How to putAll on Java hashMap contents of one to another, but not replace existing keys and values?

Question

I need to copy all keys and values from one A HashMap onto another one B, but not to replace existing keys and values.

Whats the best way to do that?

I was thinking instead iterating the keySet and checkig if it exist or not, I would

Map temp = new HashMap(); // generic later
temp.putAll(Amap);
A.clear();
A.putAll(Bmap);
A.putAll(temp);

score 122 · Accepted Answer · edited May 23 '17 at 10:31

122

It looks like you are willing to create a temporary Map, so I'd do it like this:

Map tmp = new HashMap(patch);
tmp.keySet().removeAll(target.keySet());
target.putAll(tmp);

Here, patch is the map that you are adding to the target map.

Thanks to Louis Wasserman, here's a version that takes advantage of the new methods in Java 8:

patch.forEach(target::putIfAbsent);

edited May 23 '17 at 10:31

Community

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answered Aug 25 '11 at 17:50

erickson

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In this case you don't need to do `removeAll` since they'll get overriden anyways. – Reverend Gonzo Aug 25 '11 at 18:26
@Reverend Gonzo: Overwritten by what? – erickson Aug 25 '11 at 18:34
@luksmir No, that would remove the existing entries that should be preserved. – erickson Sep 08 '14 at 22:03
Hi! Why mapWithValues.putAll(mapWithAnotherValues) doens't work? – betomoretti Nov 25 '15 at 17:02
@betomoretti Because if both maps have a value for a key, that would replace the original value with another. That would violate the requirement "not to replace existing keys and values." – erickson Nov 25 '15 at 17:28
@erickson I messed up the requirement. But , in case of that requirement doesn't exists, it's work? Sorry for my English and thanks! – betomoretti Nov 26 '15 at 14:03
@erickson Well, mapWithValues.putAll(mapWithAnotherValues) void mapWithValues. So, this it's not a good solution. In fact, doesn't work. – betomoretti Nov 26 '15 at 15:57
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@erickson a much more compact Java 8 version: `patch.forEach(target::putIfAbsent)`. – Louis Wasserman Jan 05 '16 at 20:08

score 16 · Answer 2 · answered Aug 25 '11 at 17:41

Using Guava's Maps class' utility methods to compute the difference of 2 maps you can do it in a single line, with a method signature which makes it more clear what you are trying to accomplish:

public static void main(final String[] args) {
    // Create some maps
    final Map<Integer, String> map1 = new HashMap<Integer, String>();
    map1.put(1, "Hello");
    map1.put(2, "There");
    final Map<Integer, String> map2 = new HashMap<Integer, String>();
    map2.put(2, "There");
    map2.put(3, "is");
    map2.put(4, "a");
    map2.put(5, "bird");

    // Add everything in map1 not in map2 to map2
    map2.putAll(Maps.difference(map1, map2).entriesOnlyOnLeft());
}

Reverend Gonzo · Answer 3 · 2011-08-25T17:42:45.203

8

Just iterate and add:

for(Map.Entry e : a.entrySet())
  if(!b.containsKey(e.getKey())
    b.put(e.getKey(), e.getValue());

Edit to add:

If you can make changes to a, you can also do:

a.putAll(b)

and a will have exactly what you need. (all the entries in b and all the entries in a that aren't in b)

edited Aug 25 '11 at 17:42

answered Aug 25 '11 at 17:28

Reverend Gonzo

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1.) `entrySet()` returns a `Set`. 2.) There is no method `contains(...)`for `java.util.Map`. – Fabian Barney Aug 25 '11 at 17:39
Isnt there a smarter way? Like a Collections utility or so? I hate all the for loops everywhere. – rapadura Aug 25 '11 at 17:39
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@AntonioP: All the for loops everywhere? There's one for loop and it's about as straightforward as you can get. – Reverend Gonzo Aug 25 '11 at 17:44
+1 then, just works when Maps used not with raw type, I think. Got myself a bit fuddled with the Map.Entry :-) – Fabian Barney Aug 25 '11 at 17:50
@Reverend Gonzo, ah, elsewhere in the code. :) Wont .putAll replace existing keys and values of a? Ill go with this sense it does seem the most straightforward way. – rapadura Aug 25 '11 at 18:24
@AntonioP, if you want to do something custom you can't avoid custom code (like looping). – Steve Kuo Aug 25 '11 at 18:34
I guess this is better for performance than to create a whole new map? – Simon Baars Nov 06 '17 at 14:57

score 6 · Answer 4 · answered Mar 22 '16 at 14:29

You can make it in just 1 line if you change maps order in @erickson's solution:

mapWithNotSoImportantValues.putAll( mapWithImportantValues );

