CODEWITHSUNDEEP
javahashmap
Mavin
Mavin

Reputation: 3249

How can I combine two HashMap objects containing the same types?

I have two HashMap objects defined like so:

HashMap<String, Integer> map1 = new HashMap<String, Integer>();
HashMap<String, Integer> map2 = new HashMap<String, Integer>();

I also have a third HashMap object:

HashMap<String, Integer> map3;

How can I merge map1 and map2 together into map3?

Upvotes: 315

Views: 400045

Answers (18)

Ismail Sahin
Ismail Sahin

Reputation: 2710

I use this to initialize in a single line,

final Map<String, Object> map = new HashMap<>() {{
    putAll(map1);
    putAll(map2);
}};

CAUTION: DO NOT USE ABOVE ONE! After reading below comment from @Ian Robertson, I quit using this.

Upvotes: 0

Mermaid Man
Mermaid Man

Reputation: 107

new HashMap(Map.of("k1", 1) 
{{   putAll(Map.of("k2", 2)}}

//Options 2: if you don't wanna use an Anonymous inner class, you can:
Map m;
(m = new HashMap(Map.of("k1", 1)))
         .putAll(Map.of("k2", 2));

This is the shortest way to concat two Maps into a new Map, as far as I'm aware. For brevities sake, I use Map.of a lot for small Map sizes. Polymorphically, Map.of provides methods accepting between 0 & at most 10 key/value pairs, and when that's not enough, I usually write something like this to concat two Map.of Maps.

Oracle docs for creating immutable Lists, Sets, and Maps

Upvotes: 0

Svend
Svend

Reputation: 7170

Assuming the following input:

    import java.util.stream.Stream;
    import java.util.stream.Collectors;
    import java.util.Map;
   
    ...

    var m1 = Map.of("k1", 1, "k2", 2);
    var m2 = Map.of("k3", 3, "k4", 4);

When you're sure not to have any key collisions between both input maps, a simple expression that avoids any mutations and yields an immutable result could be:

    var merged = Stream.concat(
        m1.entrySet().stream(),
        m2.entrySet().stream()
    ).collect(Collectors.toUnmodifiableMap(Map.Entry::getKey, Map.Entry::getValue));

In case key collisions are possible, we can provide a lambda to specify how to de-duplicate them. For example if we'd like to keep the largest value in case an entry is present in both input, we could:

    .collect(Collectors.toUnmodifiableMap(
        Map.Entry::getKey,
        Map.Entry::getValue,
        Math::max))  // any function  (Integer, Integer) -> Integer  is ok here

Upvotes: 2

Soudipta Dutta
Soudipta Dutta

Reputation: 2122

Method 1: Put maps in a List and then join

public class Test15 {
public static void main(String[] args) {

    Map<String, List<String>> map1 = new HashMap<>();
    map1.put("London", Arrays.asList("A", "B", "C"));
    map1.put("Wales", Arrays.asList("P1", "P2", "P3"));

    Map<String, List<String>> map2 = new HashMap<>();
    map2.put("Calcutta", Arrays.asList("Protijayi", "Gina", "Gini"));
    map2.put("London", Arrays.asList( "P4", "P5", "P6"));
    map2.put("Wales", Arrays.asList( "P111", "P5555", "P677666"));
    
    System.out.println(map1);System.out.println(map2);
    
    
    
    // put the maps in an ArrayList
    
    List<Map<String, List<String>>> maplist = new ArrayList<Map<String,List<String>>>();
    maplist.add(map1);
    maplist.add(map2);
    /*
<T,K,U> Collector<T,?,Map<K,U>> toMap(

                                  Function<? super T,? extends K> keyMapper,

                                  Function<? super T,? extends U> valueMapper,

                                  BinaryOperator<U> mergeFunction)
    */
    
 Map<String, List<String>> collect = maplist.stream()
    .flatMap(ch -> ch.entrySet().stream())
    .collect(
            Collectors.toMap(
            
            //keyMapper,
            
            Entry::getKey,
            
            //valueMapper
            Entry::getValue,
            
            // mergeFunction
     (list_a,list_b) -> Stream.concat(list_a.stream(), list_b.stream()).collect(Collectors.toList())
            
            ));
    
    
    
