Stream groupingBy: reducing to first element of list

Question

I have a List<Valuta> which can be represented (simplified) JSON-style:

[ { codice=EUR, description=Euro, ratio=1 }, { codice=USD, description=Dollars, ratio=1.1 } ]

I want to transform that in a Map<String, Valuta> like this:

{ EUR={ codice=EUR, description=Euro, ratio=1 }, USD={ codice=USD, description=Dollars, ratio=1.1 }}

I wrote this one-liner:

getValute().stream().collect(Collectors.groupingBy(Valuta::getCodice));

but this returns a Map<String, List<Valuta>> instead of what I need.

I suppose mapping() function would work for me, but don't know how.

Solution 1

Actually, you need to use Collectors.toMap here instead of Collectors.groupingBy:

Map<String, Valuta> map = 
    getValute().stream()
               .collect(Collectors.toMap(Valuta::getCodice, Function.identity()));

groupingBy is used to group elements of a Stream based on a grouping function. 2 Stream elements that will have the same result with the grouping function will be collected into a List by default.

toMap will collect the elements into a Map where the key is the result of applying a given key mapper and the value is the result of applying a value mapper. Note that toMap, by default, will throw an exception if a duplicate is encountered.

Solution 2

It's a bit late in the game, but try this:

Map<String, Valuta> map = 
    getValute().stream()
               .collect(Collectors.groupingBy(Valuta::getCodice,
                            Collectors.collectingAndThen(
                                Collectors.toList(), 
                                values -> values.get(0))));

Solution 3

You could use Collectors.toMap(keyMappingFunction, valueMappingFunction)

Map<String, Valuta> map = list
        .stream()
        .collect(Collectors.toMap(Valuta::getCodice, v -> v));

You can replace v->v with Function.identity() if you find it more readable. Note that toMap, by default, will throw an exception if a duplicate is encountered.

Solution 4

The toMap version which opts to choose the 1st value on collisions instead of throwing an exception is:

Collectors.toMap(keyMapper, valueMapper, mergeFunction) ex:

Map<String, Valuta> map = list
        .stream()
        .collect(Collectors.toMap(Valuta::getCodice, v -> v, (v1, v2) -> v1));

Solution 5

Not completely related to this question but quiet similar case .If you want to group list by multiple params and after that need return object of different type as value then you can try this solution . Hope helps to someone!

public Map<Integer, Map<String, ObjectTypeB>> setObjectTypeB(List<ObjectTypeA> typeAList) {

   Map<Integer, Map<String, ObjectTypeB>> map = typeAList.stream()
        .collect(groupingBy(ObjectTypeA::getId,
         groupingBy(ObjectTypeA::getDate, 
         collectingAndThen(mapping(this::getObjectTypeB,toList()),values -> values.get(0)))));    
    return map;
}

    public ObjectTypeB getObjectTypeB(ObjectTypeA typeA) {
       ObjectTypeB typeB = new ObjectTypeB();
       typeB.setUnits(typeA.getUnits());
       return typeB;
}

View more solutions

Comments

[user909481]

How to do the same with groupingBy? I cannot use toMap because I must expect duplicate keys from the key mapper. I see groupingBy has signature for downstream collector as 2nd argument but couldn't make it work yet.

[BaiJiFeiLong]

This is what i am finding for

[fricke]

this is what's needed in case you actually do have duplicates, but don't want to use it

[user1735921]

Lets say I want Map<String, String> instead of Map<String, Valuta>, I need Valuta.getDescription() , basically, is that possible?

[jocull]

Be aware that Collectors.toMap will throw an exception if you somehow have duplicate keys! They are not completely equivalent.

[jocull]

You might be able to use the mergeFunction of .toMap to simply drop any duplicates and avoid extra object allocations: stackoverflow.com/a/32313069/97964

[jocull]

Discovered this - you might be able to use the mergeFunction of .toMap to simply drop or replace any duplicates and avoid extra object allocations: stackoverflow.com/a/32313069/97964

[toongeorges]

Why create a list in the stream? Here your solution adapted without creating the list: getValute().stream().collect(groupingBy(Valuta::getCodice, reducing(null, identity(), (first, last) -> last)))

[rogerdpack]

These appear to be three equivalent ways to perform the default groupingBy behavior. Which is interesting but I don't think what the OP wanted (which was a single k -> v). :)

[andresp]

Does the merge function preserve ordering? i.e. is it guaranteed that v1 always appears before v2 in list?

[3ygun]

@andresp If you're working with stream() called on an ArrayList (which uses it's ordered iterator) than yes. Otherwise, it's dependent on your collection see something like stackoverflow.com/a/40583642/6480404