Java 8 List from String and List of String

Question

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?

Rob Spoor · Accepted Answer · 2022-02-14 13:52:03Z

8

Use Stream.concat inside the flatMap operation:

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

answered Feb 14, 2022 at 13:52

Rob Spoor

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4 Comments

Ashish Mishra Over a year ago

Thanks for the reply, this solution works, the only issue is if externalIds is null, then it will throw an error. I can initialize the list in bean to prevent this. Any other way do you think?

Rob Spoor Over a year ago

Allowing null collections, maps or arrays are usually bad to work with, because you need to add a lot of null checks. Using an empty collection/list/set, map or array is almost always better.

ciccioska Over a year ago

collect(Collectors.toList())can be replaced by toList()

Rob Spoor Over a year ago

@ciccioska indeed, if you're using Java 16 or later

Bohemian · Accepted Answer · 2022-02-14 14:19:04Z

1

Try this:

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

edited Feb 14, 2022 at 14:19

answered Feb 14, 2022 at 13:55

Bohemian♦

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6 Comments

Ashish Mishra Over a year ago

.flatMap(List::stream) is not working , its giving error ""Non-static method cannot be referenced from a static context""

Rob Spoor Over a year ago

I think that Bohemian tried to let the map step return a Stream<List<String>>, but it's a Stream<Stream<List<String>>>. You'd need to use flatMap(Function.identity()) to get a Stream<List<String>> back, then flatMap(List::stream) to get a Stream<String>.

Bohemian Over a year ago

@Rob I did get a little tangled there. Thanks for the hint, but you can avoid flatMap(Function.identity()) by flatMap instead of map first up. See latest edit.

Rob Spoor Over a year ago

@Bohemian you're right, the map returned a Stream, that can of course be done with flatMap immediately.

Bohemian Over a year ago

@Rob I'm going to go out on a limb and say this is more readable, because it first uplifts id to be the same type as externalIds, then takes care of business.

|

WJS · Accepted Answer · 2022-02-14 14:55:17Z

0

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();

edited Feb 14, 2022 at 14:55

answered Feb 14, 2022 at 14:40

WJS

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