Java 8 Streams - Map Multiple Object of same type to a list using streams

Question

Is it possible to do the below mentioned steps using streams in a better way ?

Set<Long> memberIds = new HashSet<>();
marksDistribution.parallelStream().forEach(marksDistribution -> {
        memberIds.add(marksDistribution.getStudentId());
        memberIds.add(marksDistribution.getTeacherId());
      });

instanceDistribution.getStudentId() and instanceDistribution.getTeacherId() are both of type Long.

It might be possible that this kind of question is asked but I am not able to understand it. In simple yes or no. If yes/no, then how and bit explanation. And if possible kindly, discuss the efficiency.

Eran · Accepted Answer · 2017-01-17 10:24:49Z

9

Yes, you can use flatMap to map a single element of your Stream into a Stream of multiple elements, and then flatten them into a single Stream :

Set<Long> memberIds = 
    marksDistribution.stream()
                     .flatMap (marksDistribution -> Stream.of(marksDistribution.getStudentId(), marksDistribution.getTeacherId()))
                     .collect(Collectors.toSet());

answered Jan 17, 2017 at 10:24

Eran

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4 Comments

balboa_21 Over a year ago

can you discuss the efficiency also compared to what I have done in question ?

Eugene Over a year ago

@Eran 1+ from me, still if u have jdk-9 you could use a flatMapping collector: list.stream().collect(Collectors.flatMapping((MarksDistribution md) -> Stream.of(md.getStudentId(), md.getTeacherId()), Collectors.toSet()));

Eran Over a year ago

@Eugene Nice, I should probably try jdk-9 sometime soon.

Holger Over a year ago

@Eugene: like with all Stream operations, I would use the Collector counterpart only when necessary. You could express almost every Stream operation as a Collector-only operation, especially with Java 9, but it would be quiet hard to read…

Alexis C. · Accepted Answer · 2017-01-17 10:32:02Z

5

You can use the 3-args version of collect:

Set<Long> memberIds = 
    marksDistribution.parallelStream()
                     .collect(HashSet::new, 
                              (s, m) -> {
                                   s.add(m.getStudentId());
                                   s.add(m.getTeacherId());
                               }, Set::addAll);

Your current version may produce wrong results, since you are adding elements in parallel in a non-thread safe collection. So it may be possible that you have multiple times the same value in the set.

answered Jan 17, 2017 at 10:32

Alexis C.

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5 Comments

balboa_21 Over a year ago

can you discuss the efficiency also compared to what I have done in question ?

Alexis C. Over a year ago

@balboa_21 The problem is that what you have done is flawed. Before thinking about efficiency, you should think about correctness. That said, using collect is the canonical way to get what you want using streams.

balboa_21 Over a year ago

c: Thanks you for the quick response. Ok, if supposedly, I am not using parallelStream() ?.

Alexis C. Over a year ago

@balboa_21 Still use collect. Otherwise you are left with a piece of code that won't work if someone decide to change in the future .stream() to .parallelStream().

Holger Over a year ago

@balboa_21: if you are not using a parallel operation, you can use forEach directly on the Collection, i.e. without a Stream operation at all.

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