CODEWITHSUNDEEP
javajava-8java-stream
mrod
mrod

Reputation: 792

Java Streams: group a List into a Map of Maps

How could I do the following with Java Streams?

Let's say I have the following classes:

class Foo {
    Bar b;
}

class Bar {
    String id;
    String date;
}

I have a List<Foo> and I want to convert it to a Map <Foo.b.id, Map<Foo.b.date, Foo>. I.e: group first by the Foo.b.id and then by Foo.b.date.

I'm struggling with the following 2-step approach, but the second one doesn't even compile:

Map<String, List<Foo>> groupById =
        myList
                .stream()
                .collect(
                        Collectors.groupingBy(
                                foo -> foo.getBar().getId()
                        )
                );

Map<String, Map<String, Foo>> output = groupById.entrySet()
        .stream()
        .map(
                entry -> entry.getKey(),
                entry -> entry.getValue()
                        .stream()
                        .collect(
                                Collectors.groupingBy(
                                        bar -> bar.getDate()
                                )
                        )
        );

Thanks in advance.

Upvotes: 41

Views: 43512

Answers (3)

weston
weston

Reputation: 54781

An alternative is to support the equality contract on your key, Bar:

class Bar {
    String id;
    String date;

    public boolean equals(Object o){
       if (o == null) return false;
       if (!o.getClass().equals(getClass())) return false;
       Bar other = (Bar)o;
       return Objects.equals(o.id, id) && Objects.equals(o.date, date);
    }

    public int hashCode(){
       return id.hashCode*31 + date.hashCode;
    }    
}

Now you can just have a Map<Bar, Foo>.

Upvotes: 1

Alex Salauyou
Alex Salauyou

Reputation: 14338

Suppose (b.id, b.date) pairs are distinct. If so, in second step you don't need grouping, just collecting to Map where key is foo.b.date and value is foo itself:

Map<String, Map<String, Foo>> map = 
       myList.stream()
             .collect(Collectors.groupingBy(f -> f.b.id))    // map {Foo.b.id -> List<Foo>}
             .entrySet().stream()
             .collect(Collectors.toMap(e -> e.getKey(),                 // id
                                       e -> e.getValue().stream()       // stream of foos
                                             .collect(Collectors.toMap(f -> f.b.date, 
                                                                       f -> f))));

Or even more simple:

Map<String, Map<String, Foo>> map = 
       myList.stream()
             .collect(Collectors.groupingBy(f -> f.b.id, 
                                            Collectors.toMap(f -> f.b.date, 
                                                             f -> f)));

Upvotes: 5

Flown
Flown

Reputation: 11740

You can group your data in one go assuming there are only distinct Foo:

Map<String, Map<String, Foo>> map = list.stream()
        .collect(Collectors.groupingBy(f -> f.b.id, 
                 Collectors.toMap(f -> f.b.date, Function.identity())));

Saving some characters by using static imports:

Map<String, Map<String, Foo>> map = list.stream()
        .collect(groupingBy(f -> f.b.id, toMap(f -> f.b.date, identity())));

Upvotes: 70

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