TreeMap sort by value

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闹比i
闹比i 2020-11-22 02:44

I want to write a comparator that will let me sort a TreeMap by value instead of the default natural ordering.

I tried something like this, but can\'t find out what

9条回答
  •  再見小時候
    2020-11-22 03:46

    You can't 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.

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

    static >
    SortedSet> entriesSortedByValues(Map map) {
        SortedSet> sortedEntries = new TreeSet>(
            new Comparator>() {
                @Override public int compare(Map.Entry e1, Map.Entry 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 map = new TreeMap();
        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 map's 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!!!)

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