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

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());

The below logic is the solution without modifying the original list

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");

List<String> str3 = str1.stream()
                        .filter(item -> !str2.contains(item))
                        .collect(Collectors.toList());

str1 // ["A", "B", "C", "D"]
str2 // ["D", "E"]
str3 // ["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.