Java sort a Hashmap on Value

Question

Learning Java as I go (Python background). Simple word count program in Java 7 code (can not use J8!).

I have a hash map of word:count pairs. Now I need to sort on count (decreasing order), and break ties with using word in alphabetical order.

Have read s/o and I tried a treemap but it does not seem to handle ties very well so I don't think that is right.

I have seen a lot of solutions posted that define a new class sortbyvalue and define a comparator. These will not work for me as I need to keep the solution all contained in the existing class.

I am looking for feedback on this idea:

iterate over the map entries (me) in the hashmap
use me.getKey = K and me.getValue = V
new Map.Entry reverse_me = (V,K) {not sure about this syntax}
add reverse_me to a List
repeat for all me in map
List.sort { this is where I am unsure, on how this will sort and no idea how to write a comparator. At this point each List element would be a (count, word) pair and the sort() should sort by count in decreasing and then by word in case of same counts in alphabetical order)

This would be the final output.

Is this a logical progression? I can tell from the many posts that there are many opinions on how to do this, but this is the one I can wrap my head around.

Also, can not use Guava.

TreeMap handles ties perfectly (but the sorting happens on the key, not the value of course, use TreeSet for value ordering), the fault was most likely with your code. HashMap on the other hand cannot hold an order (except LinkedHashMap, but even that's not sortable). — Kayaman
– Kayaman, Commented Sep 10, 2015 at 8:10
First thing first, in java, you will need to learn about Collections.sort(...), Comparator and Comparable. You will need that knowledge for solving your problem. — dotvav
– dotvav, Commented Sep 10, 2015 at 8:17

YoungHobbit · Accepted Answer · 2015-09-10 18:08:47Z

3

You can create an List of Entry set from the map. Sort the List using Collections.sort(). You can pass the custom Comparator for sorting by Key when Value(s) are same.

Set<Entry<String, Integer>> set = map.entrySet();
List<Entry<String, Integer>> list = new ArrayList<Entry<String, Integer>>(set);
Collections.sort( list, new Comparator<Map.Entry<String, Integer>>()
{
  public int compare( Map.Entry<String, Integer> o1, Map.Entry<String, Integer> o2 )
  {
    int result = (o2.getValue()).compareTo( o1.getValue() );
    if (result != 0) {
      return result;
    } else {
      return o1.getKey().compareTo(o2.getKey());
    }
  }
} );

edited Sep 10, 2015 at 18:08

answered Sep 10, 2015 at 8:20

YoungHobbit

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2 Comments

YoungHobbit Over a year ago

@fabian Updated the example. Thanks.

Max Wen Over a year ago

I used this and it worked in my application. Besides being first answer I think it's a very intuitive solution for a coder with Python, not Java, background. thx

Cristian Sevescu · Accepted Answer · 2015-09-10 08:41:18Z

1

This collection reflects the correct order only as long as the map entries are not changed

        HashMap<String, Integer> map = new HashMap<>();
        TreeSet<Map.Entry<String, Integer>> entriesSet = new TreeSet<>(new Comparator<Map.Entry<String, Integer>>(){
            @Override
            public int compare(Map.Entry<String, Integer> me1, Map.Entry<String, Integer> me2) {
                return me1.getValue().compareTo(me2.getValue());  
            }
        });
        entriesSet.addAll(map.entrySet());

answered Sep 10, 2015 at 8:41

Cristian Sevescu

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1 Comment

Max Wen Over a year ago

I already used this accepted answer mostly because it came across first. I do like this TreeSet solution though, it looks very straightforward. As I learn more Java I hope to understand this.

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