How sort a LinkedHashMap by the size of each key's value?

Question

I have a LinkedHashMap< Integer, HashSet< Integer > > and I want to sort the keys based on decreasing order of the size of their values. I currently have this fragment from another thread but I'm not sure how to change it to work with my case.

myMap.entrySet().stream()
         .sorted(Map.Entry.comparingByValue(Comparator.reverseOrder()))
         .collect(Collectors.toMap(
                  Map.Entry::getKey,
                  Map.Entry::getValue,
                  (x,y) -> {throw new AssertionError();},
                  LinkedHashMap::new
         ));

Here's what I've tried

List<Map.Entry<String, Integer>> entries =
  new ArrayList<Map.Entry<String, Integer>>(map.entrySet());
Collections.sort(entries, new Comparator<Map.Entry<String, Integer>>() {
  public int compare(Map.Entry<String, Integer> a, Map.Entry<String,Integer> b){
    return a.getValue().size().compareTo(b.getValue().size());
   }
});
Map<String, Integer> sortedMap = new LinkedHashMap<String, Integer>();
for (Map.Entry<String, Integer> entry : entries) {
  sortedMap.put(entry.getKey(), entry.getValue());
}

I just added the .size() in there. I found it from this thread. The size I'm referring to is the number of items in the HashSet. It is usually just getValue().size() but it's not working here.

Sorting LinkedHashMap

Possible duplicate of [How to sort a Map on the values in Java?](http://stackoverflow.com/questions/109383/how-to-sort-a-mapkey-value-on-the-values-in-java) — Andy Turner, Oct 19 '15 at 23:08
I tried implementing what they did with (o1.getValue().size()).compareTo(o2.getValue().size()) but it didnt' work. I'm not sure how to compare the sizes of the values is my problem. — pmdaly, Oct 19 '15 at 23:18
No trace of what you've tried in your post. Please add it by editing and referencing the other answer. Actually I'd say you didn't unserstand what's going on in that code fragment. You should try to analyze it. Please also clarify "`... the size of their values`" (does that mean String length?) Edit your post! — try-catch-finally, Oct 19 '15 at 23:24

score 1 · Accepted Answer · answered Oct 20 '15 at 00:43

What you've tried was close. Here's a fixed version:

List<Map.Entry<Integer, Set<Integer>>> entries = new ArrayList<>(
            map.entrySet());
Collections.sort(entries,
        new Comparator<Map.Entry<Integer, Set<Integer>>>() {
            public int compare(Map.Entry<Integer, Set<Integer>> a,
                    Map.Entry<Integer, Set<Integer>> b) {
                return Integer.compare(
                          a.getValue().size(),
                          b.getValue().size());
            }
        });

Map<Integer, Set<Integer>> sortedMap = new LinkedHashMap<>();
for (Map.Entry<Integer, Set<Integer>> entry : entries) {
    sortedMap.put(entry.getKey(), entry.getValue());
}

Setting up with

 LinkedHashMap<Integer, Set<Integer>> map = new LinkedHashMap<>();
 map.put(1, new HashSet<>(Arrays.asList(1,2,3)));
 map.put(2, new HashSet<>(Arrays.asList(1,2)));
 map.put(3, new HashSet<>(Arrays.asList(1)));

and running

System.out.println(sortedMap);

after sorting outputs

{3=[1], 2=[1, 2], 1=[1, 2, 3]}

How sort a LinkedHashMap by the size of each key's value?

1 Answers1