问题
I need create inverse map - select unique values and for them find keys. Seems that only way is to iterate all key/value pairs, because entrySet returns set of so value not unique? Thanks.
回答1:
The values in a map may not be unique. But if they are (in your case) you can do as you wrote in your question and create a generic method to convert it:
private static <V, K> Map<V, K> invert(Map<K, V> map) {
Map<V, K> inv = new HashMap<V, K>();
for (Entry<K, V> entry : map.entrySet())
inv.put(entry.getValue(), entry.getKey());
return inv;
}
Java 8:
public static <V, K> Map<V, K> invert(Map<K, V> map) {
return map.entrySet()
.stream()
.collect(Collectors.toMap(Entry::getValue, Entry::getKey));
}
Example of usage:
public static void main(String[] args) {
Map<String, Integer> map = new HashMap<String, Integer>();
map.put("Hello", 0);
map.put("World!", 1);
Map<Integer, String> inv = invert(map);
System.out.println(inv); // outputs something like "{0=Hello, 1=World!}"
}
Side note: the put(.., ..) method will return the the "old" value for a key. If it is not null you may throw a new IllegalArgumentException("Map values must be unique") or something like that.
回答2:
Take a look at Google Guava BiMap.
Example usage
Map<Integer, String> map = new HashMap<>();
map.put(1, "one");
map.put(2, "two");
Map<String, Integer> inverted = HashBiMap.create(map).inverse();
回答3:
To get an inverted form of a given map in java 8:
public static <K, V> Map<V, K> inverseMap(Map<K, V> sourceMap) {
return sourceMap.entrySet().stream().collect(
Collectors.toMap(Entry::getValue, Entry::getKey,
(a, b) -> a) //if sourceMap has duplicate values, keep only first
);
}
Example usage
Map<Integer, String> map = new HashMap<Integer, String>();
map.put(1, "one");
map.put(2, "two");
Map<String, Integer> inverted = inverseMap(map);
回答4:
Seems that only way is to iterate all key/value pairs, because entrySet returns set of so value not unique?
It's one way at least. Here's an example:
Map<Integer, String> map = new HashMap<Integer, String>();
map.put(1, "one");
map.put(2, "two");
Map<String, Integer> inverted = new HashMap<String, Integer>();
for (Integer i : map.keySet())
inverted.put(map.get(i), i);
In case of non-unique values, this algorithm will map the last value found to it's key. (Since the iteration order is undefined for most maps, this should be as good as any solution.)
If you really do want to keep the first value found for each key, you could change it to
if (!inverted.containsKey(map.get(i)))
inverted.put(map.get(i), i);
回答5:
I would give another approach to this problem giving an extra dimension: duplicate values in EntrySet.
public static void main(String[] args) {
HashMap<Integer, String> s = new HashMap<Integer, String>();
s.put(1, "Value1");
s.put(2, "Value2");
s.put(3, "Value2");
s.put(4, "Value1");
/*
* swap goes here
*/
HashMap<String,List<Integer>> newMap = new HashMap<String, List<Integer>>();
for (Map.Entry<Integer, String> en : s.entrySet()) {
System.out.println(en.getKey() + " " + en.getValue());
if(newMap.containsKey(en.getValue())){
newMap.get(en.getValue()).add(en.getKey());
} else {
List<Integer> tmpList = new ArrayList<Integer>();
tmpList.add(en.getKey());
newMap.put(en.getValue(), tmpList);
}
}
for(Map.Entry<String, List<Integer>> entry: newMap.entrySet()){
System.out.println(entry.getKey() + " " + entry.getValue());
}
}
T result will be that:
1 Value1
2 Value2
3 Value2
4 Value1
Value1 [1, 4]
Value2 [2, 3]
回答6:
Apache Commons Collections also provides a BidiMap interface for bi-directional maps, along with several implementations.
BidiMap JavaDoc
回答7:
You have to assume that values may be identical, since the Map contract allows it.
In my opinion the best solution lies in using a wrapper. It will contain the original value, and add an id. Its hashCode() function will rely on the id, and you provide a Getter for the original value. Code would be something like this:
public class MapKey
{
/**
* A new ID to differentiate equal values
*/
private int _id;
/**
* The original value now used as key
*/
private String _originalValue;
public MapKey(String originalValue)
{
_originalValue = originalValue;
//assuming some method for generating ids...
_id = getNextId();
}
public String getOriginalValue()
{
return _originalValue;
}
@Override
public int hashCode()
{
final int prime = 31;
int result = 1;
result = prime * result + _id;
return result;
}
@Override
public boolean equals(Object obj)
{
if (this == obj)
return true;
if (obj == null)
return false;
if (getClass() != obj.getClass())
return false;
MapKey other = (MapKey) obj;
if (_id != other._id)
return false;
return true;
}
@Override
public String toString()
{
StringBuilder sb = new StringBuilder();
sb.append("MapKey value is ");
sb.append(_originalValue);
sb.append(" with ID number ");
sb.append(_id);
return sb.toString();
}
Inverting the map would be something like this:
public Map <MapKey, Integer> invertMap(Map <Integer, String> map)
{
Map <MapKey, Integer> invertedMap = new HashMap <MapKey, Integer>();
Iterator<Entry<Integer, String>> it = map.entrySet().iterator();
while(it.hasNext())
{
//getting the old values (to be reversed)
Entry<Integer, String> entry = it.next();
Integer oldKey = entry.getKey();
String oldValue = entry.getValue();
//creating the new MapKey
MapKey newMapKey = new MapKey(oldValue);
invertedMap.put(newMapKey, oldKey);
}
return invertedMap;
}
Printing the values something like this:
for(MapKey key : invertedMap.keySet())
{
System.out.println(key.toString() + " has a new value of " + invertedMap.get(key));
}
None of this code is tested, but I believe it's the best solution since it makes use of OO inheritance design instead of "c" style checks and allows you to display all the original keys and values.
回答8:
With Guava
Multimaps.transformValues(Multimaps.index(map.entrySet(), Map.Entry::getValue),
Map.Entry::getKey)
You'll get a multimap (basically a map of lists) in return.
来源：https://stackoverflow.com/questions/7146990/java-invert-map