Java Map sort by value

Question

I was looking for ways of sorting Map<String, Integer> by values. I found this post, which solved my sorting problem, but not exactly. According to the post, I wrote the following code:

import java.util.*;

public class Sort {

    static class ValueComparator implements Comparator<String> {

        Map<String, Integer> base;

        ValueComparator(Map<String, Integer> base) {
            this.base = base;
        }

        @Override
        public int compare(String a, String b) {
            if (base.get(a) >= base.get(b)) {
                return 1;
            } else {
                return -1;
            }
        }
    }

    public static void main(String[] args) {
        HashMap<String, Integer> map = new HashMap<String, Integer>();
        ValueComparator vc = new ValueComparator(map);
        TreeMap<String, Integer> sorted = new TreeMap<String, Integer>(vc);
        map.put("A", 1);
        map.put("B", 2);
        sorted.putAll(map);
        for (String key : sorted.keySet()) {
            System.out.println(key + " : " + sorted.get(key)); // why null values here?
        }
        System.out.println(sorted.values()); // But we do have non-null values here!
    }
}

Output:

A : null
B : null
[1, 2]
BUILD SUCCESSFUL (total time: 0 seconds)

As you can see from the output, the get method always returns null. The reason is my ValueComparator.compare() method never returns 0, which I've figured out by making this post.

Someone suggested in that post the following to solve the null value problem:

        public int compare(String a, String b) {
            if (base.get(a) > base.get(b)) {
                return 1;
            }else if(base.get(a) ==  base.get(b)){
                return 0;
            }
            return -1;  
        }

I've tested this piece of code and it introduces a key merging problem. In other words, when the values are equal their corresponding keys are merged.

I've also tried the following:

            public int compare(String a, String b) {
                if (a.equals(b)) return 0;
                if (base.get(a) >= base.get(b)) {
                    return 1;
                } else return -1;
            }

It does not work either. Some of the values are still null. Besides, this workaround might potentially have logical problems.

Anyone can propose a fully working solution to my problem? I'd like the sort by value function to work and the get method to work at the same time.

Why/how would you sort a Map by value? I don't think that even makes sense... Isn't that what Lists and Sets are for? Just call _Map.values()_ and add the results to a _TreeSet_ or an ordered _List_. — jahroy, Dec 13 '12 at 04:11
@jahroy How would you relate the sorted values back to their corresponding keys? I don't want to maintain relationship between key and value manually. — Terry Li, Dec 13 '12 at 04:14
OK... so you want _the keys_ to be sorted based on the order of the values. I _guess_ I understand what you want now. — jahroy, Dec 13 '12 at 04:15
In that case, you should just use this: `return base.get(a).compareTo(base.get(b))` — jahroy, Dec 13 '12 at 04:16
@jahroy If you have better solutions than using maps, it would be nice as well :) — Terry Li, Dec 13 '12 at 04:17
I just added an answer... You want to use `Integer.compareTo()` — jahroy, Dec 13 '12 at 04:19

score 6 · Accepted Answer · answered Dec 13 '12 at 04:30

In your compare function, when the values are equal, you should then compare the keys. This will ensure that different keys having the same value will not be "merged", because it disambiguates entries that would otherwise compare equal.

For example:

    @Override
    public int compare(String a, String b) {
        Integer x = base.get(a);
        Integer y = base.get(b);
        if (x.equals(y)) {
            return a.compareTo(b);
        }
        return x.compareTo(y);
    }

(you'll need to modify the code above to match your policy for null values)

Note that your approach of sorting on values is pretty fragile, though. Your "sorted" map will not support addition of new entries, which could be pretty confusing.

score 4 · Answer 2 · answered Dec 13 '12 at 04:03

4

base.get(a) ==  base.get(b)

This code compares boxed Integers by reference.

Change it to base.get(a).equals(base.get(b)) and it should work.

answered Dec 13 '12 at 04:03

SLaks

868,454
176
1,908
1,964

I don't think it's the real problem :) – Terry Li Dec 13 '12 at 04:07
@TerryLi: No; the `==` operator does not autounbox. – SLaks Dec 13 '12 at 14:14

jahroy · Answer 3 · 2012-12-13T04:28:49.693

1

Try this:

return base.get(a).compareTo(base.get(b));

To demonstrate that you are incorrect about auto-unboxing:

Integer a = new Integer(2);
Integer b = new Integer(2);
boolean isEqual = ( a == b );
System.out.println("equal: " + isEqual);
System.out.println("a: " + a);
System.out.println("b: " + b);

My output is:

equal: false
a: 2
b: 2

edited Dec 13 '12 at 04:28

answered Dec 13 '12 at 04:17

jahroy

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It has the same key merging problem. By the way, you can compare `Integer` objects directly using `==`. – Terry Li Dec 13 '12 at 04:26
Good point! I didn't know that. But my problem persists even with the correct comparison method :( – Terry Li Dec 13 '12 at 04:31

score 1 · Answer 4 · answered Dec 13 '12 at 04:17

Try this ...

HashMap<String, Integer> h = new HashMap<String, Integer>();
h.put("z",30);
h.put("e",10);
h.put("b",20);
h.put("c",20);
List<Map.Entry> a = new ArrayList<Map.Entry>(h.entrySet());
Collections.sort(a,
         new Comparator() {
             public int compare(Object o1, Object o2) {
                 Map.Entry e1 = (Map.Entry) o1;
                 Map.Entry e2 = (Map.Entry) o2;
                 return ((Comparable) e1.getValue()).compareTo(e2.getValue());
             }
         });

for (Map.Entry e : a) {
        System.out.println(e.getKey() + " " + e.getValue());
}

Ouptut:

e 10
b 20
c 20
z 30

Java Map sort by value

4 Answers4

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