How to store enum to map using Java 8 stream API

Question

I have an enum with another enum as a parameter

public enum MyEntity{
   Entity1(EntityType.type1,
    ....


   MyEntity(EntityType type){
     this.entityType = entityType;
   }
}

I want to create a method that return the enum by type

public MyEntity getEntityTypeInfo(EntityType entityType) {
        return lookup.get(entityType);
    }

usually I would have written

private static final Map<EntityType, EntityTypeInfo> lookup = new HashMap<>();

static {
    for (MyEntity d : MyEntity.values()){
        lookup.put(d.getEntityType(), d);
    }
}

What is the best practice to write it with java stream?

Is your enum called EntityTypeInfo or MyEntity?

Eran
– Eran

2015-06-29 10:14:47 +00:00
Commented Jun 29, 2015 at 10:14 — Eran
– Eran, Commented Jun 29, 2015 at 10:14

Community · Accepted Answer · 2017-05-23 12:17:25Z

21

I guess there are some typos in your code (the method should be static in my opinion, your constructor is doing a no-op at the moment), but if I'm following you, you can create a stream from the array of enums and use the toMap collector, mapping each enum with its EntityType for the keys, and mapping the instance itself as a value:

private static final Map<EntityType, EntityTypeInfo> lookup =
    Arrays.stream(EntityTypeInfo.values())
          .collect(Collectors.toMap(EntityTypeInfo::getEntityType, e -> e));

The toMap collector does not make any guarantee about the map implementation returned (although it's currently a HashMap), but you can always use the overloaded variant if you need more control, providing a throwing merger as parameter.

You could also use another trick with a static class, and fill the map in the constructor.

edited May 23, 2017 at 12:17

CommunityBot

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answered Jun 29, 2015 at 10:18

Alexis C.

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6 Comments

dustin.schultz Over a year ago

You can improve this with Function.identity() instead of e -> e as Function.identity() always returns the same function.

Alexis C. Over a year ago

@dustin.schultz e -> e is exactly the same, as this lambda expression does not capture values, it will be a singleton that is re-used at each invocation. Furthermore, Function.identity() is implemented as return t -> t;.

dustin.schultz Over a year ago

Ah yes, doesn't capture, so uses singleton. I still think Function.identity() is very-slightly more clear. :)

Alexis C. Over a year ago

@dustin.schultz Matter of taste, I guess. I find e -> e really explicit and shorter but everyone has its own preference ;-).

aliopi Over a year ago

Stream.of() is nicer than Arrays.stream() :)

|

dobrivoje · Accepted Answer · 2020-01-24 09:08:27Z

public enum FibreSpeed {
        a30M( "30Mb Fibre Connection - Broadband Only", 100 ),
        a150M( "150Mb Fibre Connection - Broadband Only", 300 ),
        a1G( "1Gb Fibre Connection - Broadband Only", 500 ),
        b30M( "30Mb Fibre Connection - Broadband & Phone", 700 ),
        b150M( "150Mb Fibre Connection - Broadband & Phone", 900 ),
        b1G( "1Gb Fibre Connection - Broadband & Phone", 1000 );

        public String speed;
        public int    weight;

        FibreSpeed(String speed, int weight) {
            this.speed = speed;
            this.weight = weight;
        }

        public static Map<String, Integer> SPEEDS = Stream.of( values() ).collect( Collectors.toMap( k -> k.speed, v -> v.weight ) );
    }

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