How to remove specific value from string array in java?

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Removing an element from an Array (Java)

How to remove specific String array value for example

String[] str_array = {"item1","item2","item3"};

i want to remove "item2" from str_array pls help me i want output like

String[] str_array = {"item1","item3"};

You needed to copy to a new array if you want the length to reduce also. Use ArrayList if you want dynamic size array — nhahtdh, Oct 10 '12 at 05:10
i know remove value from arraylist we use "list.remove()" but i need is there any property to remove string in string array — Vicky, Oct 10 '12 at 05:14
The arrays by themselves don't have these methods. If you want to do it in a single and effective line, you should follow PeterLawrey's advice: http://stackoverflow.com/a/644719/1065197 — Luiggi Mendoza, Oct 10 '12 at 05:16

Vikdor · Accepted Answer · 2020-04-02T19:18:18.893

47

I would do it as follows:

String[] str_array = {"item1","item2","item3"};
List<String> list = new ArrayList<String>(Arrays.asList(str_array));
list.remove("item2");
str_array = list.toArray(new String[0]);

edited Apr 02 '20 at 19:18

answered Oct 10 '12 at 05:13

Vikdor

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Inefficient compared to use of `System.arraycopy`, especially when used multiple times. – FThompson Oct 10 '12 at 05:19
@Vulcan, thanks for your comment, would you mind updating my answer to show how to use System.arraycopy here? Thanks. – Vikdor Oct 10 '12 at 05:21
@Vikdor I'll post an answer that uses it, one moment. – FThompson Oct 10 '12 at 05:22
@Vikdor Sorry for the delay, got distracted with something. I've posted the answer though. – FThompson Oct 10 '12 at 05:42
this really useful... This answer helped me loads – Afrah_Rahman Dec 11 '19 at 05:54

FThompson · Answer 2 · 2012-10-10T06:27:01.287

7

If you must use arrays, System.arraycopy is the most efficient, scalable solution. However, if you must remove one element from an array several times, you should use an implementation of List rather than an array.

The following utilizes System.arraycopy in order to achieve the desired effect.

public static Object[] remove(Object[] array, Object element) {
    if (array.length > 0) {
        int index = -1;
        for (int i = 0; i < array.length; i++) {
            if (array[i].equals(element)) {
                index = i;
                break;
            }
        }
        if (index >= 0) {
            Object[] copy = (Object[]) Array.newInstance(array.getClass()
                    .getComponentType(), array.length - 1);
            if (copy.length > 0) {
                System.arraycopy(array, 0, copy, 0, index);
                System.arraycopy(array, index + 1, copy, index, copy.length - index);
            }
            return copy;
        }
    }
    return array;
}

Also, you can increase the method's efficiency if you know that your array consists of only Comparable objects. You can use Arrays.sort to sort them before passing them through the remove method, modified to use Arrays.binarySearch to find index rather than a for loop, raising that portion of the method's efficiency from O(n) to O(nlogn).

edited Oct 10 '12 at 06:27

answered Oct 10 '12 at 05:41

FThompson

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It's not very nice to get an `Object[]` back when you want a `String[]`. [`Array.newInstance`](http://docs.oracle.com/javase/6/docs/api/java/lang/reflect/Array.html#newInstance%28java.lang.Class,%20int%29) lets you construct an array of arbitrary type. – gustafc Oct 10 '12 at 06:13
@gustafc Good point. The downside to that is slightly slower execution time though. – FThompson Oct 10 '12 at 06:22
It's also worth noting that *O(n log n)* isn't necessarily faster than *O(n)* as Big O only tells you how the cost of handling *n+1* elements is compared to handling *n* elements with the same algorithm. The constant cost associated with sorting may well outweigh the benefits of binary search (especially for small arrays). – gustafc Oct 10 '12 at 06:22
`Array.newInstance` *might* be slower, but it is not necessarily so. HotSpot can optimize it to be just as fast as a regular array instantiation in most cases. Either way, you end up with a more robust and correct program. – gustafc Oct 10 '12 at 06:28
Ah, that's good. It was slightly slower in a small test I just conducted (average of 0.40ms for new Object[x], and average 0.45ms for Array.newInstance), but the margin is so small that I'd say it's irrelevant and may not even exist. – FThompson Oct 10 '12 at 06:30
2

Indeed. Also, if that part of the code becomes a bottleneck, arrays probably aren't the most appropriate data structure for your program. – gustafc Oct 10 '12 at 06:36

score 4 · Answer 3 · answered Oct 10 '12 at 05:28

Other Option is to copy array to other array accept than remove item.

 public static String[] removeItemFromArray(String[] input, String item) {
    if (input == null) {
        return null;
    } else if (input.length <= 0) {
        return input;
    } else {
        String[] output = new String[input.length - 1];
        int count = 0;
        for (String i : input) {
            if (!i.equals(item)) {
                output[count++] = i;
            }
        }
        return output;
    }
}

Note, this will remove *all* instances of `item` from the input array rather than just one. — FThompson, Oct 10 '12 at 05:46

How to remove specific value from string array in java?

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