Merging lists under same objects in a list using Java streams

Question

I have two objects like following:

public class A {
    private Integer id;
    private String name;
    private List list;

    public A(Integer id         


        

Answer 1

Assuming class A has a copy constructor that effectively copies the List<B> list attribute and a method that merges two instances of A:

public A(A another) {
    this.id = another.id;
    this.name = another.name;
    this.list = new ArrayList<>(another.list);
}

public A merge(A another) {
    list.addAll(another.list):
    return this;
}

You could achieve what you want as follows:

Map<Integer, A> result = listOfA.stream()
    .collect(Collectors.toMap(A::getId, A::new, A::merge));

Collection<A> result = map.values();

This uses Collectors.toMap, which expects a function that extracts the key of the map from the elements of the stream (here this would be A::getId, which extracts the id of A), a function that transforms each element of the stream to the values of the map (here it would be A::new, which references the copy constructor) and a merge function that combines two values of the map that have the same key (here this would be A::merge, which is only called when the map already contains an entry for the same key).

If you need a List<A> instead of a Collection<A>, simply do:

List<A> result = new ArrayList<>(map.values());

Answer 2

If you can use vanilla Java here is a very easy solution. The list is iterated at once.

Map<Integer, A> m = new HashMap<>();
for (A a : list) {
    if (m.containsKey(a.getId()))
        m.get(a.getId()).getList().addAll(a.getList());
    else
         m.put(a.getId(), new A(a.getId(), a.getName(), a.getList()));
}
List<A> output =  new ArrayList<>(m.values());

Answer 3

If you don't want to use extra functions you can do the following, it's readable and easy to understand, first group by id, create a new object with the first element in the list and then join all the B's classes to finally collect the A's.

List<A> result = list.stream()
    .collect(Collectors.groupingBy(A::getId))
    .values().stream()
    .map(grouped -> new A(grouped.get(0).getId(), grouped.get(0).getName(),
            grouped.stream().map(A::getList).flatMap(List::stream)
                .collect(Collectors.toList())))
    .collect(Collectors.toList());

Another way is to use a binary operator and the Collectors.groupingBy method. Here you use the java 8 optional class to create the new A the first time when fst is null.

BinaryOperator<A> joiner = (fst, snd) -> Optional.ofNullable(fst)
    .map(cur -> { cur.getList().addAll(snd.getList()); return cur; })
    .orElseGet(() -> new A(snd.getId(), snd.getName(), new ArrayList<>(snd.getList())));

Collection<A> result = list.stream()
    .collect(Collectors.groupingBy(A::getId, Collectors.reducing(null, joiner)))
    .values();

If you don't like to use return in short lambdas (doesn't look that well) the only option is a filter because java does not provide another method like stream's peek (note: some IDEs highlight to 'simplify' the expression and mutations shouldn't be made in filter [but i think in maps neither]).

BinaryOperator<A> joiner = (fst, snd) -> Optional.ofNullable(fst)
    .filter(cur -> cur.getList().addAll(snd.getList()) || true)
    .orElseGet(() -> new A(snd.getId(), snd.getName(), new ArrayList<>(snd.getList())));

You can also use this joiner as a generic method and create a left to right reducer with a consumer that allows to join the new mutable object created with the initializer function.

public class Reducer {
    public static <A> Collector<A, ?, A> reduce(Function<A, A> initializer, 
                                                BiConsumer<A, A> combiner) {
        return Collectors.reducing(null, (fst, snd) -> Optional.ofNullable(fst)
            .map(cur -> { combiner.accept(cur, snd); return cur; })
            .orElseGet(() -> initializer.apply(snd)));
    }
    public static <A> Collector<A, ?, A> reduce(Supplier<A> supplier, 
                                                BiConsumer<A, A> combiner) {
        return reduce((ign) -> supplier.get(), combiner);
    }
}

And use it like

Collection<A> result = list.stream()
    .collect(Collectors.groupingBy(A::getId, Reducer.reduce(
        (cur) -> new A(cur.getId(), cur.getName(), new ArrayList<>(cur.getList())),
        (fst, snd) -> fst.getList().addAll(snd.getList())
    ))).values();

Or like if you have an empty constructor that initializes the collections

Collection<A> result = list.stream()
    .collect(Collectors.groupingBy(A::getId, Reducer.reduce(A::new,
        (fst, snd) -> {
            fst.getList().addAll(snd.getList());
            fst.setId(snd.getId());
            fst.setName(snd.getName());
        }
    ))).values();

Finally, if you already have the copy constructor or the merge method mentioned in the other answers you can simplify the code even more or use the Collectors.toMap method.

Answer 4

Collection<A> merge(List<A> list) {
    return list.stream()
            .collect(Collectors.toMap(a -> a.id, Function.identity(), this::merge))
            .values();
}

A merge(A a1, A a2) {
    if (!a1.name.equals(a2.name)) {
        throw new IllegalArgumentException("We assumed same id means same name");
    }
    return new A(a1.id, a1.name, union(a1.list, a2.list));
}

List<B> union(List<B> l1, List<B> l2) {
    List<B> result = new ArrayList<>(l1);
    result.addAll(l2);
    return result;
}