How to sort a HashMap by value but keep the order of duplicates?

Question

I'm trying to sort the tableProbability map into a new one called sorted. In tableProbability the values are the following:

Key	Value
M	0.1
U	0.3
L	0.3
T	0.2
I	0.1

I have the following code that sorts the Map:

LinkedHashMap<Character, Double> sorted = new LinkedHashMap<>();
tableProbability.entrySet()
        .stream()
        .sorted(Map.Entry.comparingByValue(Comparator.reverseOrder()))
        .forEachOrdered(x -> sorted.put(x.getKey(), x.getValue()));

But what I end up getting is the following Map:

Key	Value
L	0.3
U	0.3
T	0.2
I	0.1
M	0.1

And what I am supposed to get is:

Key	Value
U	0.3
L	0.3
T	0.2
M	0.1
I	0.1

Is there any way to retain the duplicate order or at least when it finds a duplicate to put it past the one with the equal value?

score 2 · Accepted Answer · 2021-02-17 17:24:06Z

Your code works fine, but you can simplify it as follows:

Source map:

LinkedHashMap<Character, Double> tableProbability =
        new LinkedHashMap<>() {{
            put('M', 0.1);
            put('U', 0.3);
            put('L', 0.3);
            put('T', 0.2);
            put('I', 0.1);
        }};

System.out.println(tableProbability);
// {M=0.1, U=0.3, L=0.3, T=0.2, I=0.1}

This code works fine:

LinkedHashMap<Character, Double> sorted = new LinkedHashMap<>();
tableProbability.entrySet()
        .stream()
        .sorted(Map.Entry.comparingByValue(Comparator.reverseOrder()))
        .forEachOrdered(x -> sorted.put(x.getKey(), x.getValue()));

System.out.println(sorted);
// {U=0.3, L=0.3, T=0.2, M=0.1, I=0.1}

Simplified version:

LinkedHashMap<Character, Double> sorted2 = tableProbability
        .entrySet().stream()
        .sorted(Map.Entry.comparingByValue(Comparator.reverseOrder()))
        .collect(LinkedHashMap::new,
                (col, e) -> col.put(e.getKey(), e.getValue()),
                HashMap::putAll);

System.out.println(sorted2);
// {U=0.3, L=0.3, T=0.2, M=0.1, I=0.1}

^{See also: Ordering Map<String, Integer> by List<String> using streams}

WJS · Accepted Answer · 2021-01-07 15:23:09Z

0

You may want to do it this way. It is common operation. If you want to return a TreeMap, you can specify it below. And normally one assigns to the interface type. For a TreeMap it would be NavigableMap.

Map<Character, Double> sorted =
        tableProbability.entrySet().stream()
                .sorted(Map.Entry.comparingByValue(
                        Comparator.reverseOrder()))
                .collect(Collectors.toMap(Entry::getKey,
                        Entry::getValue,
                        (a,b)->a, // merge, not used here but
                                  // syntactically required
                        LinkedHashMap::new // type of map to return
                        ));

answered Jan 7, 2021 at 15:23

WJS

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Comments

Bohemian · Accepted Answer · 2021-01-23 06:20:10Z

0

Use a compound comparator:

.sorted(Map.Entry.comparingByValue(Comparator.reverseOrder())
    .andThen(Map.Entry.comparingByKey())
)

answered Jan 23, 2021 at 6:20

Bohemian♦

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