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Java 8 List<V> into Map<K, V>

I want to translate a List of objects into a Map using Java 8's streams and lambdas.

This is how I would write it in Java 7 and below.

private Map<String, Choice> nameMap(List<Choice> choices) {
        final Map<String, Choice> hashMap = new HashMap<>();
        for (final Choice choice : choices) {
            hashMap.put(choice.getName(), choice);
        }
        return hashMap;
}

I can accomplish this easily using Java 8 and Guava but I would like to know how to do this without Guava.

In Guava:

private Map<String, Choice> nameMap(List<Choice> choices) {
    return Maps.uniqueIndex(choices, new Function<Choice, String>() {

        @Override
        public String apply(final Choice input) {
            return input.getName();
        }
    });
}

And Guava with Java 8 lambdas.

private Map<String, Choice> nameMap(List<Choice> choices) {
    return Maps.uniqueIndex(choices, Choice::getName);
}

A
Alexis C.

Based on Collectors documentation it's as simple as:

Map<String, Choice> result =
    choices.stream().collect(Collectors.toMap(Choice::getName,
                                              Function.identity()));

As a side note, even after Java 8, the JDK still can't compete in a brevity contest. The Guava alternative looks so much readable: Maps.uniqueIndex(choices, Choice::getName).
Using (statically imported) Seq from the JOOL library (which I'd recommend to anyone using Java 8), you can also improve the brevity with: seq(choices).toMap(Choice::getName)
Are there any benefits from using Function.identity? I mean, it -> it is shorter
@shabunc I don't know of any benefit and actually use it -> it myself. Function.identity() is used here mostly because it's used in the referenced documentation and that was all I knew about lambdas at the time of writing
@zapl, oh, actually it turns out there are reasons behind this - stackoverflow.com/questions/28032827/…
s
stkent

If your key is NOT guaranteed to be unique for all elements in the list, you should convert it to a Map<String, List<Choice>> instead of a Map<String, Choice>

Map<String, List<Choice>> result =
 choices.stream().collect(Collectors.groupingBy(Choice::getName));

This actually gives you Map> which deals with the possibility of non-unique keys, but isn't what the OP requested. In Guava, Multimaps.index(choices, Choice::getName) is probably a better option if this is what you want anyhow.
or rather use Guava's Multimap which comes quite handy in scenarios where same key maps to multiple values. There are various utility methods readily available in Guava to use such data structures rather than creating a Map>
@RichardNichols why is the guava Multimaps method a better option? It can be an inconvenience since it doesn't return a Map Object.
@RichardNichols it might no be what OP requested but I was looking for exactly this and am so happy that this answer exists!
O
Oleksandr Pyrohov

Use getName() as the key and Choice itself as the value of the map:

Map<String, Choice> result =
    choices.stream().collect(Collectors.toMap(Choice::getName, c -> c));

Please write some description so that user can understand.
It's really too bad there isn't more details here, because I like this answer best.
Collectors.toMap(Choice::getName,c->c) (2 chars shorter)
It's equal to choices.stream().collect(Collectors.toMap(choice -> choice.getName(),choice -> choice)); First function for key, second function for value
I know how easy it is to see and understand c -> c but Function.identity() carries more semantic information. I usually use a static import so that i can just use identity()
S
Sahil Chhabra

Most of the answers listed, miss a case when the list has duplicate items. In that case there answer will throw IllegalStateException. Refer the below code to handle list duplicates as well:

public Map<String, Choice> convertListToMap(List<Choice> choices) {
    return choices.stream()
        .collect(Collectors.toMap(Choice::getName, choice -> choice,
            (oldValue, newValue) -> newValue));
  }

E
Emre Colak

Here's another one in case you don't want to use Collectors.toMap()

Map<String, Choice> result =
   choices.stream().collect(HashMap<String, Choice>::new, 
                           (m, c) -> m.put(c.getName(), c),
                           (m, u) -> {});

Which is better to use then Collectors.toMap() or our own HashMap as you showed in above example?
This example provided an example for how to place something else in the map. I wanted a value not provided by a method call. Thanks!
The third argument function is not correct. There you should provide some function to merge two Hashmaps, something like Hashmap::putAll
R
Renukeswar

One more option in simple way

Map<String,Choice> map = new HashMap<>();
choices.forEach(e->map.put(e.getName(),e));

There is no viable difference in using this or java 7 type.
The SO asked with Java 8 Streams.
S
Sahil Chhabra

For example, if you want convert object fields to map:

Example object:

class Item{
        private String code;
        private String name;

        public Item(String code, String name) {
            this.code = code;
            this.name = name;
        }

        //getters and setters
    }

And operation convert List To Map:

List<Item> list = new ArrayList<>();
list.add(new Item("code1", "name1"));
list.add(new Item("code2", "name2"));

Map<String,String> map = list.stream()
     .collect(Collectors.toMap(Item::getCode, Item::getName));

J
John McClean

If you don't mind using 3rd party libraries, AOL's cyclops-react lib (disclosure I am a contributor) has extensions for all JDK Collection types, including List and Map.

ListX<Choices> choices;
Map<String, Choice> map = choices.toMap(c-> c.getName(),c->c);

V
Vikas Suryawanshi

You can create a Stream of the indices using an IntStream and then convert them to a Map :

Map<Integer,Item> map = 
IntStream.range(0,items.size())
         .boxed()
         .collect(Collectors.toMap (i -> i, i -> items.get(i)));

This is not a good option because you do a get() call for every element, hence increasing the complexity of the operation ( o(n * k) if items is an hashmap).
Isn't get(i) on a hashmap O(1)?
@IvovanderVeeken the get(i) in the code snippet is on the list, not the map.
@Zaki I was talking about Nicolas' remark. I don't see the n*k complexity if items is a hashmap instead of a list.
i
iZian

I was trying to do this and found that, using the answers above, when using Functions.identity() for the key to the Map, then I had issues with using a local method like this::localMethodName to actually work because of typing issues.

