Merge Two List which have different size, In one with stream api on certain conditions

Question

Suppose I have two list which i want to get merged into one

Merge Condition : compare with id

Case 1 : 
List 1 = [{id=1,name=null},{id=2,name=null},{id=3,name=c}]
List 2 = [{id=1,name=a},{id=2,name=b}]

After the merge, it should be [{id=1,name=a},{id=2,name=b},{id=3,name=c}]

Case 1 : 
List 1 = [{id=1,name=null}]
List 2 = [{id=1,name=a},{id=2,name=b}]

After the merge, it should be [{id=1,name=a},{id=2,name=b}]

My code:

 for (Object ObjofList1: list1) {
             List<DummyObject> do = new ArrayLIst();
            SomeObject sm = list2.stream()
                    .filter(list2object -> list2object.getId().equals(ObjofList1.getId()))
                    .findAny()
                    .orElse(ObjofList1);
                          do.add(dealerSalesTempObject);
          
        }

It is working fine if list2 has more elements in it, but if list1 has more element then we won't be able to merge because of less iteration.

My question is i don't want to see which one is shorter or longer but at last all objects should be in one list..

Answer 1

Introducing Stream to such problem just makes things more complicated for no real reason. Based on your example you just want to create a Set containing elements from both Lists. Something like this would be the easiest way to achieve it:

Set<Object> resultSet = new HashSet<>(list1);
result.addAll(list2);
List<Object> resultList = new ArrayList<>(result); // optionally if u want a List

This solution assumes that your class implements equals() and hashcode() methods.

Answer 2

You can merge these two lists using streams, and then use distinct, in order to remove duplicate elements, and finally reconvert it to a list:

Stream.concat(list1.stream(), list2.stream()).distinct().collect(Collectors.toList());

NOTE: you need to implement equals and hashcode in your DummyObject, it's required to remove the duplicated elements when generating the Set.

HashCode and Equals for your case (Considering the infos in the comments):

@Override
public boolean equals(Object o) {
    if (this == o) return true;
    if (o == null || getClass() != o.getClass()) return false;

    StockProcessData that = (StockProcessData) o;

    if (id != null ? !id.equals(that.id) : that.id != null) return false;
    if (unContractedStock != null ? !unContractedStock.equals(that.unContractedStock) : that.unContractedStock != null)
        return false;
    return contractedStock != null ? contractedStock.equals(that.contractedStock) : that.contractedStock == null;
}

@Override
public int hashCode() {
    int result = id != null ? id.hashCode() : 0;
    result = 31 * result + (unContractedStock != null ? unContractedStock.hashCode() : 0);
    result = 31 * result + (contractedStock != null ? contractedStock.hashCode() : 0);
    return result;
}

Answer 3

You have updated your question and mentioned the use case of merging objects with the same id into a single object. The way I can think of would be something like this:

List<DummyObject> list3 = new ArrayList<>(
    Stream
        .concat(list1.stream(), list2.stream())
        .collect(Collectors.toMap(
            DummyObject::getId, 
            Function.identity(), 
            (a, b) -> {
                 // here you can merge a and b
            }
         ))
         .values()
    );

---

Old answer

No need for Streams, a simple SortedSet is enough:

SortedSet<DummyObject> result = new TreeSet<>(Comparator.comparing(DummyObject::getId));
result.addAll(list1);
result.addAll(list2);

The SortedSet can be used when you don't want to override the hashCode() and equals() methods. Additionally the result is sorted by id.

If you need a List afterwards you can convert it easily back:

List<DummyObject> list3 = new ArrayList<>(result);

But using a Set may be more declarative, as it clearly indicates that no duplicates are available.