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From: Lew <lewbloch@gmail.com>
Newsgroups: comp.lang.java.programmer
Subject: Re: How associate two HashMaps
Date: Tue, 8 May 2012 09:27:58 -0700 (PDT)
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javax.swing.JSnarker wrote:
> Stefan Ram wrote:
>> Ricardo Nuno writes:
>>> What do I need to associate a value, from one hashMap to a key from a
>>> new HashMap
>>
>>    It is hard to parse this sentence with the comma.
>>
>>    =BBTo associate a value to a key=AB is a little bit vague.
>=20
> My guess is that he wants a bidirectional map. For that you'd need a=20
> custom class delegating to two internal HashMap instances. Something like
>=20
> public class BidiMap<K,V> extends AbstractMap<K,V> {

In Apache Commons Collections 'BidiMap' is an interface with a double handf=
ul=20
of implementing classes:
<http://commons.apache.org/collections/api-release/org/apache/commons/colle=
ctions/BidiMap.html>

>      private Map<K,V> map1, map2;
>      private BidiMap<V,K> reversed;
>=20
>      public BidiMap () {
>          map1 =3D new HashMap<K,V>();
>          map2 =3D new HashMap<K,V>();
>          reversed =3D new BidiMap<K,V>(map2, map1, this);
>      }
>=20
>      private BidiMap (Map<K,V> map1, Map<K,V> map2, BidiMap<V,K> reversed=
) {
>          this.map1 =3D map1; this.map2 =3D map2; this.reversed =3D revers=
ed;
>      }
>=20
>      private BidiMap<V,K> getReversedMap () {
>          return reversed;
>      }
>=20
>      /* Map methods go here. View methods delegate to map1. Modifying
>         methods operate on both maps, but reverse key and value when
>         modifying map2. Also, if a value is already in map2, looks
>         it up in map2 to find the corresponding key and removes
>         that key from map1 before adding the new mappings to both maps.
>      */
> }
>=20
> Changes to the reversed map will be reflected in the original and vice=20
> versa.

JSnarker's example is a good one for understanding the mechanics of the=20
bidirectional map. The Apache Commons libraries are the go-to source for=20
most convenience types not in the standard API.

If you need a different kind of association, well, there are all kinds of w=
ays=20
to do that, as Stefan pointed out.

Another possibility is that you're looking for a way to associate more than=
 one value with the same key. There are a number of ways to do that:

Here's one (not tried or even compiled here, so might need tweaking):

public interface Pair<L, R>
{
  L getLeft();
  R getRight();
}

public class BasePair<L, R> implements Pair<L, R>
{
  private final L left;
  private final R right;
  public BasePair(L left, R right)
  {
    this.left =3D left;
    this.right =3D right;
  }

  public L getLeft() { return left; }
  public R getRight() { return right; }
}

public class MultiAssociator<K, L, R>
{
  private final Map<K, Pair<L, R>> associator =3D new HashMap<>();

  public Pair<L, R> put(K key, Pair<L, R> pair)
  {
    if (key =3D=3D null || pair =3D=3D null)=20
    {
      throw new IllegalArgumentException("Argument cannot be null");
    }
    return associator.put(key, pair);
  }

  public Pair<L, R> put(K key, L left, R right)
  {
    return associator.put(key, new Pair<>(left, right));
  }

  public Pair<L, R> put(K key, L left)
  {
    if (key =3D=3D null)=20
    {
      throw new IllegalArgumentException("Argument cannot be null");
    }
    Pair<L, R> value =3D associator.get(key);
    Pair<L, R> repla =3D value =3D=3D null? new Pair<>(left, null)=20
                : new Pair<>(left, value.getRight());
    return associator.put(key, repla);
  }

  public Pair<L, R> put(K key, R right)
  {
    if (key =3D=3D null)=20
    {
      throw new IllegalArgumentException("Argument cannot be null");
    }
    Pair<L, R> value =3D associator.get(key);
    Pair<L, R> repla =3D value =3D=3D null? new Pair(null, right)=20
                : new Pair(value.getLeft(), right);
    return associator.put(key, repla);
  }

  public Pair<L, R> get(K key)
  {
    return associator.get(key);
  }
}

--=20
Lew