177

Is there any fast (and nice looking) way to remove an element from an array in Java?

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  • 6
    Even if the question is duplicate, the answer in the other question is neither fast nor nice looking. It transforms the array into an arraylist (by hand). Commented Sep 22, 2017 at 14:37

15 Answers 15

273

You could use commons lang's ArrayUtils.

array = ArrayUtils.removeElement(array, element)

commons.apache.org library:Javadocs

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8 Comments

@Clive Guava appears to only work on collections.
does this shrink the array as well?
@SupunWijerathne it has to, to change the size/length.
It is not working in Java8 and I am not getting any method with the name removeElement
@AtulAgrawal You need to include the library in the link on the last line.
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48

Your question isn't very clear. From your own answer, I can tell better what you are trying to do:

public static String[] removeElements(String[] input, String deleteMe) {
    List result = new LinkedList();

    for(String item : input)
        if(!deleteMe.equals(item))
            result.add(item);

    return result.toArray(input);
}

NB: This is untested. Error checking is left as an exercise to the reader (I'd throw IllegalArgumentException if either input or deleteMe is null; an empty list on null list input doesn't make sense. Removing null Strings from the array might make sense, but I'll leave that as an exercise too; currently, it will throw an NPE when it tries to call equals on deleteMe if deleteMe is null.)

Choices I made here:

I used a LinkedList. Iteration should be just as fast, and you avoid any resizes, or allocating too big of a list if you end up deleting lots of elements. You could use an ArrayList, and set the initial size to the length of input. It likely wouldn't make much of a difference.

1 Comment

Note, you'll want to use List<String> result. When I do this in the current compiler, the toArray command gives a type error (the other solution is to cast the result.)
46

The best choice would be to use a collection, but if that is out for some reason, use arraycopy. You can use it to copy from and to the same array at a slightly different offset.

For example:

public void removeElement(Object[] arr, int removedIdx) {
    System.arraycopy(arr, removedIdx + 1, arr, removedIdx, arr.length - 1 - removedIdx);
}

Edit in response to comment:

It's not another good way, it's really the only acceptable way--any tools that allow this functionality (like Java.ArrayList or the apache utils) will use this method under the covers. Also, you REALLY should be using ArrayList (or linked list if you delete from the middle a lot) so this shouldn't even be an issue unless you are doing it as homework.

To allocate a collection (creates a new array), then delete an element (which the collection will do using arraycopy) then call toArray on it (creates a SECOND new array) for every delete brings us to the point where it's not an optimizing issue, it's criminally bad programming.

Suppose you had an array taking up, say, 100mb of ram. Now you want to iterate over it and delete 20 elements.

Give it a try...

I know you ASSUME that it's not going to be that big, or that if you were deleting that many at once you'd code it differently, but I've fixed an awful lot of code where someone made assumptions like that.

5 Comments

Following a "deletion" (i.e. shifting the array left by one element) won't there be a duplicate of the end element? i.e. a.length will be the same following the deletion, no? I'm not saying I dislike the idea, just that one needs to be aware of this.
+1. This works for my purposes. (I fixed the small issue you had in your sample. Hope you don't mind.)
Yes, this will just shift the elements left and there will be last element still present. We have to use new array to copy.
BTW, this is what org.apache.commons.lang.ArrayUtils does too.
It's assumed that if you are adding and deleting elements from an array you are also tracking the "Last" item in the array, so copying shouldn't be necessary.
42

You can't remove an element from the basic Java array. Take a look at various Collections and ArrayList instead.

8 Comments

i know, i just want a beautiful looking way with arraylists or sth. like that, any hint for that?
+1: Use LinkedList, life is simpler.
LinkedList is rarely a good idea. The List intrrface gives you random access, but LinkedList gives O(n) access times instead of O(1).
You can remove an element from an array via System.arrayCopy for example, but you cannot alter the size. A list is a much better solution however.
@Tom: Whether LinkedList is the correct choice depends on other factors too. "Random access", i.e. accessing a linked list via an index, is O(n).
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16

Nice looking solution would be to use a List instead of array in the first place.

List.remove(index)

If you have to use arrays, two calls to System.arraycopy will most likely be the fastest.

Foo[] result = new Foo[source.length - 1];
System.arraycopy(source, 0, result, 0, index);
if (source.length != index) {
    System.arraycopy(source, index + 1, result, index, source.length - index - 1);
}

(Arrays.asList is also a good candidate for working with arrays, but it doesn't seem to support remove.)

