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POHow to efficiently (performance) remove many items from List in Java?
 

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  1. POHow to efficiently (performance) remove many items from List in Java?
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    <p>I have quite large List named items (>= 1,000,000 items) and some condition denoted by &lt;cond&gt; that selects items to be deleted and &lt;cond&gt; is true for many (maybe half) of items on my list.</p> <p>My goal is to efficiently remove items selected by &lt;cond&gt; and retain all other items, source list may be modified, new list may be created - best way to do it should be chosen considering performance.</p> <p>Here is my testing code:</p> <pre><code> System.out.println("preparing items"); List&lt;Integer&gt; items = new ArrayList&lt;Integer&gt;(); // Integer is for demo for (int i = 0; i &lt; 1000000; i++) { items.add(i * 3); // just for demo } System.out.println("deleting items"); long startMillis = System.currentTimeMillis(); items = removeMany(items); long endMillis = System.currentTimeMillis(); System.out.println("after remove: items.size=" + items.size() + " and it took " + (endMillis - startMillis) + " milli(s)"); </code></pre> <p>and naive implementation:</p> <pre><code>public static &lt;T&gt; List&lt;T&gt; removeMany(List&lt;T&gt; items) { int i = 0; Iterator&lt;T&gt; iter = items.iterator(); while (iter.hasNext()) { T item = iter.next(); // &lt;cond&gt; goes here if (/*&lt;cond&gt;: */i % 2 == 0) { iter.remove(); } i++; } return items; } </code></pre> <p>As you can see I used item index modulo 2 == 0 as remove condition (&lt;cond&gt;) - just for demonstation purposes.</p> <p>What better version of <code>removeMany</code> may be provided and why this better version is actually better?</p>
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    1. COWhat performance metrics matter - just speed, or is memory use important? Is the list short lived? Is every entry of the shorter (post-remove) list guaranteed to be accessed? I wonder if creating a new list iterator that stores the removal condition as a return condition might be an efficient solution for a certain class of problems. Instead of removing from the list, you could have the iterator's the next() method skip items that don't match the condition. This would have the benifit of testing only the entries upon which you operate, with the penalty of wasting lots of memory.
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    2. COjust as in above example: input is a list, output is a list (the same with selected items removed or new one with retained items) and speed is my most important metric.
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    3. COthanks for your answers! i have just given my answer which compiles different proposed approaches and tests them in practice. i hope my code has no errors and my final conclusions are helpful.
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