How to sort a HashMap in Java [duplicate]

Question

Do you have to use a HashMap? If you only need the Map Interface use a TreeMap

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If you want to sort by comparing values in the HashMap. You have to write code to do this, if you want to do it once you can sort the values of your HashMap:

Map<String, Person> people = new HashMap<>();
Person jim = new Person("Jim", 25);
Person scott = new Person("Scott", 28);
Person anna = new Person("Anna", 23);

people.put(jim.getName(), jim);
people.put(scott.getName(), scott);
people.put(anna.getName(), anna);

// not yet sorted
List<Person> peopleByAge = new ArrayList<>(people.values());

Collections.sort(peopleByAge, Comparator.comparing(Person::getAge));

for (Person p : peopleByAge) {
    System.out.println(p.getName() + "\t" + p.getAge());
}

If you want to access this sorted list often, then you could insert your elements into a HashMap<TreeSet<Person>>, though the semantics of sets and lists are a bit different.

Sorted List by hasmap keys:

SortedSet<String> keys = new TreeSet<String>(myHashMap.keySet());

Sorted List by hashmap values:

SortedSet<String> values = new TreeSet<String>(myHashMap.values());

In case of duplicated map values:

List<String> mapValues = new ArrayList<String>(myHashMap.values());
Collections.sort(mapValues);

Good Luck!

http://snipplr.com/view/2789/sorting-map-keys-by-comparing-its-values/

get the keys

List keys = new ArrayList(yourMap.keySet());

Sort them

 Collections.sort(keys)

print them.

In any case, you can't have sorted values in HashMap (according to API This class makes no guarantees as to the order of the map; in particular, it does not guarantee that the order will remain constant over time ].

Though you can push all these values to LinkedHashMap, for later use as well.

Seems like you might want a treemap.

http://docs.oracle.com/javase/7/docs/api/java/util/TreeMap.html

You can pass in a custom comparator to it if that applies.

In Java 8:

Comparator<Entry<String, Item>> valueComparator = 
    (e1, e2) -> e1.getValue().getField().compareTo(e2.getValue().getField());

Map<String, Item> sortedMap = 
    unsortedMap.entrySet().stream().
    sorted(valueComparator).
    collect(Collectors.toMap(Entry::getKey, Entry::getValue,
                             (e1, e2) -> e1, LinkedHashMap::new));

Using Guava:

Map<String, Item> map = ...;
Function<Item, Integer> getField = new Function<Item, Integer>() {
    public Integer apply(Item item) {
        return item.getField(); // the field to sort on
    }
};
comparatorFunction = Functions.compose(getField, Functions.forMap(map));
comparator = Ordering.natural().onResultOf(comparatorFunction);
Map<String, Item> sortedMap = ImmutableSortedMap.copyOf(map, comparator);

Custom compare function which includes functionality for the Turkish alphabet or other different languages than english.

public <K extends Comparable,V extends Comparable> LinkedHashMap<K,V> sortByKeys(LinkedHashMap<K,V> map){
    List<K> keys = new LinkedList<K>(map.keySet());
    Collections.sort(keys, (Comparator<? super K>) new Comparator<String>() {
        @Override
        public int compare(String first, String second) {
            Collator collator = Collator.getInstance(Locale.getDefault());
            //Collator collator = Collator.getInstance(new Locale("tr", "TR"));
            return collator.compare(first, second);
        }
    });

    LinkedHashMap<K,V> sortedMap = new LinkedHashMap<K,V>();
    for(K key: keys){
        sortedMap.put(key, map.get(key));
    }

    return sortedMap;
}

here is the using example as the following

LinkedHashMap<String, Boolean> ligList = new LinkedHashMap<String, Boolean>();
ligList = sortByKeys(ligList);

Without any more information, it's hard to know exactly what you want. However, when choosing what data structure to use, you need to take into account what you need it for. Hashmaps are not designed for sorting - they are designed for easy retrieval. So in your case, you'd probably have to extract each element from the hashmap, and put them into a data structure more conducive to sorting, such as a heap or a set, and then sort them there.