连接两个哈希映射而不从两个映射中删除常见的entires

Question

I have two hashmaps, in particular vocabs of two languages say english and german.I would like to concatenate both these map to return a single map.I tried :

 hashmap.putall()

But, removed some of the entries which are common in both maps and replace it by single entry only.But i want to keep both the vocabs intact just concatenate those. Is there any method to do it? if not any other way to do. I would prefer any methods in hashmap.

[EDIT]

To make more clear, lets see two maps

      at the 500    um die 500
      0   1   2     0   1   2

resutls into

  at the  500 um die 500
  0   1    2   3  4   5               

Answer 1

You'll have to write your own custom "putAll()` method then. Something like this would work:

HashMap<String> both = new HashMap<String>(english);

for(String key : german.keySet()) {
    if(english.containsKey(key)) {
        both.put(key, english.get(key)+german.get(key));
    }
}

This first copies the English HashMap . Then puts in all the German words, concatenating if there is a duplicate key. You might want some kind of separator character like a / in between so you can later extract the two.

Answer 2

There isn't anything like that in the Java main library itself, you will have to use something provided by third parties like Google Guava's Multimap , it does exactly what you want, or build something like this manually.

You can download the Guava library at the project's website . Using a multimap is the same as using a map, as in:

Multimap<String,String> both = new ArrayListMultimap <String,String>();
both.putAll( german );
both.putAll( english);

for ( Entry<String,String> entry : both.entrySet() ) {
  System.out.printf( "%s -> %s%n", entry.getKey(), entry.getValue() );
}

This code will print all key-value pairs including the ones that are present on both maps. So, if you have me->me at both german and english they would be printed twice.

Answer 3

Similar to @tskuzzy's answer

Map<String, String> both = new HashMap<String, String>();

both.putAll(german);
both.putAll(english);
for (String e : english.keySet()) 
    if (german.containsKey(e))
        both.put(e, english.get(e) + german.get(e));

Answer 4

You cannot do that directly with any Map implementation, since in a map, each key is unique.

A possible workaround is to use Map<Key, List<Value>> , and then do the concatenation of your maps manually. The advantage of using a List for the concatenated map, is that it will be easy to retrieve each of the individual values without any extra fiddling.

Something like that would work:

public Map<Key, List<Value>> concat(Map<Key, Value> first, Map<Key, Value> second){
    Map<Key, List<Value>> concat = new HashMap<Key, List<Value>>();
    putMulti(first, concat);
    putMulti(second, concat);
    return concat;
}

private void putMulti(Map<Key, Value> content, Map<Key, List<Value>> dest){
    for(Map.Entry<Key, Value> entry : content){
        List<Value> vals = dest.get(entry.getKey());
        if(vals == null){
            vals = new ArrayList<Value>();
            dest.put(entry.getKey(), vals);
        }
        vals.add(entry.getValue());
    }
}

Answer 5

Slight improvisation of @tskuzzy and @Peter's answer here. Just define your own StrangeHashMap by extending HashMap .

public class StrangeHashMap extends HashMap<String, String> {
    @Override
    public String put(String key, String value) {
        if(this.containsKey(key)) {
            return super.put(key, super.get(key) + value);
        } else {
            return super.put(key, value);
        }
    }
}

You can use it as so:

Map<String, String> map1 = new HashMap<String, String>();
map1.put("key1", "Value1");
map1.put("key2", "Value2");

Map<String, String> map2 = new HashMap<String, String>();
map2.put("key1", "Value2");
map2.put("key3", "Value3");

Map<String, String> all = new StrangeHashMap();

all.putAll(map1);
all.putAll(map2);

System.out.println(all);

The above prints the below for me:

{key3=Value3, key2=Value2, key1=Value1Value2}

Answer 6

Given the new elements in the question, it seems that what you actually need to use is lists. In this case, you can just do:

List<String> english = ...;
List<String> german = ...;
List<String> concat = new ArrayList<String>(english.size() + german.size());
concat.addAll(english);
concat.addAll(german);

And there you are. You can still use concat.get(n) to retreive the value nth value in the concatenated list.