java – TreeMap sort by value

java – TreeMap sort by value

You cant have the TreeMap itself sort on the values, since that defies the SortedMap specification:

A Map that further provides a total ordering on its keys.

However, using an external collection, you can always sort Map.entrySet() however you wish, either by keys, values, or even a combination(!!) of the two.

Heres a generic method that returns a SortedSet of Map.Entry, given a Map whose values are Comparable:

static <K,V extends Comparable<? super V>>
SortedSet<Map.Entry<K,V>> entriesSortedByValues(Map<K,V> map) {
    SortedSet<Map.Entry<K,V>> sortedEntries = new TreeSet<Map.Entry<K,V>>(
        new Comparator<Map.Entry<K,V>>() {
            @Override public int compare(Map.Entry<K,V> e1, Map.Entry<K,V> e2) {
                int res = e1.getValue().compareTo(e2.getValue());
                return res != 0 ? res : 1;
            }
        }
    );
    sortedEntries.addAll(map.entrySet());
    return sortedEntries;
}

Now you can do the following:

    Map<String,Integer> map = new TreeMap<String,Integer>();
    map.put(A, 3);
    map.put(B, 2);
    map.put(C, 1);   

    System.out.println(map);
    // prints {A=3, B=2, C=1}
    System.out.println(entriesSortedByValues(map));
    // prints [C=1, B=2, A=3]

Note that funky stuff will happen if you try to modify either the SortedSet itself, or the Map.Entry within, because this is no longer a view of the original map like entrySet() is.

Generally speaking, the need to sort a maps entries by its values is atypical.

---

Note on == for Integer

Your original comparator compares Integer using ==. This is almost always wrong, since == with Integer operands is a reference equality, not value equality.

    System.out.println(new Integer(0) == new Integer(0)); // prints false!!!

Related questions

• When comparing two Integers in Java does auto-unboxing occur? (NO!!!)
• Is it guaranteed that new Integer(i) == i in Java? (YES!!!)

polygenelubricants answer is almost perfect. It has one important bug though. It will not handle map entries where the values are the same.

This code:…

Map<String, Integer> nonSortedMap = new HashMap<String, Integer>();
nonSortedMap.put(ape, 1);
nonSortedMap.put(pig, 3);
nonSortedMap.put(cow, 1);
nonSortedMap.put(frog, 2);

for (Entry<String, Integer> entry  : entriesSortedByValues(nonSortedMap)) {
    System.out.println(entry.getKey()+:+entry.getValue());
}

Would output:

ape:1
frog:2
pig:3

Note how our cow dissapeared as it shared the value 1 with our ape :O!

This modification of the code solves that issue:

static <K,V extends Comparable<? super V>> SortedSet<Map.Entry<K,V>> entriesSortedByValues(Map<K,V> map) {
        SortedSet<Map.Entry<K,V>> sortedEntries = new TreeSet<Map.Entry<K,V>>(
            new Comparator<Map.Entry<K,V>>() {
                @Override public int compare(Map.Entry<K,V> e1, Map.Entry<K,V> e2) {
                    int res = e1.getValue().compareTo(e2.getValue());
                    return res != 0 ? res : 1; // Special fix to preserve items with equal values
                }
            }
        );
        sortedEntries.addAll(map.entrySet());
        return sortedEntries;
    }

java – TreeMap sort by value

In Java 8:

LinkedHashMap<Integer, String> sortedMap = map.entrySet().stream()
  .sorted(Map.Entry.comparingByValue(/* Optional: Comparator.reverseOrder() */))
  .collect(Collectors.toMap(Map.Entry::getKey,
                            Map.Entry::getValue,
                            (e1, e2) -> e1, LinkedHashMap::new));