How to create a list of instances efficiently in java using a another list of instances

Question

I have a list of instances where I need to create another list of instances using the instances in the first list. As the example shown below, I can use a foreach or a for loop. Is there a better way to do this more efficiently?

List<Mesage> messages; 
List<ArchMessage> archMessages = new ArrayList<>();
for(Message message : messages) {
       archMessages.add(new ArchMessage(message));
}

What is the relationship between Message and ArcMessage classes? — PM 77-1
– PM 77-1, Commented Dec 26, 2017 at 18:36
You haven't defined your problem/solution space clearly enough. What do you mean by efficiently? More readable? More maintainable? Best performance? Most scalable? — scottb
– scottb, Commented Dec 26, 2017 at 18:37
@scottb what I meant by efficiently is the best performance and scalable. — Hasangi
– Hasangi, Commented Jan 9, 2018 at 8:54

Eran · Accepted Answer · 2017-12-26 18:32:57Z

2

You can use Java 8 Streams:

List<ArchMessage> archMessages =
    messages.stream()
            .map(message -> new ArchMessage(message))
            .collect(Collectors.toList());

or

List<ArchMessage> archMessages =
    messages.stream()
            .map(ArchMessage::new)
            .collect(Collectors.toList());

answered Dec 26, 2017 at 18:32

Eran

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Ele · Accepted Answer · 2017-12-26 18:40:46Z

0

If you're unable to use streams from Java 8, Apache Commons Collections library provides the following:

http://commons.apache.org/proper/commons-collections/

Collection<ArchMessage> archMessages = CollectionUtils.collect(messages, new Transformer() {
    public Object transform(Object o) {
        return new ArchMessage((Mesage) o);
    }
});

answered Dec 26, 2017 at 18:40

Ele

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Ousmane D. · Accepted Answer · 2017-12-26 18:51:40Z

0

That's as efficient as it gets but let's say your code could benefit from parallelism then using the streams API would be ideal as we can get parallelism for free by using parallelStream e.g.

 List<ArchMessage> resultSet = messages.parallelStream()
                                       .map(ArchMessage::new)
                                       .collect(Collectors.toList());

Other than that your current approach seems good and readable.

edited Dec 26, 2017 at 18:51

answered Dec 26, 2017 at 18:37

Ousmane D.

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score 0 · Accepted Answer · 2018-02-13 18:48:57Z

0

Feedback to your recherche: I would recommend to don't use a foreach. It will throw your efficient down cause of an inner request at each. I would recommend to use a for each time..

edited Feb 13, 2018 at 18:48

answered Dec 26, 2017 at 18:53

user8040474

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