Code:
Map<Integer, HashSet<String>> test = new TreeMap<>();
test.put(1, new HashSet<>());
test.put(2, new HashSet<>());
test.put(3, new HashSet<>());
test.put(4, new HashSet<>());
test.get(1).add("1");
test.get(2).add("2");
test.get(3).add("2");
test.get(4).add("3, 33");
//get value of treemap and get rid of the duplicate by using distinct and printout
//at the end
test.values().stream().distinct().forEach(i -> System.out.println(i));
output:
[1]
[2]
[3, 33]
My question is how I can printout the key and value at the same time without having duplicate value?
Expected Result:
1= [1]
2= [2]
3= [3, 33]
I even try below code, yet it gives me the treemap with the duplicate values:
Code:
List<Map.Entry<Integer, HashSet<String>>> list = new ArrayList<>();
list.addAll(test.entrySet());
list.stream().distinct().forEach( i -> System.out.println(i));
Output:
1=[1]
2=[2]
3=[2]
4=[3, 33]
test.entrySet().stream()
.collect(
Collectors.toMap(
Map.Entry::getValue,
x -> x,
(a, b) -> a
)
).values()
.forEach(System.out::println);
Edit:
Explanation: this snippet will take the stream of entries and put them into a map of value to entry while discarding duplicates (see javadoc for Collectors#toMap). It then takes the values of that map as a collection. The result is the collection of map entries that are distinct by Map.Entry::getValue
.
Edit 2:
From your comments I think I understand what you are trying to do. You are using this TreeSet as a 1-based list and you want keys to collapse as you remove duplicate values. Is that correct? Maybe you can explain why you are doing this instead of just using a list.
Streams aren't well-suited for this sort of approach, so this will not be pretty, but here you go: Stream the values, eliminate duplicates, collect into a list, then turn the list back into a map.
test.values().stream()
.distinct()
.collect(
Collectors.collectingAndThen(
Collectors.toList(),
lst -> IntStream.range(0, lst.size()).boxed().collect(
Collectors.toMap(i -> i + 1, i -> lst.get(i))
)
)
).entrySet().forEach(System.out::println);
output:
1=[1]
2=[2]
3=[3, 33]