Java: Composite key in hashmaps

问题:

I would like to store a group of objects in a hashmap , where the key shall be a composite of two string values. is there a way to achieve this?

i can simply concatenate the two strings , but im sure there is a better way to do this.

回答1:

You could have a custom object containing the two strings:

class StringKey {
    private String str1;
    private String str2;
}

Problem is, you need to determine the equality test and the hash code for two such objects.

Equality could be the match on both strings and the hashcode could be the hashcode of the concatenated members (this is debatable):

class StringKey {
    private String str1;
    private String str2;

    @Override
    public boolean equals(Object obj) {
        if(obj != null && obj instanceof StringKey) {
            StringKey s = (StringKey)obj;
            return str1.equals(s.str1) && str2.equals(s.str2);
        }
        return false;
    }

    @Override
    public int hashCode() {
        return (str1 + str2).hashCode();
    }
}

回答2:

public int hashCode() {
    return (str1 + str2).hashCode();
}

This seems to be a terrible way to generate the hashCode: Creating a new string instance every time the hash code is computed is terrible! (Even generating the string instance once and caching the result is poor practice.)

There are a lot of suggestions here:

How do I calculate a good hash code for a list of strings?

public int hashCode() {
    final int prime = 31;
    int result = 1;
    for ( String s : strings ) {
        result = result * prime + s.hashCode();
    }
    return result;
}

For a pair of strings, that becomes:

return string1.hashCode() * 31 + string2.hashCode();

That is a very basic implementation. Lots of advice through the link to suggest better tuned strategies.

回答3:

Why not create a (say) Pair object, which contains the two strings as members, and then use this as the key ?

e.g.

public class Pair {
   private final String str1;
   private final String str2;

   // this object should be immutable to reliably perform subsequent lookups
}

Don't forget about equals() and hashCode(). See this blog entry for more on HashMaps and keys, including a background on the immutability requirements. If your key isn't immutable, then you can change its components and a subsequent lookup will fail to locate it (this is why immutable objects such as String are good candidates for a key)

You're right that concatenation isn't ideal. For some circumstances it'll work, but it's often an unreliable and fragile solution (e.g. is AB/C a different key from A/BC ?).

回答4:

You don't need to reinvent the wheel. Simply use the Guava's HashBasedTable<R,C,V> implementation of Table<R,C,V> interface, for your need. Here is an example

Table<String, String, Integer> table = HashBasedTable.create();

table.put("key-1", "lock-1", 50);
table.put("lock-1", "key-1", 100);

System.out.println(table.get("key-1", "lock-1")); //prints 50
System.out.println(table.get("lock-1", "key-1")); //prints 100

table.put("key-1", "lock-1", 150); //replaces 50 with 150

Happy coding!

回答5:

I have a similar case. All I do is concatenate the two strings separated by a tilde ( ~ ).

So when the client calls the service function to get the object from the map, it looks like this:

MyObject getMyObject(String key1, String key2) {
    String cacheKey = key1 + "~" + key2;
    return map.get(cachekey);
}

It is simple, but it works.

回答6:

I see that many people use nested maps. That is, to map Key1 -> Key2 -> Value (I use the computer science/ aka haskell curring notation for (Key1 x Key2) -> Value mapping which has two arguments and produces a value), you first supply the first key -- this returns you a (partial) map Key2 -> Value, which you unfold in the next step.

For instance,

Map<File, Map<Integer, String>> table = new HashMap(); // maps (File, Int) -> Distance

add(k1, k2, value) {
  table2 = table1.get(k1);
  if (table2 == null) table2 = table1.add(k1, new HashMap())
  table2.add(k2, value)
}

get(k1, k2) {
  table2 = table1.get(k1);
  return table2.get(k2)
}

I am not sure that it is better or not than the plain composite key construction. You may comment on that.

回答7:

Reading about the spaguetti/cactus stack I came up with a variant which may serve for this purpose, including the possibility of mapping your keys in any order so that map.lookup("a","b") and map.lookup("b","a") returns the same element. It also works with any number of keys not just two.

I use it as a stack for experimenting with dataflow programming but here is a quick and dirty version which works as a multi key map (it should be improved: Sets instead of arrays should be used to avoid looking up duplicated ocurrences of a key)

public class MultiKeyMap <K,E> {
    class Mapping {
        E element;
        int numKeys;
        public Mapping(E element,int numKeys){
            this.element = element;
            this.numKeys = numKeys;
        }
    }
    class KeySlot{
        Mapping parent;
        public KeySlot(Mapping mapping) {
            parent = mapping;
        }
    }
    class KeySlotList extends LinkedList<KeySlot>{}
    class MultiMap extends HashMap<K,KeySlotList>{}
    class MappingTrackMap extends HashMap<Mapping,Integer>{}

    MultiMap map = new MultiMap();

    public void put(E element, K ...keys){
        Mapping mapping = new Mapping(element,keys.length);
        for(int i=0;i<keys.length;i++){
            KeySlot k = new KeySlot(mapping);
            KeySlotList l = map.get(keys[i]);
            if(l==null){
                l = new KeySlotList();
                map.put(keys[i], l);
            }
            l.add(k);
        }
    }
    public E lookup(K ...keys){
        MappingTrackMap tmp  = new MappingTrackMap();
        for(K key:keys){
            KeySlotList l = map.get(key);
            if(l==null)return null;
            for(KeySlot keySlot:l){
                Mapping parent = keySlot.parent;
                Integer count = tmp.get(parent);
                if(parent.numKeys!=keys.length)continue;
                if(count == null){
                    count = parent.numKeys-1;
                }else{
                    count--;
                }
                if(count == 0){
                    return parent.element;
                }else{
                    tmp.put(parent, count);
                }               
            }
        }
        return null;
    }
    public static void main(String[] args) {
        MultiKeyMap<String,String> m = new MultiKeyMap<String,String>();
        m.put("brazil", "yellow", "green");
        m.put("canada", "red", "white");
        m.put("USA", "red" ,"white" ,"blue");
        m.put("argentina", "white","blue");

        System.out.println(m.lookup("red","white"));  // canada
        System.out.println(m.lookup("white","red"));  // canada
        System.out.println(m.lookup("white","red","blue")); // USA
    }
}

回答8:

public static String fakeMapKey(final String... arrayKey) {
    String[] keys = arrayKey;

    if (keys == null || keys.length == 0)
        return null;

    if (keys.length == 1)
        return keys[0];

    String key = "";
    for (int i = 0; i < keys.length; i++)
        key += "{" + i + "}" + (i == keys.length - 1 ? "" : "{" + keys.length + "}");

    keys = Arrays.copyOf(keys, keys.length + 1);

    keys[keys.length - 1] = FAKE_KEY_SEPARATOR;

    return  MessageFormat.format(key, (Object[]) keys);}

public static string FAKE_KEY_SEPARATOR = "~";


INPUT:
fakeMapKey("keyPart1","keyPart2","keyPart3");


OUTPUT: keyPart1~keyPart2~keyPart3