迭代时从集合中删除元素

Question

AFAIK, there are two approaches:

1. Iterate over a copy of the collection
2. Use the iterator of the actual collection

For instance,

 List<Foo> fooListCopy = new ArrayList<Foo>(fooList); for(Foo foo: fooListCopy){ // modify actual fooList }

and

Iterator<Foo> itr = fooList.iterator(); while(itr.hasNext()){ // modify actual fooList using itr.remove() }

Are there any reasons to prefer one approach over the other (eg preferring the first approach for the simple reason of readability)?

Answer 1

Let me give a few examples with some alternatives to avoid a ConcurrentModificationException .

Suppose we have the following collection of books

List<Book> books = new ArrayList<Book>(); books.add(new Book(new ISBN("0-201-63361-2"))); books.add(new Book(new ISBN("0-201-63361-3"))); books.add(new Book(new ISBN("0-201-63361-4")));

Collect and Remove

The first technique consists in collecting all the objects that we want to delete (eg using an enhanced for loop) and after we finish iterating, we remove all found objects.

 ISBN isbn = new ISBN("0-201-63361-2"); List<Book> found = new ArrayList<Book>(); for(Book book: books){ if(book.getIsbn().equals(isbn)){ found.add(book); } } books.removeAll(found);

This is supposing that the operation you want to do is "delete".

If you want to "add" this approach would also work, but I would assume you would iterate over a different collection to determine what elements you want to add to a second collection and then issue an addAll method at the end.

Using ListIterator

If you are working with lists, another technique consists in using a ListIterator which has support for removal and addition of items during the iteration itself.

 ListIterator<Book> iter = books.listIterator(); while(iter.hasNext()){ if(iter.next().getIsbn().equals(isbn)){ iter.remove(); } }

Again, I used the "remove" method in the example above which is what your question seemed to imply, but you may also use its add method to add new elements during iteration.

Using JDK >= 8

For those working with Java 8 or superior versions, there are a couple of other techniques you could use to take advantage of it.

You could use the new removeIf method in the Collection base class:

 ISBN other = new ISBN("0-201-63361-2"); books.removeIf(b -> b.getIsbn().equals(other));

Or use the new stream API:

 ISBN other = new ISBN("0-201-63361-2"); List<Book> filtered = books.stream().filter(b -> b.getIsbn().equals(other)).collect(Collectors.toList());

In this last case, to filter elements out of a collection, you reassign the original reference to the filtered collection (ie books = filtered ) or used the filtered collection to removeAll the found elements from the original collection (ie books.removeAll(filtered) ).

Use Sublist or Subset

There are other alternatives as well. If the list is sorted, and you want to remove consecutive elements you can create a sublist and then clear it:

 books.subList(0,5).clear();

Since the sublist is backed by the original list this would be an efficient way of removing this subcollection of elements.

Something similar could be achieved with sorted sets using NavigableSet.subSet method, or any of the slicing methods offered there.

Considerations:

What method you use might depend on what you are intending to do

• The collect and removeAl technique works with any Collection (Collection, List, Set, etc).
• The ListIterator technique obviously only works with lists, provided that their given ListIterator implementation offers support for add and remove operations.
• The Iterator approach would work with any type of collection, but it only supports remove operations.
• With the ListIterator / Iterator approach the obvious advantage is not having to copy anything since we remove as we iterate. So, this is very efficient.
• The JDK 8 streams example don't actually removed anything, but looked for the desired elements, and then we replaced the original collection reference with the new one, and let the old one be garbage collected. So, we iterate only once over the collection and that would be efficient.
• In the collect and removeAll approach the disadvantage is that we have to iterate twice. First we iterate in the foor-loop looking for an object that matches our removal criteria, and once we have found it, we ask to remove it from the original collection, which would imply a second iteration work to look for this item in order to remove it.
• I think it is worth mentioning that the remove method of the Iterator interface is marked as "optional" in Javadocs, which means that there could be Iterator implementations that throw UnsupportedOperationException if we invoke the remove method. As such, I'd say this approach is less safe than others if we cannot guarantee the iterator support for removal of elements.

Answer 2

Old Timer Favorite (it still works):

 List<String> list; for(int i = list.size() - 1; i >= 0; --i) { if(list.get(i).contains("bad")) { list.remove(i); } }

Benefits:

1. It only iterates over the list once
2. No extra objects created, or other unneeded complexity
3. No problems with trying to use the index of a removed item, because... well, think about it

Answer 3

In Java 8, there is another approach. Collection#removeIf

eg:

 List<Integer> list = new ArrayList<>(); list.add(1); list.add(2); list.add(3); list.removeIf(i -> i > 2);

Answer 4

Are there any reasons to prefer one approach over the other

The first approach will work, but has the obvious overhead of copying the list.

The second approach will not work because many containers don't permit modification during iteration. This includes ArrayList .

If the only modification is to remove the current element, you can make the second approach work by using itr.remove() (that is, use the iterator 's remove() method, not the container 's). This would be my preferred method for iterators that support remove() .

Answer 5

Only second approach will work. You can modify collection during iteration using iterator.remove() only. All other attempts will cause ConcurrentModificationException .

Answer 6

You can't do the second, because even if you use the remove() method on Iterator , you'll get an Exception thrown .

Personally, I would prefer the first for all Collection instances, despite the additional overheard of creating the new Collection , I find it less prone to error during edit by other developers. On some Collection implementations, the Iterator remove() is supported, on other it isn't. You can read more in the docs for Iterator .

The third alternative, is to create a new Collection , iterate over the original, and add all the members of the first Collection to the second Collection that are not up for deletion. Depending on the size of the Collection and the number of deletes, this could significantly save on memory, when compared to the first approach.

Answer 7

I would choose the second as you don't have to do a copy of the memory and the Iterator works faster. So you save memory and time.

Answer 8

There is also a simple solution to "iterate" a Collection and removing each items.

List<String> list = new ArrayList<>();
//Fill the list

It simply conciste on looping until the list is empty, and on each iteration, we remove the first element with remove(0) .

while(!list.isEmpty()){
    String s = list.remove(0);
    // do you thing
}

I don't believe this has any improvement compare to the Iterator , it still required to have a mutable list but I like the simplicity of this solution.

Answer 9

You can see this sample; If we think remove odd value from a list:

 public static void main(String[] args) { Predicate<Integer> isOdd = v -> v % 2 == 0; List<Integer> listArr = Arrays.asList(5, 7, 90, 11, 55, 60); listArr = listArr.stream().filter(isOdd).collect(Collectors.toList()); listArr.forEach(System.out::println); }

Answer 10

why not this?

 for( int i = 0; i < Foo.size(); i++ ) { if( Foo.get(i).equals( some test ) ) { Foo.remove(i); } }

And if it's a map, not a list, you can use keyset()