在TreeMap上应用Distinct函数 [英] Apply Distinct Function on TreeMap

问题描述

代码：

   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));

输出：

[1]
[2]
[3, 33]

我的问题是如何在没有重复值的情况下同时打印出键和值？

预期结果：

  1= [1]
  2= [2]
  3= [3, 33]

我甚至尝试下面的代码，但它给了我带有重复值的树形图：

代码：

   List<Map.Entry<Integer, HashSet<String>>> list = new ArrayList<>();
   list.addAll(test.entrySet());
   list.stream().distinct().forEach( i -> System.out.println(i));

输出：

1=[1]
2=[2]
3=[2]
4=[3, 33]

推荐答案

test.entrySet().stream()
        .collect(
                Collectors.toMap(
                        Map.Entry::getValue,
                        x -> x,
                        (a, b) -> a
                )
        ).values()
        .forEach(System.out::println);

修改：

说明：此代码段将获取条目流并将它们放入要输入的值的映射中，同时丢弃重复项（请参阅javadoc以获取收藏家#toMap ）。然后，它将该映射的值作为集合。结果是收集了由 Map.Entry :: getValue 分隔的映射条目。

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.

编辑2：

从您的评论我想我明白你是什么试图做。您正在使用此TreeSet作为基于1的列表，并且您希望在删除重复值时折叠键。那是对的吗？也许你可以解释为什么你这样做而不只是使用列表。

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不适合这种方法，所以这不会很漂亮，但在这里你去：流式传输数值，消除重复数据，收集到列表中，然后将列表转回地图。

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]

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