In this case you replace values in mapWithNotSoImportantValues with value from mapWithImportantValues with the same keys.

Nikolas Charalambidis · Answer 5 · 2021-01-22T11:18:23.133

Java 8 solution using `Map#merge`

As of java-8 you can use Map#merge(K key, V value, BiFunction remappingFunction) which merges a value into the Map using remappingFunction in case the key is already found in the Map you want to put the pair into.

// using lambda
newMap.forEach((key, value) -> map.merge(key, value, (oldValue, newValue) -> oldValue));

// using for-loop
for (Map.Entry<Integer, String> entry: newMap.entrySet()) {
    map.merge(entry.getKey(), entry.getValue(), (oldValue, newValue) -> oldValue);
}

The code iterates the newMap entries (key and value) and each one is merged into map through the method merge. The remappingFunction is triggered in case of duplicated key and in that case it says that the former (original) oldValue value will be used and not rewritten.

With this solution, you don't need a temporary Map.

Let's have an example of merging newMap entries into map and keeping the original values in case of the duplicated antry.

Map<Integer, String> newMap = new HashMap<>();
newMap.put(2, "EVIL VALUE");                         // this will NOT be merged into
newMap.put(4, "four");                               // this WILL be merged into
newMap.put(5, "five");                               // this WILL be merged into

Map<Integer, String> map = new HashMap<>();
map.put(1, "one");
map.put(2, "two");
map.put(3, "three");

newMap.forEach((k, v) -> map.merge(k, v, (oldValue, newValue) -> oldValue));

map.forEach((k, v) -> System.out.println(k + " " + v));

1 one
2 two
3 three
4 four
5 five

score 2 · Answer 6 · answered Dec 28 '16 at 18:49

public class MyMap {

public static void main(String[] args) {

    Map<String, String> map1 = new HashMap<String, String>();
    map1.put("key1", "value1");
    map1.put("key2", "value2");
    map1.put("key3", "value3");
    map1.put(null, null);

    Map<String, String> map2 = new HashMap<String, String>();
    map2.put("key4", "value4");
    map2.put("key5", "value5");
    map2.put("key6", "value6");
    map2.put("key3", "replaced-value-of-key3-in-map2");
    // used only if map1 can be changes/updates with the same keys present in map2.
    map1.putAll(map2);

    // use below if you are not supposed to modify the map1.
    for (Map.Entry e : map2.entrySet())
        if (!map1.containsKey(e.getKey()))
            map1.put(e.getKey().toString(), e.getValue().toString());
    System.out.println(map1);
}}

score 2 · Answer 7 · answered Feb 12 '17 at 21:25

2

With Java 8 there is this API method to accomplish your requirement.

map.putIfAbsent(key, value)

If the specified key is not already associated with a value (or is mapped to null) associates it with the given value and returns null, else returns the current value.

answered Feb 12 '17 at 21:25

newday

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can we have putAllIfAbsent ? thats what I need ? – user1735921 Sep 20 '18 at 10:40

score 2 · Answer 8 · answered Oct 02 '20 at 10:04

2

As others have said, you can use putIfAbsent. Iterate over each entry in the map that you want to insert, and invoke this method on the original map:

mapToInsert.forEach(originalMap::putIfAbsent);

answered Oct 02 '20 at 10:04

whistling_marmot

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How to putAll on Java hashMap contents of one to another, but not replace existing keys and values?

8 Answers8

Java 8 solution using `Map#merge`

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How to putAll on Java hashMap contents of one to another, but not replace existing keys and values?

8 Answers8

Java 8 solution using Map#merge

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Java 8 solution using `Map#merge`