Java Hashmap：如何从值中获取键？

Question

If I have the value "foo" , and a HashMap<String> ftw for which ftw.containsValue("foo") returns true , how can I get the corresponding key? Do I have to loop through the hashmap? What is the best way to do that?

Answer 1

If your data structure has many-to-one mapping between keys and values you should iterate over entries and pick all suitable keys:

public static <T, E> Set<T> getKeysByValue(Map<T, E> map, E value) {
    Set<T> keys = new HashSet<T>();
    for (Entry<T, E> entry : map.entrySet()) {
        if (Objects.equals(value, entry.getValue())) {
            keys.add(entry.getKey());
        }
    }
    return keys;
}

In case of one-to-one relationship, you can return the first matched key:

public static <T, E> T getKeyByValue(Map<T, E> map, E value) {
    for (Entry<T, E> entry : map.entrySet()) {
        if (Objects.equals(value, entry.getValue())) {
            return entry.getKey();
        }
    }
    return null;
}

In Java 8:

public static <T, E> Set<T> getKeysByValue(Map<T, E> map, E value) {
    return map.entrySet()
              .stream()
              .filter(entry -> Objects.equals(entry.getValue(), value))
              .map(Map.Entry::getKey)
              .collect(Collectors.toSet());
}

Also, for Guava users, BiMap may be useful. For example:

BiMap<Token, Character> tokenToChar = 
    ImmutableBiMap.of(Token.LEFT_BRACKET, '[', Token.LEFT_PARENTHESIS, '(');
Token token = tokenToChar.inverse().get('(');
Character c = tokenToChar.get(token);

Answer 2

If you choose to use the Commons Collections library instead of the standard Java Collections framework, you can achieve this with ease.

The BidiMap interface in the Collections library is a bi-directional map, allowing you to map a key to a value (like normal maps), and also to map a value to a key, thus allowing you to perform lookups in both directions. Obtaining a key for a value is supported by the getKey() method.

There is a caveat though, bidi maps cannot have multiple values mapped to keys, and hence unless your data set has 1:1 mappings between keys and values, you cannot use bidi maps.

---

If you want to rely on the Java Collections API, you will have to ensure the 1:1 relationship between keys and values at the time of inserting the value into the map. This is easier said than done.

Once you can ensure that, use the entrySet() method to obtain the set of entries (mappings) in the Map. Once you have obtained the set whose type is Map.Entry , iterate through the entries, comparing the stored value against the expected, and obtain the corresponding key .

---

Support for bidi maps with generics can be found in Google Guava and the refactored Commons-Collections libraries (the latter is not an Apache project). Thanks to Esko for pointing out the missing generic support in Apache Commons Collections. Using collections with generics makes more maintainable code.

---

Since version 4.0 the official Apache Commons Collections™ library supports generics .

See the summary page of the " org.apache.commons.collections4.bidimap " package for the list of available implementations of the BidiMap , OrderedBidiMap and SortedBidiMap interfaces that now support Java generics .

Answer 3

public class NewClass1 {

    public static void main(String[] args) {
       Map<Integer, String> testMap = new HashMap<Integer, String>();
        testMap.put(10, "a");
        testMap.put(20, "b");
        testMap.put(30, "c");
        testMap.put(40, "d");
        for (Entry<Integer, String> entry : testMap.entrySet()) {
            if (entry.getValue().equals("c")) {
                System.out.println(entry.getKey());
            }
        }
    }
}

Some additional info... May be useful to you

Above method may not be good if your hashmap is really big. If your hashmap contain unique key to unique value mapping, you can maintain one more hashmap that contain mapping from Value to Key.

That is you have to maintain two hashmaps

1. Key to value

2. Value to key 

In that case you can use second hashmap to get key.

Answer 4

You could insert both the key,value pair and its inverse into your map structure

map.put("theKey", "theValue");
map.put("theValue", "theKey");

Using map.get("theValue") will then return "theKey".

It's a quick and dirty way that I've made constant maps, which will only work for a select few datasets:

• Contains only 1 to 1 pairs
• Set of values is disjoint from the set of keys (1->2, 2->3 breaks it)

Answer 5

I think your choices are

• Use a map implementation built for this, like the BiMap from google collections. Note that the google collections BiMap requires uniqueless of values, as well as keys, but it provides high performance in both directions performance
• Manually maintain two maps - one for key -> value, and another map for value -> key
• Iterate through the entrySet() and to find the keys which match the value. This is the slowest method, since it requires iterating through the entire collection, while the other two methods don't require that.