    System.out.println("Final Result(Map after join) => " + collect);
    /*
    {Wales=[P1, P2, P3], London=[A, B, C]}
{Calcutta=[Protijayi, Gina, Gini], Wales=[P111, P5555, P677666], London=[P4, P5, P6]}
Final Result(Map after join) => {Calcutta=[Protijayi, Gina, Gini], Wales=[P1, P2, P3, P111, P5555, P677666], London=[A, B, C, P4, P5, P6]}
*/
    
}//main


}

Method 2 : Normal Map merge

public class Test15 {
public static void main(String[] args) {

    Map<String, List<String>> map1 = new HashMap<>();
    map1.put("London", Arrays.asList("A", "B", "C"));
    map1.put("Wales", Arrays.asList("P1", "P2", "P3"));

    Map<String, List<String>> map2 = new HashMap<>();
    map2.put("Calcutta", Arrays.asList("Protijayi", "Gina", "Gini"));
    map2.put("London", Arrays.asList( "P4", "P5", "P6"));
    map2.put("Wales", Arrays.asList( "P111", "P5555", "P677666"));
    
    System.out.println(map1);System.out.println(map2);
    
    
    

    /*
<T,K,U> Collector<T,?,Map<K,U>> toMap(

                                  Function<? super T,? extends K> keyMapper,

                                  Function<? super T,? extends U> valueMapper,

                                  BinaryOperator<U> mergeFunction)
    */
    
    
Map<String, List<String>> collect = Stream.of(map1,map2)
    .flatMap(ch -> ch.entrySet().stream())
    .collect(
            Collectors.toMap(
            
            //keyMapper,
            
            Entry::getKey,
            
            //valueMapper
            Entry::getValue,
            
            // mergeFunction
     (list_a,list_b) -> Stream.concat(list_a.stream(), list_b.stream()).collect(Collectors.toList())
            
            ));
    
    
    
    System.out.println("Final Result(Map after join) => " + collect);
    /*
    {Wales=[P1, P2, P3], London=[A, B, C]}
{Calcutta=[Protijayi, Gina, Gini], Wales=[P111, P5555, P677666], London=[P4, P5, P6]}
Final Result(Map after join) => {Calcutta=[Protijayi, Gina, Gini], Wales=[P1, P2, P3, P111, P5555, P677666], London=[A, B, C, P4, P5, P6]}

*/
    
}//main


}

In Python, HashMap is called Dictionary and we can merge them very easily.

x = {'Roopa': 1, 'Tabu': 2}
y = {'Roopi': 3, 'Soudipta': 4}


z = {**x,**y}
print(z)

{'Roopa': 1, 'Tabu': 2, 'Roopi': 3, 'Soudipta': 4}

Upvotes: 0

P Mittal
P Mittal

Reputation: 172

You can use putAll function for Map as explained in the code below

HashMap<String, Integer> map1 = new HashMap<String, Integer>();
map1.put("a", 1);
map1.put("b", 2);
map1.put("c", 3);
HashMap<String, Integer> map2 = new HashMap<String, Integer>();
map1.put("aa", 11);
map1.put("bb", 12);
HashMap<String, Integer> map3 = new HashMap<String, Integer>();
map3.putAll(map1);
map3.putAll(map2);
map3.keySet().stream().forEach(System.out::println);
map3.values().stream().forEach(System.out::println);

Upvotes: 3

Ishan Bhatt
Ishan Bhatt

Reputation: 10231

Very late but let me share what I did when I had the same issue.

Map<String, List<String>> map1 = new HashMap<>();
map1.put("India", Arrays.asList("Virat", "Mahi", "Rohit"));
map1.put("NZ", Arrays.asList("P1","P2","P3"));

Map<String, List<String>> map2 = new HashMap<>();
map2.put("India", Arrays.asList("Virat", "Mahi", "Rohit"));
map2.put("NZ", Arrays.asList("P1","P2","P4"));

Map<String, List<String>> collect4 = Stream.of(map1, map2)
                .flatMap(map -> map.entrySet().stream())
                .collect(
                        Collectors.toMap(
                                Map.Entry::getKey,
                                Map.Entry::getValue,
                                (strings, strings2) -> {
                                    List<String> newList = new ArrayList<>();
                                    newList.addAll(strings);
                                    newList.addAll(strings2);
                                    return newList;
                                }
                        )
                );
collect4.forEach((s, strings) -> System.out.println(s+"->"+strings));

