java 8 stream collector mapping List<List> to List

Question

Lets say I have this class:

public class Employee{

  private int id;
  private List<Car> cars;
//getters , equals and hashcode by id
}

public class Car {
  private String name;

}

I have a List of Employees (same id may be duplicated):

List<Employee> emps =  ..

Map<Employee, List<List<Car>>> resultMap = emps.stream().collect(
            Collectors.groupingBy(Function.identity(),
                    Collectors.mapping(Employee::getCars, Collectors.toList());

This gives me Map<Employee, List<List<Car>>>, how can I get a Map<Employee, List<Car> (like a flat List)?

Eran · Accepted Answer · 2017-05-10 12:19:01Z

7

I don't see why you are using groupingBy when you are not doing any grouping at all. It seems all you need is to create a Map where the key in a Employee and the value is the Employee's cars:

Map<Employee, List<Car> map =
    emps.stream().collect(Collectors.toMap(Function.identity(),Employee::getCars);

If you are grouping in order to join two instances of the same Employee, you can still use toMap with a merge function:

Map<Employee, List<Car> map =
    emps.stream()
        .collect(Collectors.toMap(Function.identity(),
                                  Employee::getCars,
                                  (v1,v2)-> {v1.addAll(v2); return v1;},
                                  HashMap::new);

Note that this will mutate some of the original Lists returned by Employee::getCars, so you might want to create a new List instead of adding the elements of one list to the other.

edited May 10, 2017 at 12:19

answered May 10, 2017 at 12:01

Eran

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3 Comments

Jeremy Grand Over a year ago

Yeah, sorry I did not properly read the case. Your solution is fine. I was meaning to use the flatMap to convert the list of lists into a single list but as it is inside the toMap collector, your solution is way better.

user1321466 Over a year ago

Could you please help one more time? what if in Employee class we have Set<Car> cars, but I want to get Map<Employee, List<Car>>

Eran Over a year ago

@user1321466 In that case you can replace Employee::getCars with e->new ArrayList<Car>(e.getCars()).

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