Answer 6

Using Java 8:

ftw.forEach((key, value) -> {
    if (value.equals("foo")) {
        System.out.print(key);
    }
});

Answer 7

Decorate map with your own implementation

class MyMap<K,V> extends HashMap<K, V>{

    Map<V,K> reverseMap = new HashMap<V,K>();

    @Override
    public V put(K key, V value) {
        // TODO Auto-generated method stub
        reverseMap.put(value, key);
        return super.put(key, value);
    }

    public K getKey(V value){
        return reverseMap.get(value);
    }
}

Answer 8

There is no unambiguous answer, because multiple keys can map to the same value. If you are enforcing unique-ness with your own code, the best solution is to create a class that uses two Hashmaps to track the mappings in both directions.

Answer 9

要查找映射到该值的所有键，请使用map.entrySet()遍历哈希图中的所有对。

Answer 10

I think this is best solution, original address: Java2s

    import java.util.HashMap;
    import java.util.Map;

        public class Main {

          public static void main(String[] argv) {
            Map<String, String> map = new HashMap<String, String>();
            map.put("1","one");
            map.put("2","two");
            map.put("3","three");
            map.put("4","four");

            System.out.println(getKeyFromValue(map,"three"));
          }


// hm is the map you are trying to get value from it
          public static Object getKeyFromValue(Map hm, Object value) {
            for (Object o : hm.keySet()) {
              if (hm.get(o).equals(value)) {
                return o;
              }
            }
            return null;
          }
        }

An easy usage: if you put all data in hasMap and you have item = "Automobile", so you are looking its key in hashMap. that is good solution.

getKeyFromValue(hashMap, item);
System.out.println("getKeyFromValue(hashMap, item): "+getKeyFromValue(hashMap, item));

Answer 11

If you build the map in your own code, try putting the key and value in the map together:

public class KeyValue {
    public Object key;
    public Object value;
    public KeyValue(Object key, Object value) { ... }
}

map.put(key, new KeyValue(key, value));

Then when you have a value, you also have the key.

Answer 12

I'm afraid you'll just have to iterate your map. Shortest I could come up with:

Iterator<Map.Entry<String,String>> iter = map.entrySet().iterator();
while (iter.hasNext()) {
    Map.Entry<String,String> entry = iter.next();
    if (entry.getValue().equals(value_you_look_for)) {
        String key_you_look_for = entry.getKey();
    }
}

Answer 13

for(int key: hm.keySet()) {
    if(hm.get(key).equals(value)) {
        System.out.println(key); 
    }
}

Answer 14

听起来最好的方法是使用map.entrySet()迭代条目，因为map.containsValue()可能无论如何都map.containsValue() 。

Answer 15

For Android development targeting API < 19, Vitalii Fedorenko one-to-one relationship solution doesn't work because Objects.equals isn't implemented. Here's a simple alternative:

public <K, V> K getKeyByValue(Map<K, V> map, V value) {
    for (Map.Entry<K, V> entry : map.entrySet()) {
            if (value.equals(entry.getValue())) {
            return entry.getKey();
        }
    }
    return null;
}

Answer 16

You can use the below:

public class HashmapKeyExist {
    public static void main(String[] args) {
        HashMap<String, String> hmap = new HashMap<String, String>();
        hmap.put("1", "Bala");
        hmap.put("2", "Test");

        Boolean cantain = hmap.containsValue("Bala");
        if(hmap.containsKey("2") && hmap.containsValue("Test"))
        {
            System.out.println("Yes");
        }
        if(cantain == true)
        {
            System.out.println("Yes"); 
        }

        Set setkeys = hmap.keySet();
        Iterator it = setkeys.iterator();

        while(it.hasNext())
        {
            String key = (String) it.next();
            if (hmap.get(key).equals("Bala"))
            {
                System.out.println(key);
            }
        }
    }
}

Answer 17

I think keySet() may be well to find the keys mapping to the value, and have a better coding style than entrySet() .

Ex:

Suppose you have a HashMap map , ArrayList res , a value you want to find all the key mapping to , then store keys to the res .