It gives the following output

NZ->[P1, P2, P3, P1, P2, P4]
India->[Virat, Mahi, Rohit, Virat, Mahi, Rohit]

Upvotes: 7

Karunesh
Karunesh

Reputation: 11

    HashMap<Integer,String> hs1 = new HashMap<>();
    hs1.put(1,"ram");
    hs1.put(2,"sita");
    hs1.put(3,"laxman");
    hs1.put(4,"hanuman");
    hs1.put(5,"geeta");

    HashMap<Integer,String> hs2 = new HashMap<>();
    hs2.put(5,"rat");
    hs2.put(6,"lion");
    hs2.put(7,"tiger");
    hs2.put(8,"fish");
    hs2.put(9,"hen");

    HashMap<Integer,String> hs3 = new HashMap<>();//Map is which we add

    hs3.putAll(hs1);
    hs3.putAll(hs2);

    System.out.println(" hs1 : " + hs1);
    System.out.println(" hs2 : " + hs2);
    System.out.println(" hs3 : " + hs3);

Duplicate items will not be added(that is duplicate keys) as when we will print hs3 we will get only one value for key 5 which will be last value added and it will be rat. **[Set has a property of not allowing the duplicate key but values can be duplicate]

Upvotes: 1

Omer Vishlitzky
Omer Vishlitzky

Reputation: 76

you can use HashMap<String, List<Integer>> to merge both hashmaps and avoid losing elements paired with the same key.

HashMap<String, Integer> map1 = new HashMap<>();
HashMap<String, Integer> map2 = new HashMap<>();
map1.put("key1", 1);
map1.put("key2", 2);
map1.put("key3", 3);
map2.put("key1", 4);
map2.put("key2", 5);
map2.put("key3", 6);
HashMap<String, List<Integer>> map3 = new HashMap<>();
map1.forEach((str, num) -> map3.put(str, new ArrayList<>(Arrays.asList(num))));
//checking for each key if its already in the map, and if so, you just add the integer to the list paired with this key
for (Map.Entry<String, Integer> entry : map2.entrySet()) {
    Integer value = entry.getValue();
    String key = entry.getKey();
    if (map3.containsKey(key)) {
        map3.get(key).add(value);
    } else {
        map3.put(key, new ArrayList<>(Arrays.asList(value)));
    }
}
map3.forEach((str, list) -> System.out.println("{" + str + ": " + list + "}"));

output:

{key1: [1, 4]}
{key2: [2, 5]}
{key3: [3, 6]}

Upvotes: 3

Thomas Decaux
Thomas Decaux

Reputation: 22651

A small snippet I use very often to create map from other maps:

static public <K, V> Map<K, V> merge(Map<K, V>... args) {
    final Map<K, V> buffer = new HashMap<>();

    for (Map m : args) {
        buffer.putAll(m);
    }

    return buffer;
}

Upvotes: 4

Vadzim
Vadzim

Reputation: 26160

Java 8 alternative one-liner for merging two maps:

defaultMap.forEach((k, v) -> destMap.putIfAbsent(k, v));

The same with method reference:

defaultMap.forEach(destMap::putIfAbsent);

Or idemponent for original maps solution with third map:

Map<String, Integer> map3 = new HashMap<String, Integer>(map2);
map1.forEach(map3::putIfAbsent);

And here is a way to merge two maps into fast immutable one with Guava that does least possible intermediate copy operations:

ImmutableMap.Builder<String, Integer> builder = ImmutableMap.<String, Integer>builder();
builder.putAll(map1);
map2.forEach((k, v) -> {if (!map1.containsKey(k)) builder.put(k, v);});
ImmutableMap<String, Integer> map3 = builder.build();

See also Merge two maps with Java 8 for cases when values present in both maps need to be combined with mapping function.

