Java 8 lambda get and remove element from list

To Remove element from the list

objectA.removeIf(x -> conditions);

eg:

objectA.removeIf(x -> blockedWorkerIds.contains(x));

List<String> str1 = new ArrayList<String>();
str1.add("A");
str1.add("B");
str1.add("C");
str1.add("D");

List<String> str2 = new ArrayList<String>();
str2.add("D");
str2.add("E");

str1.removeIf(x -> str2.contains(x)); 

str1.forEach(System.out::println);

OUTPUT: A B C

Although the thread is quite old, still thought to provide solution - using Java8.

Make the use of removeIf function. Time complexity is O(n)

producersProcedureActive.removeIf(producer -> producer.getPod().equals(pod));

API reference: removeIf docs

Assumption: producersProcedureActive is a List

NOTE: With this approach you won't be able to get the hold of the deleted item.

Consider using vanilla java iterators to perform the task:

public static <T> T findAndRemoveFirst(Iterable<? extends T> collection, Predicate<? super T> test) {
    T value = null;
    for (Iterator<? extends T> it = collection.iterator(); it.hasNext();)
        if (test.test(value = it.next())) {
            it.remove();
            return value;
        }
    return null;
}

Advantages:

1. It is plain and obvious.
2. It traverses only once and only up to the matching element.
3. You can do it on any Iterable even without stream() support (at least those implementing remove() on their iterator).

Disadvantages:

1. You cannot do it in place as a single expression (auxiliary method or variable required)

As for the

Is it possible to combine get and remove in a lambda expression?

other answers clearly show that it is possible, but you should be aware of

1. Search and removal may traverse the list twice
2. ConcurrentModificationException may be thrown when removing element from the list being iterated

The direct solution would be to invoke ifPresent(consumer) on the Optional returned by findFirst(). This consumer will be invoked when the optional is not empty. The benefit also is that it won't throw an exception if the find operation returned an empty optional, like your current code would do; instead, nothing will happen.

If you want to return the removed value, you can map the Optional to the result of calling remove:

producersProcedureActive.stream()
                        .filter(producer -> producer.getPod().equals(pod))
                        .findFirst()
                        .map(p -> {
                            producersProcedureActive.remove(p);
                            return p;
                        });

But note that the remove(Object) operation will again traverse the list to find the element to remove. If you have a list with random access, like an ArrayList, it would be better to make a Stream over the indexes of the list and find the first index matching the predicate:

IntStream.range(0, producersProcedureActive.size())
         .filter(i -> producersProcedureActive.get(i).getPod().equals(pod))
         .boxed()
         .findFirst()
         .map(i -> producersProcedureActive.remove((int) i));

With this solution, the remove(int) operation operates directly on the index.

Use can use filter of Java 8, and create another list if you don't want to change the old list:

List<ProducerDTO> result = producersProcedureActive
                            .stream()
                            .filter(producer -> producer.getPod().equals(pod))
                            .collect(Collectors.toList());