You can write code below:

    for (int key : map.keySet()) {
        if (map.get(key) == value) {
            res.add(key);
        }
    }

rather than use entrySet() below:

    for (Map.Entry s : map.entrySet()) {
        if ((int)s.getValue() == value) {
            res.add((int)s.getKey());
        }
    }

Hope it helps :)

Answer 18

Yes, you have to loop through the hashmap, unless you implement something along the lines of what these various answers suggest. Rather than fiddling with the entrySet, I'd just get the keySet(), iterate over that set, and keep the (first) key that gets you your matching value. If you need all the keys that match that value, obviously you have to do the whole thing.

As Jonas suggests, this might already be what the containsValue method is doing, so you might just skip that test all-together, and just do the iteration every time (or maybe the compiler will already eliminate the redundancy, who knows).

Also, relative to the other answers, if your reverse map looks like

Map<Value, Set<Key>>

you can deal with non-unique key->value mappings, if you need that capability (untangling them aside). That would incorporate fine into any of the solutions people suggest here using two maps.

Answer 19

public static class SmartHashMap <T1 extends Object, T2 extends Object> {
    public HashMap<T1, T2> keyValue;
    public HashMap<T2, T1> valueKey;

    public SmartHashMap(){
        this.keyValue = new HashMap<T1, T2>();
        this.valueKey = new HashMap<T2, T1>();
    }

    public void add(T1 key, T2 value){
        this.keyValue.put(key, value);
        this.valueKey.put(value, key);
    }

    public T2 getValue(T1 key){
        return this.keyValue.get(key);
    }

    public T1 getKey(T2 value){
        return this.valueKey.get(value);
    }

}

Answer 20

In java8

map.entrySet().stream().filter(entry -> entry.getValue().equals(value))
    .forEach(entry -> System.out.println(entry.getKey()));

Answer 21

You can get the key using values using following code..

ArrayList valuesList = new ArrayList();
Set keySet = initalMap.keySet();
ArrayList keyList = new ArrayList(keySet);

for(int i = 0 ; i < keyList.size() ; i++ ) {
    valuesList.add(initalMap.get(keyList.get(i)));
}

Collections.sort(valuesList);
Map finalMap = new TreeMap();
for(int i = 0 ; i < valuesList.size() ; i++ ) {
    String value = (String) valuesList.get(i);

    for( int j = 0 ; j < keyList.size() ; j++ ) {
        if(initalMap.get(keyList.get(j)).equals(value)) {
            finalMap.put(keyList.get(j),value);
        }   
    }
}
System.out.println("fianl map ---------------------->  " + finalMap);

Answer 22

public static String getKey(Map<String, Integer> mapref, String value) {
    String key = "";
    for (Map.Entry<String, Integer> map : mapref.entrySet()) {
        if (map.getValue().toString().equals(value)) {
            key = map.getKey();
        }
    }
    return key;
}

Answer 23

import java.util.ArrayList;
import java.util.HashMap;
import java.util.Iterator;
import java.util.List;
import java.util.Set;

public class M{
public static void main(String[] args) {

        HashMap<String, List<String>> resultHashMap = new HashMap<String, List<String>>();

        Set<String> newKeyList = resultHashMap.keySet();


        for (Iterator<String> iterator = originalHashMap.keySet().iterator(); iterator.hasNext();) {
            String hashKey = (String) iterator.next();

            if (!newKeyList.contains(originalHashMap.get(hashKey))) {
                List<String> loArrayList = new ArrayList<String>();
                loArrayList.add(hashKey);
                resultHashMap.put(originalHashMap.get(hashKey), loArrayList);
            } else {
                List<String> loArrayList = resultHashMap.get(originalHashMap
                        .get(hashKey));
                loArrayList.add(hashKey);
                resultHashMap.put(originalHashMap.get(hashKey), loArrayList);
            }
        }

        System.out.println("Original HashMap : " + originalHashMap);
        System.out.println("Result HashMap : " + resultHashMap);
    }
}

Answer 24

import java.util.HashMap;
import java.util.HashSet;
import java.util.Set;

public class ValueKeysMap<K, V> extends HashMap <K,V>{
    HashMap<V, Set<K>> ValueKeysMap = new HashMap<V, Set<K>>();

    @Override
    public boolean containsValue(Object value) {
        return ValueKeysMap.containsKey(value);
    }

    @Override
    public V put(K key, V value) {
        if (containsValue(value)) {
            Set<K> keys = ValueKeysMap.get(value);
            keys.add(key);
        } else {
            Set<K> keys = new HashSet<K>();
            keys.add(key);
            ValueKeysMap.put(value, keys);
        }
        return super.put(key, value);
    }

