sorting a hashmap who's key is another hashmap

Question

so this is my hashmap

   public HashMap<Integer, HashMap<String, Integer>> girls =  
          new HashMap<Integer, HashMap<String, **Integer**>>();;

I want to sort the bolded by value. For clarification the outer hashMaps key stands for a year a girl child was born and the inner hashmap stands for a name mapped to the popularity ranking of the name.

So let's say that in 2015, the name Abigail was given to 47373 babies and it was the most popular name in that year, I'd want to return the number 1 bc it's the number one name. Is there any way to sort a hashmap in this way?

how would I turn the inner hashmaps values into an arraylist that I could then easily sort? Any help?

Possible duplicate of [Sort a Map by values (Java)](http://stackoverflow.com/questions/109383/sort-a-mapkey-value-by-values-java) — Jacob G., May 11 '17 at 21:35
*sorting a hashmap who's key is another hashmap* I don't see anything like that in your question. — shmosel, May 11 '17 at 21:36
http://stackoverflow.com/questions/43920316/returning-the-string-associated-with-the-year-and-rank sorry, but it seems that you still have some issues understanding the difference between keys and values. I suggest you to read at least the javadoc https://docs.oracle.com/javase/7/docs/api/java/util/HashMap.html — Spock, May 11 '17 at 21:37
And a `HashMap` is typically going to be better off as a `TreeMap`. — chrylis -cautiouslyoptimistic-, May 11 '17 at 21:37
@Raphael yeah or any year that's specified it can be the 2nd, 3rd, 4th, etc up to 10 — emmynaki, May 11 '17 at 21:41

score 0 · Answer 1 · answered May 11 '17 at 21:59

There is no easy/elegant way to sort a Map by value in its data structure.

HashMaps are unsorted by definition.
LinkedHashMaps are sorted by insertion order.
TreeMaps are sorted by key.

If you really need to, you could write an algorithm which builds up you data structure using a LinkedHashMap as the "inner" structure and make sure the largest value is inserted first.

Alternatively, you could write a small class

class NameFrequency
{
  String name;
  int frequency;
}

and make your data structure a HashMap<Integer, TreeSet<NameFrequency>> and define a comparator for the TreeSet which orders those objects the way you like.

Or, finally, you could leave your data structure as it is and only order it when accessing it:

girls.get(2015).entrySet().stream()
  .sorted((entry1, entry2) -> entry2.getValue() - entry1.getValue())
  .forEachOrdered(entry -> System.out.println(entry.getKey() + ": " + entry.getValue()));

"NameFrequency"... that does sound better than my "NameCount". Careful with TreeSet, though. Just today I came across a question that surprised me given that it has different criteria for equality than Set. Check the link in my answer for why that might end up wrong. In Java8 using lambdas would indeed be the more elegant solution. — G_H, May 11 '17 at 22:02

score 0 · Answer 2 · edited May 23 '17 at 11:47

You're better off just creating a class for a name and number of occurrences.

import java.util.Objects;

public class NameCount implements Comparable<NameCount> {

    private final String name;
    private int count;

    public NameCount(String name) {
        this.name = name;
        count = 0;
    }

    public NameCount(String name, int count) {
        this.name = name;
        this.count = count;
    }

    public int getCount() {
        return count;
    }

    public void setCount(int count) {
        this.count = count;
    }

    public void incrementCount() {
        count++;
    }

    @Override
    public int hashCode() {
        return Objects.hashCode(this.name);
    }

    @Override
    public boolean equals(Object obj) {
        if(obj == null) return false;
        if(getClass() != obj.getClass()) return false;
        final NameCount other = (NameCount)obj;
        if(!Objects.equals(this.name, other.name)) return false;
        return true;
    }

    @Override
    public int compareTo(NameCount o) {
        return o.count - count;
    }

}

You can then define your map as Map<Integer, List<NameCount>>. Note how the above class defines equality and hash code based only on the name, so if you want to see if a name is in a list, you can just create a NameCount for it and use contains. The compareTo implementation orders from higher count to lower, so when getting the List<NameCount> for a given year, you can then use Collections.sort(list) on it and ask for the index for a NameCount with the same name.

public void test(Map<Integer, List<NameCount>> map) {

    int year = 2017;
    List<NameCount> list = map.get(year);
    // Do null-check on list first when using this...
    Collections.sort(list);
    NameCount check = new NameCount("Abigail");
    int rank = list.indexOf(check) + 1;

}

It might seem to make more sense to use TreeSet map values to guarantee unique name entries and keep them sorted all the time, but note that TreeSet defines equality based on comparison, not equals, and it wouldn't let you get the index.

sorting a hashmap who's key is another hashmap

2 Answers2

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