Can I filter a Stream<T> by element's class an get a Stream<U> in one step? [duplicate]

Question

Lets say I have

class Dog extends Animal {}
class Cat extends Animal {}

And I have a list of Animals Using Guava FluentIterable I can filter and convert in one step

List<Cat> cats = FluentIterable.from(animals)
    .filter(Cat.class)
    .toList();

Using Java8 I need to do

List<Cat> cats = animals.stream()
     .filter(c -> c instanceof Cat)
     .map(c -> (Cat) c)
     .collect(Collectors.toList());

There is no way I can make the filter & map in one step, right?

Answer 1

The map step is unnecessary at runtime (it simply does nothing), you need it just to bypass the type checking during the compilation. Alternatively you can use dirty unchecked cast:

List<Cat> cats = ((Stream<Cat>) (Stream<?>) animals.stream().filter(
        c -> c instanceof Cat)).collect(Collectors.toList());

Unfortunately there's no standard way to do this in single step, but you may use third-party libraries. For example, in my StreamEx library there's a select method which solves this problem:

List<Cat> cats = StreamEx.of(animals).select(Cat.class).toList();