    @Override
    public V remove(Object key) {
        V value = super.remove(key);
        Set<K> keys = ValueKeysMap.get(value);
        keys.remove(key);
        if(keys.size() == 0) {
           ValueKeysMap.remove(value);
        }
        return value;
    }

    public Set<K> getKeys4ThisValue(V value){
        Set<K> keys = ValueKeysMap.get(value);
        return keys;
    }

    public boolean valueContainsThisKey(K key, V value){
        if (containsValue(value)) {
            Set<K> keys = ValueKeysMap.get(value);
            return keys.contains(key);
        }
        return false;
    }

    /*
     * Take care of argument constructor and other api's like putAll
     */
}

Answer 25

/**
 * This method gets the Key for the given Value
 * @param paramName
 * @return
 */
private String getKeyForValueFromMap(String paramName) {
    String keyForValue = null;
    if(paramName!=null)) {
        Set<Entry<String,String>> entrySet = myMap().entrySet();
        if(entrySet!=null && entrySet.size>0) {
            for(Entry<String,String> entry : entrySet) {
                if(entry!=null && paramName.equalsIgnoreCase(entry.getValue())) {
                    keyForValue = entry.getKey();
                }
            }
        }
    }
    return keyForValue;
}

Answer 26

My 2 cents. You can get the keys in an array and then loop through the array. This will affect performance of this code block if the map is pretty big , where in you are getting the keys in an array first which might consume some time and then you are looping. Otherwise for smaller maps it should be ok.

String[] keys =  yourMap.keySet().toArray(new String[0]);

for(int i = 0 ; i < keys.length ; i++){
    //This is your key    
    String key = keys[i];

    //This is your value
    yourMap.get(key)            
}

Answer 27

While this does not directly answer the question, it is related.

This way you don't need to keep creating/iterating. Just create a reverse map once and get what you need.

/**
 * Both key and value types must define equals() and hashCode() for this to work.
 * This takes into account that all keys are unique but all values may not be.
 *
 * @param map
 * @param <K>
 * @param <V>
 * @return
 */
public static <K, V> Map<V, List<K>> reverseMap(Map<K,V> map) {
    if(map == null) return null;

    Map<V, List<K>> reverseMap = new ArrayMap<>();

    for(Map.Entry<K,V> entry : map.entrySet()) {
        appendValueToMapList(reverseMap, entry.getValue(), entry.getKey());
    }

    return reverseMap;
}


/**
 * Takes into account that the list may already have values.
 * 
 * @param map
 * @param key
 * @param value
 * @param <K>
 * @param <V>
 * @return
 */
public static <K, V> Map<K, List<V>> appendValueToMapList(Map<K, List<V>> map, K key, V value) {
    if(map == null || key == null || value == null) return map;

    List<V> list = map.get(key);

    if(list == null) {
        List<V> newList = new ArrayList<>();
        newList.add(value);
        map.put(key, newList);
    }
    else {
        list.add(value);
    }

    return map;
}

Answer 28

Simplest utility method to fetch a key of a given value from a Map:

public static void fetchValue(Map<String, Integer> map, Integer i)
{   
Stream stream = map.entrySet().stream().filter(val-> val.getValue().equals(i)).map(Map.Entry::getKey);
stream.forEach(System.out::println);    
}

detailed explaination:

1. Method fetchValue accepts the map, which has String as key and Integer as value.

2. Then we use entryset().stream() to convert result into a stream.

3. Next we use filter (intermediate operation) which gives us a value that is equal to the second argument.

4. Finally, we use forEach(final operation) to print our end result.

Answer 29

Use a thin wrapper: HMap

import java.util.Collections;
import java.util.HashMap;
import java.util.Map;

public class HMap<K, V> {

   private final Map<K, Map<K, V>> map;

   public HMap() {
      map = new HashMap<K, Map<K, V>>();
   }

   public HMap(final int initialCapacity) {
      map = new HashMap<K, Map<K, V>>(initialCapacity);
   }

   public boolean containsKey(final Object key) {
      return map.containsKey(key);
   }

   public V get(final Object key) {
      final Map<K, V> entry = map.get(key);
      if (entry != null)
         return entry.values().iterator().next();
      return null;
   }

   public K getKey(final Object key) {
      final Map<K, V> entry = map.get(key);
      if (entry != null)
         return entry.keySet().iterator().next();
      return null;
   }

   public V put(final K key, final V value) {
      final Map<K, V> entry = map
            .put(key, Collections.singletonMap(key, value));
      if (entry != null)
         return entry.values().iterator().next();
      return null;
   }
}

Answer 30

It's important to note that since this question, Apache Collections supports Generic BidiMaps . So a few of the top voted answers are no longer accurate on that point.