2 Comments

+1: Use LinkedList or ArrayList.
Should the final copy be if (result.length != index)... rather than if (source.length != index)...?
13

I think the question was asking for a solution without the use of the Collections API. One uses arrays either for low level details, where performance matters, or for a loosely coupled SOA integration. In the later, it is OK to convert them to Collections and pass them to the business logic as that.

For the low level performance stuff, it is usually already obfuscated by the quick-and-dirty imperative state-mingling by for loops, etc. In that case converting back and forth between Collections and arrays is cumbersome, unreadable, and even resource intensive.

By the way, TopCoder, anyone? Always those array parameters! So be prepared to be able to handle them when in the Arena.

Below is my interpretation of the problem, and a solution. It is different in functionality from both of the one given by Bill K and jelovirt. Also, it handles gracefully the case when the element is not in the array.

Hope that helps!

public char[] remove(char[] symbols, char c)
{
    for (int i = 0; i < symbols.length; i++)
    {
        if (symbols[i] == c)
        {
            char[] copy = new char[symbols.length-1];
            System.arraycopy(symbols, 0, copy, 0, i);
            System.arraycopy(symbols, i+1, copy, i, symbols.length-i-1);
            return copy;
        }
    }
    return symbols;
}

1 Comment

Great. Too many responses answering a different question to OP's.
4

You could use the ArrayUtils API to remove it in a "nice looking way". It implements many operations (remove, find, add, contains,etc) on Arrays.
Take a look. It has made my life simpler.

Comments

3

okay, thx a lot now i use sth like this:

public static String[] removeElements(String[] input, String deleteMe) {
    if (input != null) {
        List<String> list = new ArrayList<String>(Arrays.asList(input));
        for (int i = 0; i < list.size(); i++) {
            if (list.get(i).equals(deleteMe)) {
                list.remove(i);
            }
        }
        return list.toArray(new String[0]);
    } else {
        return new String[0];
    }
}

2 Comments

If you really need to leave the inital array unchanged, you'd better create an empty list and fill it with the right elements rather than doing it this way.
I'm not sure this is what people had in mind when they suggested using collections, but at any rate, be careful with those list indices. It looks like you're skipping the element immediately following any removal (try {"a", "b", "deleteMe", "deleteMe", "c"}).
3

Some more pre-conditions are needed for the ones written by Bill K and dadinn

Object[] newArray = new Object[src.length - 1];
if (i > 0){
    System.arraycopy(src, 0, newArray, 0, i);
}

if (newArray.length > i){
    System.arraycopy(src, i + 1, newArray, i, newArray.length - i);
}

return newArray;

Comments

3

You can not change the length of an array, but you can change the values the index holds by copying new values and store them to a existing index number. 1=mike , 2=jeff // 10 = george 11 goes to 1 overwriting mike .

Object[] array = new Object[10];
int count = -1;

public void myFunction(String string) {
    count++;
    if(count == array.length) { 
        count = 0;  // overwrite first
    }
    array[count] = string;    
}

1 Comment

I think pointing out, that the length of an array can not be changed is an important detail!
0

Copy your original array into another array, without the element to be removed.

A simplier way to do that is to use a List, Set... and use the remove() method.

Comments

0

Swap the item to be removed with the last item, if resizing the array down is not an interest.

1 Comment

This would break things if the array was sorted prior to the remove.
0

I hope you use the java collection / java commons collections!

With an java.util.ArrayList you can do things like the following:

yourArrayList.remove(someObject);

yourArrayList.add(someObject);

10 Comments

An array is not a collection...
But the most collections are arrays! See: en.wikipedia.org/wiki/Array
Yep, but this question is java tagged and, in java, an array is not a collection...
I don't start to fight a religious war about what is a collection of elements and what isn't. Writing Java with a lot of procedural elements is bad! Take profit from the OO fatures! You can create nearly every collection from the Java Array construct.
I don't see why this guy is being modded down. If you need to be able to easily remove an element from an ordered group, then it's pretty clear that perhaps an array is the wrong kind of group to use in the first place. So List is a good suggestion, and Set might be better, depending on the app.
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-4

Use an ArrayList:

alist.remove(1); //removes the element at position 1

Comments

-6

Sure, create another array :)

Comments

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