Why is this usage of Stream::flatMap wrong?

Question

I expected to be able to use Stream::flatMap like this

public static List<String> duplicate(String s) {

    List<String> l = new ArrayList<String>();
    l.add(s);
    l.add(s);

    return l;
}


listOfStrings.stream().flatMap(str -> duplicate(str)).collect(Collectors.toList());

But I get the following compiler error

Test.java:25: error: incompatible types: cannot infer type-variable(s) R listOfStrings.stream().flatMap(str -> duplicate(str)).collect(Collectors.toList());

(argument mismatch; bad return type in lambda expression List cannot be converted to Stream)
where R,T are type-variables: R extends Object declared in method flatMap(Function>) T extends Object declared in interface Stream

In scala I can do what I believe to be equivalent

scala> List(1,2,3).flatMap(duplicate(_))
res0: List[Int] = List(1, 1, 2, 2, 3, 3)

Why is this not a valid usage of flatMap in java?

Toreutic

check this answer among others on how and why to use flapMap() with java 8 Streams: https://mcmap.net/q/53272/-what-39-s-the-difference-between-map-and-flatmap-methods-in-java-8 – Stockbroker

Answer 1

The lambda expression in flatMap needs to return a Stream, as can be seen by the argument of flatMap which is of type Function<? super T, ? extends Stream<? extends R>>.

The following code will compile and run fine:

listOfStrings.stream()
             .flatMap(str -> duplicate(str).stream()) // note the .stream() here
             .collect(Collectors.toList());

because the lambda expression str -> duplicate(str).stream() is of type Function<String, Stream<String>>.

Shulamite

The return type for <R> Stream<R> map(Function<? super T,? extends R> mapper) is the same. Why doesn't map require .stream() but flatmap does? – Syndicalism

Answer 2

If you want to duplicate each object in the stream several times, you don't need to waste memory on this with additional ArrayList. There are several shorter and faster alternatives.

• Generate new stream using Stream.generate, then limit it:

  listOfStrings.stream()
               .flatMap(str -> Stream.generate(() -> str).limit(2))
               .collect(Collectors.toList());
  

• Generate sequence of numbers via IntStream.range and map them to the same string:

  listOfStrings.stream()
               .flatMap(str -> IntStream.range(0, 2).mapToObj(i -> str))
               .collect(Collectors.toList());
  

• Use good old Collections.nCopies:

  listOfStrings.stream()
               .flatMap(str -> Collections.nCopies(2, str).stream())
               .collect(Collectors.toList());
  

If you are sure that you will always duplicate exactly two times, there's the shortest alternative:

listOfStrings.stream()
             .flatMap(str -> Stream.of(str, str))
             .collect(Collectors.toList());

Herrah