For a Serialized BidiMap that also supports duplicate values ( 1-to-many scenario ) also consider MapDB.org .

Answer 31

If I have the value "foo" , and a HashMap<String> ftw for which ftw.containsValue("foo") returns true , how can I get the corresponding key? Do I have to loop through the hashmap? What is the best way to do that?

Answer 32

try this:

static String getKeyFromValue(LinkedHashMap<String, String> map,String value) {
    for (int x=0;x<map.size();x++){
        if( String.valueOf( (new ArrayList<String>(map.values())).get(x) ).equals(value))
            return String.valueOf((new ArrayList<String>(map.keySet())).get(x));
    }
    return null;
}

Answer 33

As far as I know keys and values of a HashMap are not mixed when you represent them as arrays:

hashmap.values().toArray()

and

hashmap.keySet().toArray()

So the following code (since java 8) should work as expected:

public Object getKeyByFirstValue(Object value) {
    int keyNumber =  Arrays.asList(hashmap.values().toArray()).indexOf(value);
    return hashmap.keySet().toArray()[keyNumber];
}

However, ( WARNING! ) it works 2-3 times slower than iteration.

Answer 34

Let the value be maxValue .

Set keySet = map.keySet();

keySet.stream().filter(x->map.get(x)==maxValue).forEach(x-> System.out.println(x));

Answer 35

lambda w/o use of external libraries
can deal with multiple values for one key (in difference to the BidiMap)

public static List<String> getKeysByValue(Map<String, String> map, String value) {
  List<String> list = map.keySet().stream()
      .collect(groupingBy(k -> map.get(k))).get(value);
  return (list == null ? Collections.emptyList() : list);
}

gets a List containing the key(s) mapping value
for an 1:1 mapping the returned list is empty or contains 1 value

Answer 36

let see my example

Map<String, String> mapPeopleAndCountry = new HashMap<>();
mapPeopleAndCountry.put("Matis", "Lithuania");
mapPeopleAndCountry.put("Carlos", "Honduras");
mapPeopleAndCountry.put("Teboho", "Lesotho");
mapPeopleAndCountry.put("Marielos", "Honduras");


List<String> peopleInHonduras = mapPeopleAndCountry.keySet()
    .stream()
    .filter(r -> mapPeopleAndCountry.get(r)
                .equals("Honduras"))
    .stream(Collectors.toList());

// will return ["Carlos", "Marielos"]

Note: untested, may contains typo

Answer 37

Found too many answers. Some were really great. But I was particularly looking for a way, so that I can get the value using loops.

So here is finally what I did: For a HashMap 1-to-1 relation:

    Map<String, String> map = new HashMap<String, String>();
    
    map.put("abc", "123");
    map.put("xyz", "456");
    
    for(Entry<String, String> entry : map.entrySet()) {
        if(entry.getValue().equalsIgnoreCase("456")) {
            System.out.println(entry.getKey());
        }
    }

Output: "xyz"

For a HashMap 1-to-many relation:

    Map<String, ArrayList<String>> service = new HashMap<String, ArrayList<String>>();
    
    service.put("abc", new ArrayList<String>());
    service.get("abc").add("a");
    service.get("abc").add("b");
    service.get("abc").add("c");
    
    service.put("xyz", new ArrayList<String>());
    service.get("xyz").add("x");
    service.get("xyz").add("y");
    service.get("xyz").add("z");
    
    for(Entry<String, ArrayList<String>> entry : service.entrySet()) {
        ArrayList<String> values = entry.getValue();
        for(String value : values) {
            if(value.equalsIgnoreCase("x")) {
                System.out.println(entry.getKey());
            }
        }
        
    }

Output: xyz

-Thanks

Answer 38

你也可以做这个工作：第一：放置映射（键，值）第二：更新你需要删除表达式的键第三：并用 oldValue 放置一个新键

Answer 39

Iterator<Map.Entry<String,String>> iterator = map.entrySet().iterator();
while (iterator.hasNext()) {
    Map.Entry<String,String> entry = iterator.next();
    if (entry.getValue().equals(value_you_look_for)) {
        String key_you_look_for = entry.getKey();
}
}