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

JavaLambdaJava 8Java Stream

Java Problem Overview


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);
}

Java Solutions


Solution 1 - Java

Based on Collectors documentation it's as simple as:

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

Solution 2 - Java

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));

Solution 3 - Java

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));

Solution 4 - Java

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) -> {});

Solution 5 - Java

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));
  }

Solution 6 - Java

> One more option in simple way

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

Solution 7 - Java

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));

Solution 8 - Java

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);

Solution 9 - Java

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)));

Solution 10 - Java

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));

Solution 11 - Java

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()
    }
}

Solution 12 - Java

I use this syntax

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

Solution 13 - Java

Map<String, Set<String>> collect = Arrays.asList(Locale.getAvailableLocales()).stream().collect(Collectors
				.toMap(l -> l.getDisplayCountry(), l -> Collections.singleton(l.getDisplayLanguage())));

Solution 14 - Java

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));

Solution 15 - Java

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));
}

Solution 16 - Java

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)));

Solution 17 - Java

Here is solution by StreamEx

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

Solution 18 - Java

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));

Solution 19 - Java

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));
}

Solution 20 - Java

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.

Solution 21 - Java

As an alternative to guava one can use kotlin-stdlib

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

Solution 22 - Java

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}

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