转换地图 <Key, List<Value> &gt;到地图 <Key, Value>

Question

What I have:

    SortedSet<Person> ss = new TreeSet<>();

    ss.add(new Person("John", "Doe", 20));
    ss.add(new Person("Max", "Power", 26));
    ss.add(new Person("Bort", "Bort", 30));
    ss.add(new Person("Scorpio", "McGreat", 56));

    Map<Integer, List<Person>> list = ss.stream().collect(Collectors.groupingBy(p -> p.name.length()));

I need to transform the Map to Map<Integer, Person> I know I need to use flatMap for that purpose but I don't know how.

What I have tried so far: Get the value-set and flatmap it.

Map <Integer, Person> list2 = list.values()
                                    .stream()
                                    .flatMap(f -> f.stream())
                                    .collect(Collectors.groupingBy(f -> f.name.length()); //doesn't work

Question: As far as I understand Java returns the values as lists when you create Maps via streams, how can I flatMap those lists ?

Additional info: Implementation of compareTo

@Override
public int compareTo(Person o) {

    int cmp = this.surname.compareTo(o.surname);
    if(cmp == 0){
        cmp = this.name.compareTo(o.name);
    }

    return cmp;
}

DISCLAIMER:

I know the use-case is a little bit odd since I sort according to length and also use compareTo by comparing the length of the name. IMHO this doesn't matter for this question since I will always get a Map<Key, List<Value>> . The question IS: How do I get Map<Key,Value>

Answer 1

In fact your requirement does not make sense. You want to group by a property, which implies that it's not necessarily a bijective function. Thus the type of the values of the resulting map is a list by default. If you know that the function you apply is bijective, you can supply another downstream collector / or use the toMap collector, having a throwing merger as parameter (the best option IMO).

So for example:

Map<Integer, Person> list = 
    ss.stream()
      .collect(Collectors.toMap(p -> p.surname.length(), 
                                p -> p, 
                                (p1, p2) -> {throw new IllegalStateException("Duplicate key with length " + p1.surname.length());}));

outputs in your case:

{3=(John, Doe, 20), 4=(Bort, Bort, 30), 5=(Max, Power, 26), 7=(Scorpio, McGreat, 56)}