Upvotes: 44

ZhekaKozlov
ZhekaKozlov

Reputation: 39526

Generic solution for combining two maps which can possibly share common keys:

In-place:

public static <K, V> void mergeInPlace(Map<K, V> map1, Map<K, V> map2,
        BinaryOperator<V> combiner) {
    map2.forEach((k, v) -> map1.merge(k, v, combiner::apply));
}

Returning a new map:

public static <K, V> Map<K, V> merge(Map<K, V> map1, Map<K, V> map2,
        BinaryOperator<V> combiner) {
    Map<K, V> map3 = new HashMap<>(map1);
    map2.forEach((k, v) -> map3.merge(k, v, combiner::apply));
    return map3;
}

Upvotes: 13

user330315
user330315

Reputation: 

map3 = new HashMap<>();

map3.putAll(map1);
map3.putAll(map2);

Upvotes: 436

Ashish Shetkar
Ashish Shetkar

Reputation: 1467

you can use - addAll method

http://download.oracle.com/javase/6/docs/api/java/util/HashMap.html

But there is always this issue that - if your two hash maps have any key same - then it will override the value of the key from first hash map with the value of the key from second hash map.

For being on safer side - change the key values - you can use prefix or suffix on the keys - ( different prefix/suffix for first hash map and different prefix/suffix for second hash map )

Upvotes: -1

Vitalii Fedorenko
Vitalii Fedorenko

Reputation: 114400

One-liner using Java 8 Stream API:

map3 = Stream.of(map1, map2).flatMap(m -> m.entrySet().stream())
       .collect(Collectors.toMap(Entry::getKey, Entry::getValue))

Among the benefits of this method is ability to pass a merge function, which will deal with values that have the same key, for example:

map3 = Stream.of(map1, map2).flatMap(m -> m.entrySet().stream())
       .collect(Collectors.toMap(Entry::getKey, Entry::getValue, Math::max))

Upvotes: 87

Jeffrey Bosboom
Jeffrey Bosboom

Reputation: 13653

If you know you don't have duplicate keys, or you want values in map2 to overwrite values from map1 for duplicate keys, you can just write

map3 = new HashMap<>(map1);
map3.putAll(map2);

If you need more control over how values are combined, you can use Map.merge, added in Java 8, which uses a user-provided BiFunction to merge values for duplicate keys. merge operates on individual keys and values, so you'll need to use a loop or Map.forEach. Here we concatenate strings for duplicate keys:

map3 = new HashMap<>(map1);
for (Map.Entry<String, String> e : map2.entrySet())
    map3.merge(e.getKey(), e.getValue(), String::concat);
//or instead of the above loop
map2.forEach((k, v) -> map3.merge(k, v, String::concat));

If you know you don't have duplicate keys and want to enforce it, you can use a merge function that throws an AssertionError:

map2.forEach((k, v) ->
    map3.merge(k, v, (v1, v2) ->
        {throw new AssertionError("duplicate values for key: "+k);}));

Taking a step back from this specific question, the Java 8 streams library provides toMap and groupingBy Collectors. If you're repeatedly merging maps in a loop, you may be able to restructure your computation to use streams, which can both clarify your code and enable easy parallelism using a parallel stream and concurrent collector.

Upvotes: 131

hvgotcodes
hvgotcodes

Reputation: 120138

HashMap has a putAll method.

http://download.oracle.com/javase/6/docs/api/java/util/HashMap.html

Upvotes: 31

Grzegorz Rożniecki
Grzegorz Rożniecki

Reputation: 27995

If you don't need mutability for your final map, there is Guava's ImmutableMap with its Builder and putAll method which, in contrast to Java's Map interface method, can be chained.

Example of use:

Map<String, Integer> mergeMyTwoMaps(Map<String, Integer> map1, Map<String, Integer> map2) {
  return ImmutableMap.<String, Integer>builder()
      .putAll(map1)
      .putAll(map2)
      .build();
}

Of course, this method can be more generic, use varargs and loop to putAll Maps from arguments etc. but I wanted to show a concept.

Also, ImmutableMap and its Builder have few limitations (or maybe features?):

  • they are null hostile (throw NullPointerException - if any key or value in map is null)
  • Builder don't accept duplicate keys (throws IllegalArgumentException if duplicate keys were added).

Upvotes: 36

fastcodejava
fastcodejava

Reputation: 41077

You could use Collection.addAll() for other types, e.g. List, Set, etc. For Map, you can use putAll.

Upvotes: 24

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