Java 8 Stream API

Question

I have this simple instruction

Stream.concat(manager.getChild().stream(), 
            manager1.getChild().stream())
            .map(dev -> dev.getSalary())
            .reduce(0, Integer::max);

that concats two List and return the developer that earning more. This returns the maximum salary of the objects in the stream, but how can I retrieve the object that has the maximum salary?

Remove the map and reduce and use max instead with an appropriate Comparator that compares by salary. — Jesper
– Jesper, Commented Mar 9, 2017 at 13:15
I try to use this Employee rich = Stream.concat(manager.getChild().stream(), manager1.getChild().stream()) .max((dev1, dev2) -> Integer.max(dev1.getSalary(), dev2.getSalary())).get(); but not work very well. — Dennis A. Boanini
– Dennis A. Boanini, Commented Mar 9, 2017 at 13:32

ZhekaKozlov · Accepted Answer · 2017-03-09 13:32:48Z

5

Use Stream.max(Comparator<? super T> comparator) method:

Stream.concat(manager.getChild().stream(),
    manager1.getChild().stream())
    .max(Comparator.comparingInt(dev -> dev.getSalary())

answered Mar 9, 2017 at 13:32

ZhekaKozlov

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Comments

Bohemian · Accepted Answer · 2017-03-09 13:33:06Z

1

Unfortunately you can't reach back to a previous step to get the stream element that provided the maximum salary; you have to compare the elements by an attribute.

To find the subordinate with the highest salary:

Employee max = Stream.of(manager, manager1)
  .map(Employee::getChild)
  .flatMap(Collection::stream)
  .max(Comparators.comparing(Employee::getSalary))
  .orElse(null);

I have assumed a class of Employee so method references can be used, and also refactored the code to a IMHO more usable/standard approach of starting by streaming the managers.

answered Mar 9, 2017 at 13:33

Bohemian♦

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2 Comments

ZhekaKozlov Over a year ago

flatMap is overkill here

Bohemian Over a year ago

@ZhekaKozlov but it's more flexible: You can plug this flow on to any stream of managers, eg a List<Employee> via .stream(), or an Employee[] via Arrays.stream() etc. Conversely, hardcoding the stream calls is not scaleable or flexible and breaks DRY

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