Why do I get an UnsupportedOperationException when trying to remove an element from a List?

JavaExceptionListArraylist

Java Problem Overview


I have this code:

public static String SelectRandomFromTemplate(String template,int count) {
   String[] split = template.split("|");
   List<String> list=Arrays.asList(split);
   Random r = new Random();
   while( list.size() > count ) {
      list.remove(r.nextInt(list.size()));
   }
   return StringUtils.join(list, ", ");
}

I get this:

06-03 15:05:29.614: ERROR/AndroidRuntime(7737): java.lang.UnsupportedOperationException
06-03 15:05:29.614: ERROR/AndroidRuntime(7737):     at java.util.AbstractList.remove(AbstractList.java:645)

How would be this the correct way? Java.15

Java Solutions


Solution 1 - Java

Quite a few problems with your code:

On Arrays.asList returning a fixed-size list

From the API:

> Arrays.asList: Returns a fixed-size list backed by the specified array.

You can't add to it; you can't remove from it. You can't structurally modify the List.

Fix

Create a LinkedList, which supports faster remove.

List<String> list = new LinkedList<String>(Arrays.asList(split));

On split taking regex

From the API:

> String.split(String regex): Splits this string around matches of the given regular expression.

| is a regex metacharacter; if you want to split on a literal |, you must escape it to \|, which as a Java string literal is "\\|".

Fix:
template.split("\\|")

On better algorithm

Instead of calling remove one at a time with random indices, it's better to generate enough random numbers in the range, and then traversing the List once with a listIterator(), calling remove() at appropriate indices. There are questions on stackoverflow on how to generate random but distinct numbers in a given range.

With this, your algorithm would be O(N).

Solution 2 - Java

This one has burned me many times. Arrays.asList creates an unmodifiable list. From the Javadoc: Returns a fixed-size list backed by the specified array.

Create a new list with the same content:

newList.addAll(Arrays.asList(newArray));

This will create a little extra garbage, but you will be able to mutate it.

Solution 3 - Java

Probably because you're working with unmodifiable wrapper.

Change this line:

List<String> list = Arrays.asList(split);

to this line:

List<String> list = new LinkedList<>(Arrays.asList(split));

Solution 4 - Java

The list returned by Arrays.asList() might be immutable. Could you try

List<String> list = new ArrayList<>(Arrays.asList(split));

Solution 5 - Java

I think that replacing:

List<String> list = Arrays.asList(split);

with

List<String> list = new ArrayList<String>(Arrays.asList(split));

resolves the problem.

Solution 6 - Java

Just read the JavaDoc for the asList method:

> Returns a {@code List} of the objects > in the specified array. The size of > the {@code List} cannot be modified, > i.e. adding and removing are > unsupported, but the elements can be > set. Setting an element modifies the > underlying array.

This is from Java 6 but it looks like it is the same for the android java.

EDIT

The type of the resulting list is Arrays.ArrayList, which is a private class inside Arrays.class. Practically speaking, it is nothing but a List-view on the array that you've passed with Arrays.asList. With a consequence: if you change the array, the list is changed too. And because an array is not resizeable, remove and add operation must be unsupported.

Solution 7 - Java

Arrays.asList() returns a list that doesn't allow operations affecting its size (note that this is not the same as "unmodifiable").

You could do new ArrayList<String>(Arrays.asList(split)); to create a real copy, but seeing what you are trying to do, here is an additional suggestion (you have a O(n^2) algorithm right below that).

You want to remove list.size() - count (lets call this k) random elements from the list. Just pick as many random elements and swap them to the end k positions of the list, then delete that whole range (e.g. using subList() and clear() on that). That would turn it to a lean and mean O(n) algorithm (O(k) is more precise).

Update: As noted below, this algorithm only makes sense if the elements are unordered, e.g. if the List represents a Bag. If, on the other hand, the List has a meaningful order, this algorithm would not preserve it (polygenelubricants' algorithm instead would).

