I want to generate a list of strings, comprising id and external ids, from a list of Bean.

public class User {
    private String id;
    private List<String> externalIds;
}

I got it using the below code, but here I need to do the stream twice.

List<User> references = new ArrayList();
Stream.concat(references.stream().map(User::getId),
references.stream().map(User::getExternalIds).flatMap(Collection::stream))
            .collect(Collectors.toList());

Is there any better way to rewrite this code?

CodePudding user response：

Use Stream.concat inside the flatMap operation:

references.stream()
        .flatMap(user -> Stream.concat(
            Stream.of(user.getId()),
            user.getExternalIds().stream()
        ))
        .collect(Collectors.toList())

CodePudding user response：

Try this:

references.stream()
  .flatMap(u -> Stream.of(List.of(u.getId()), u.getExternalIds()))
  .flatMap(List::stream)
  .collect(Collectors.toList());

CodePudding user response：

references.stream()

then:

map() which is to create a list including id & externalIds

then:

flatMap()

then:

collect()

Another way: in User, make a new get method to return a list with both id & externalIds

then:

references.stream().flatMap(User::getThatNewMethodIds()).collect(...)

CodePudding user response：

If you're using Java 16 you can do it using mapMulti. This essentially intercepts each item in the stream and replaces it with other items. It avoids re-streaming and flatmapping. Since the nature of the stream(i.e. type) is being changed from User to String, the witness before mapMulti is required.

• add the user id in the stream.
• then add each external id in the stream for that user

List<String> results = references.stream()
        .<String>mapMulti((user, consumer) -> {
            consumer.accept(user.getId());
            for (String id : user.getExternalIds()) {
                consumer.accept(id);
            }
        }).toList();