Java streams, create new object based on two different objects

Question

What is the best way to create a new object based on two different objects.

I would like to use java streams.

My two start objects

public class EventA{
    Long id;
    String name;
    ...
    Long locationID;
}

public class EventB{
    Long id
    String Name;
    ...
    Long locationID;
}

And my result class

public class Result{
    Long locationID;
    String eventAName;
    String eventBName;

    public Result(...){...}
}

I have two object arrays like

List<EventA> eventAList;
List<EventB> eventBList;

I like to get an array of Result objects. Every EventA name should be copied to the resultList. If an EventB at the same location exists I would like to save the name in eventBName.

All I have done so far is

List<Result> resultList = eventAList.stream().map(e -> new Result(e.locationID, e.name, null)).collect(Collectors.toList());

I don't know how to pass the value from EventB to the constructor

Can't you just have a common ancestor to EventA and EventB, since they display practically identical properties? You'd then hjave a List<EventCommon> to stream... — Mena
– Mena, Commented Jun 23, 2016 at 8:54
No, it's just to simplify my example. In fact they are completely different — Simon Schüpbach
– Simon Schüpbach, Commented Jun 23, 2016 at 8:55
What do you mean by "at the same location" ? Same locationID or "at the same index in eventAList and eventBList" ? — Florent Bayle
– Florent Bayle, Commented Jun 23, 2016 at 9:08
resultList should be as big as eventAList. If the location matches (same locationID) I would like to save the name of EventB inside eventBName of the result object. — Simon Schüpbach
– Simon Schüpbach, Commented Jun 23, 2016 at 9:12

Florent Bayle · Accepted Answer · 2016-06-23 09:25:51Z

3

When creating your Result, you can use a stream to iterate on the values in eventBList to retain only the ones with the same locationID as your eventAList value, then take the value you found, and map() it to it's Name value, or null if it doesn't exists:

List<Result> resultList = eventAList.stream().map(a -> new Result(a.locationID, a.name,
    eventBList.stream().filter(b -> b.locationID.equals(a.locationID)).findAny().map(b -> b.Name).orElse(null)
)).collect(Collectors.toList());

For better performances, you can use a temporary Map:

final Map<Long, String> eventBMap = eventBList.stream().collect(Collectors.toMap(b -> b.locationID, b -> b.Name));

List<Result> resultList = eventAList.stream().map(a -> new Result(a.locationID, a.name,
    eventBMap.get(a.locationID)
)).collect(Collectors.toList());

answered Jun 23, 2016 at 9:25

Florent Bayle

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Comments

Simon Schüpbach · Accepted Answer · 2016-06-23 09:26:55Z

1

I found a working way

I adjust the constructor of the Result class to

public Result(Long locationID, String eventAName, EventB eventB){
    this.locationID = locationID;
    this.eventAName = eventAName;
    this.eventBName = eventB.name;
}

and then inside my java stream

List<Result> resultList = eventAList.stream().map(ea -> new Result(ea.locationID, ea.name, eventBList.stream().filter(eb -> eb.locationID.equals(ea.locationID)).findFirst().orElse(new EventB()).get()).collect(Collectors.toList());

answered Jun 23, 2016 at 9:26

Simon Schüpbach

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Comments

George Karanikas · Accepted Answer · 2016-06-23 09:28:28Z

You can do something like the following and work on enhancements afterwards (e.g. create a map for eventBlist by locationId as key in order to have faster search)

Function<EventA, SimpleEntry<EventA, Optional<EventB>>> mapToSimpleEntry = eventA -> new SimpleEntry<>(eventA,
    eventBList.stream()
    .filter(e -> Objects.equals(e.getLocationID(), eventA.getLocationID()))
    .findFirst());

Function<SimpleEntry<EventA, Optional<EventB>>, Result> mapToResult = simpleEntry -> {
    EventA eventA = simpleEntry.getKey();
    Optional<EventB> eventB = simpleEntry.getValue();
    return new Result(eventA.getLocationID(), eventA.getName(), eventB.map(EventB::getName).orElse(null));
};

eventAList.stream()
    .map(mapToSimpleEntry)
    .map(mapToResult)
    .collect(Collectors.toList());

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