without any exception. But the result may not what you expect.
Besides "ArrayList#clear()" or Iterator way to remove all elements from one array. You can also try below codes, just remove "x++" from your loop condition.
ArrayList<String> list = new ArrayList<String>();
int listSize = list.size();
for(int x = 0; x< listSize;){
list.remove(x);
listSize = list.size();
}
Why not do it in a "while" loop?
ArrayList<String> list = new ArrayList<String>();
int listSize = list.size();
while(list.size() != 0){
list.remove(x);
}
Technically, the computer science course would likely tell you to do it something like this:
public void solve() {
ArrayList<String> list = new ArrayList<String>();
Iterator iter = list.iterator();
while(iter.hasNext()) {
System.out.println("Next value is: " + iter.next());
iter.remove();
}
}
I'm trying to write a simple game where an enemy chases the player on a grid. I'm using the simple algorithm for pathfinding from the Wikipedia page on pathfinding. This involves creating two lists with each list item containing 3 integers. Here's test code I'm trying out to build and display such a list.
When I run the following code, it prints out the same numbers for each array in the ArrayList. Why does it do this?
public class ListTest {
public static void main(String[] args) {
ArrayList<Integer[]> list = new ArrayList<Integer[]>();
Integer[] point = new Integer[3];
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 3; j++) {
point[j] = (int)(Math.random() * 10);
}
//Doesn't this line add filled Integer[] point to the
//end of ArrayList list?
list.add(point);
//Added this line to confirm that Integer[] point is actually
//being filled with 3 random ints.
System.out.println(point[0] + "," + point[1] + "," + point[2]);
}
System.out.println();
//My current understanding is that this section should step through
//ArrayList list and retrieve each Integer[] point added above. It runs, but only
//the values of the last Integer[] point from above are displayed 10 times.
Iterator it = list.iterator();
while (it.hasNext()) {
point = (Integer[])it.next();
for (int i = 0; i < 3; i++) {
System.out.print(point[i] + ",");
}
System.out.println();
}
}
}
First of all, several of the other answers are misleading and/or incorrect. Note that an array is an object. So you can use them as elements in a list, no matter whether the arrays themselves contain primitive types or object references.
Next, declaring a variable as List<int[]> list is preferred over declaring it as ArrayList<int[]>. This allows you to easily change the List to a LinkedList or some other implementation without breaking the rest of your code because it is guaranteed to use only methods available in the List interface. For more information, you should research "programming to the interface."
Now to answer your real question, which was only added as a comment. Let's look at a few lines of your code:
Integer[] point = new Integer[3];
This line creates an array of Integers, obviously.
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 3; j++) {
point[j] = (int)(Math.random() * 10);
}
//Doesn't this line add filled Integer[] point to the
//end of ArrayList list?
list.add(point);
//...
}
Here you assign values to the elements of the array and then add a reference to the array to your List. Each time the loop iterates, you assign new values to the same array and add another reference to the same array to the List. This means that the List has 10 references to the same array which has been repeatedly written over.
Iterator it = list.iterator();
while (it.hasNext()) {
point = (Integer[])it.next();
for (int i = 0; i < 3; i++) {
System.out.print(point[i] + ",");
}
System.out.println();
}
}
Now this loop prints out the same array 10 times. The values in the array are the last ones set at the end of the previous loop.
To fix the problem, you simply need to be sure to create 10 different arrays.
One last issue: If you declare it as Iterator<Integer[]> it (or Iterator<int[]> it), you do not need to cast the return value of it.next(). In fact this is preferred because it is type-safe.
Finally, I want to ask what the ints in each array represent? You might want to revisit your program design and create a class that holds these three ints, either as an array or as three member variables.
I would highly recommend to enclose the integer array of 3 numbers into a meaningful class, that would hold, display and control an array of 3 integers.
Then in your main, you can have an growing ArrayList of objects of that class.
You have an extra ) here:
element = (int[])it.next()); //with the extra parenthesis the code will not compile
should be:
element = (int[])it.next();
Besides the problem in the other answer, you cal it.next() two times, that cause the iterator move forward two times, obviously that's not what you want. The code like this:
element = (int[])it.next());
String el = (String)element;
But actually, I don't see you used el. Although it's legal, it seems meaningless.
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!!!