Update 2: So in retrospect, a better (linear, maintaining order, but with O(n) random numbers) algorithm would be something like this:

LinkedList<String> elements = ...; //to avoid the slow ArrayList.remove()
int k = elements.size() - count; //elements to select/delete
int remaining = elements.size(); //elements remaining to be iterated
for (Iterator i = elements.iterator(); k > 0 && i.hasNext(); remaining--) {
  i.next();
  if (random.nextInt(remaining) < k) {
     //or (random.nextDouble() < (double)k/remaining)
     i.remove();
     k--;
  }
}

Solution 8 - Java

This UnsupportedOperationException comes when you try to perform some operation on collection where its not allowed and in your case, When you call Arrays.asList it does not return a java.util.ArrayList. It returns a java.util.Arrays$ArrayList which is an immutable list. You cannot add to it and you cannot remove from it.

Solution 9 - Java

I've got another solution for that problem:

List<String> list = Arrays.asList(split);
List<String> newList = new ArrayList<>(list);

work on newList ;)

Solution 10 - Java

Replace

List<String> list=Arrays.asList(split);

to

List<String> list = New ArrayList<>();
list.addAll(Arrays.asList(split));

or

List<String> list = new ArrayList<>(Arrays.asList(split));

or

List<String> list = new ArrayList<String>(Arrays.asList(split));

or (Better for Remove elements)

List<String> list = new LinkedList<>(Arrays.asList(split));

Solution 11 - Java

The issue is you're creating a List using Arrays.asList() method with fixed Length meaning that

> Since the returned List is a fixed-size List, we can’t add/remove elements.

See the below block of code that I am using

> This iteration will give an Exception Since it is an iteration list Created by asList() so remove and add are not possible, it is a fixed array

List<String> words = Arrays.asList("pen", "pencil", "sky", "blue", "sky", "dog"); 
for (String word : words) {
	if ("sky".equals(word)) {
		words.remove(word);
	}
}	

> This will work fine since we are taking a new ArrayList we can perform modifications while iterating

List<String> words1 = new ArrayList<String>(Arrays.asList("pen", "pencil", "sky", "blue", "sky", "dog"));
for (String word : words) {
	if ("sky".equals(word)) {
		words.remove(word);
	}
}

Solution 12 - Java

Yes, on Arrays.asList, returning a fixed-size list.

Other than using a linked list, simply use addAll method list.

Example:

String idList = "123,222,333,444";

List<String> parentRecepeIdList = new ArrayList<String>();

parentRecepeIdList.addAll(Arrays.asList(idList.split(","))); 

parentRecepeIdList.add("555");

Solution 13 - Java

You can't remove, nor can you add to a fixed-size-list of Arrays.

But you can create your sublist from that list.

list = list.subList(0, list.size() - (list.size() - count));

public static String SelectRandomFromTemplate(String template, int count) {
   String[] split = template.split("\\|");
   List<String> list = Arrays.asList(split);
   Random r = new Random();
   while( list.size() > count ) {
      list = list.subList(0, list.size() - (list.size() - count));
   }
   return StringUtils.join(list, ", ");
}

*Other way is

ArrayList<String> al = new ArrayList<String>(Arrays.asList(template));

this will create ArrayList which is not fixed size like Arrays.asList

Solution 14 - Java

Arraylist narraylist=Arrays.asList(); // Returns immutable arraylist To make it mutable solution would be: Arraylist narraylist=new ArrayList(Arrays.asList());

Solution 15 - Java

Following is snippet of code from Arrays

public static <T> List<T> asList(T... a) {
        return new ArrayList<>(a);
    }

    /**
     * @serial include
     */
    private static class ArrayList<E> extends AbstractList<E>
        implements RandomAccess, java.io.Serializable
    {
        private static final long serialVersionUID = -2764017481108945198L;
        private final E[] a;

so what happens is that when asList method is called then it returns list of its own private static class version which does not override add funcion from AbstractList to store element in array. So by default add method in abstract list throws exception.

So it is not regular array list.

Solution 16 - Java

Arrays.asList() uses fixed size array internally.
You can't dynamically add or remove from thisArrays.asList()

Use this

Arraylist<String> narraylist=new ArrayList(Arrays.asList());

In narraylist you can easily add or remove items.

Solution 17 - Java

Creating a new list and populating valid values in new list worked for me.

Code throwing error -

List<String> list = new ArrayList<>();
   for (String s: list) {
     if(s is null or blank) {
        list.remove(s);
     }
   }
desiredObject.setValue(list);

After fix -

 List<String> list = new ArrayList<>();
 List<String> newList= new ArrayList<>();
 for (String s: list) {
   if(s is null or blank) {
      continue;
   }
   newList.add(s);
 }
 desiredObject.setValue(newList);

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