Functions.identity() actually does something to the typing in this case so the method would only work by returning Object and accepting a param of Object

To solve this, I ended up ditching Functions.identity() and using s->s instead.

So my code, in my case to list all directories inside a directory, and for each one use the name of the directory as the key to the map and then call a method with the directory name and return a collection of items, looks like:

Map<String, Collection<ItemType>> items = Arrays.stream(itemFilesDir.listFiles(File::isDirectory))
.map(File::getName)
.collect(Collectors.toMap(s->s, this::retrieveBrandItems));

g
grep

I will write how to convert list to map using generics and inversion of control. Just universal method!

Maybe we have list of Integers or list of objects. So the question is the following: what should be key of the map?

create interface

public interface KeyFinder<K, E> {
    K getKey(E e);
}

now using inversion of control:

  static <K, E> Map<K, E> listToMap(List<E> list, KeyFinder<K, E> finder) {
        return  list.stream().collect(Collectors.toMap(e -> finder.getKey(e) , e -> e));
    }

For example, if we have objects of book , this class is to choose key for the map

public class BookKeyFinder implements KeyFinder<Long, Book> {
    @Override
    public Long getKey(Book e) {
        return e.getPrice()
    }
}

u
user2069723

I use this syntax

Map<Integer, List<Choice>> choiceMap = 
choices.stream().collect(Collectors.groupingBy(choice -> choice.getName()));

groupingBy creates a Map<K,List<V>>, not a Map<K,V>.
Dup of ulises answer. And, String getName(); (not Integer)
T
Tunaki
Map<String, Set<String>> collect = Arrays.asList(Locale.getAvailableLocales()).stream().collect(Collectors
                .toMap(l -> l.getDisplayCountry(), l -> Collections.singleton(l.getDisplayLanguage())));

r
raja emani

This can be done in 2 ways. Let person be the class we are going to use to demonstrate it.

public class Person {

    private String name;
    private int age;

    public String getAge() {
        return age;
    }
}

Let persons be the list of Persons to be converted to the map

1.Using Simple foreach and a Lambda Expression on the List

Map<Integer,List<Person>> mapPersons = new HashMap<>();
persons.forEach(p->mapPersons.put(p.getAge(),p));

2.Using Collectors on Stream defined on the given List.

 Map<Integer,List<Person>> mapPersons = 
           persons.stream().collect(Collectors.groupingBy(Person::getAge));

K
Konrad Borowski

It's possible to use streams to do this. To remove the need to explicitly use Collectors, it's possible to import toMap statically (as recommended by Effective Java, third edition).

import static java.util.stream.Collectors.toMap;

private static Map<String, Choice> nameMap(List<Choice> choices) {
    return choices.stream().collect(toMap(Choice::getName, it -> it));
}

L
L. G.

Another possibility only present in comments yet:

Map<String, Choice> result =
choices.stream().collect(Collectors.toMap(c -> c.getName(), c -> c)));

Useful if you want to use a parameter of a sub-object as Key:

Map<String, Choice> result =
choices.stream().collect(Collectors.toMap(c -> c.getUser().getName(), c -> c)));

u
user_3380739

Here is solution by StreamEx

StreamEx.of(choices).toMap(Choice::getName, c -> c);

R
Rajeev Akotkar
Map<String,Choice> map=list.stream().collect(Collectors.toMap(Choice::getName, s->s));

Even serves this purpose for me,

Map<String,Choice> map=  list1.stream().collect(()-> new HashMap<String,Choice>(), 
            (r,s) -> r.put(s.getString(),s),(r,s) -> r.putAll(s));

I
Ihor Sakailiuk

If every new value for the same key name has to be overridden:

public Map < String, Choice > convertListToMap(List < Choice > choices) {
    return choices.stream()
        .collect(Collectors.toMap(Choice::getName,
            Function.identity(),
            (oldValue, newValue) - > newValue));
}

If all choices have to be grouped in a list for a name:

public Map < String, Choice > convertListToMap(List < Choice > choices) {
    return choices.stream().collect(Collectors.groupingBy(Choice::getName));
}

D
Dino
List<V> choices; // your list
Map<K,V> result = choices.stream().collect(Collectors.toMap(choice::getKey(),choice));
//assuming class "V" has a method to get the key, this method must handle case of duplicates too and provide a unique key.

F
Frank Neblung

As an alternative to guava one can use kotlin-stdlib

private Map<String, Choice> nameMap(List<Choice> choices) {
    return CollectionsKt.associateBy(choices, Choice::getName);
}

K
Karthikeyan
String array[] = {"ASDFASDFASDF","AA", "BBB", "CCCC", "DD", "EEDDDAD"};
    List<String> list = Arrays.asList(array);
    Map<Integer, String> map = list.stream()
            .collect(Collectors.toMap(s -> s.length(), s -> s, (x, y) -> {
                System.out.println("Dublicate key" + x);
                return x;
            },()-> new TreeMap<>((s1,s2)->s2.compareTo(s1))));
    System.out.println(map);

Dublicate key AA

{12=ASDFASDFASDF, 7=EEDDDAD, 4=CCCC, 3=BBB, 2=AA}

What are you trying to do here ?? Did you read the question ?