如何使用lambda获取哈希映射中值的键数

Question

I have a hash map

Map<Integer, List<String>> Directmap = new HashMap<Integer, List<String>>() {{
    put(0, Arrays.asList(a, b));
    put(1, Arrays.asList(b, c));
    put(2, Arrays.asList(d));
    put(3, Arrays.asList(d, e));
    put(4, Arrays.asList(e));
    put(5, Arrays.asList());
}};

Directmap: {0=[a, b], 1=[b, c], 2=[d], 3=[d, e], 4=[e], 5=[]}

I want to count the number of keys for each value. For example: "a" has one key, "b" has two keys, ..., "e" has two keys namely 3 and 4.

I tried something like this:

Map<Object, Long> ex = Directmap.entrySet().stream()
            .collect(Collectors.groupingBy(e -> e.getKey(), Collectors.counting()));

I want the values with number of keys as the output. Like this :

a=[1], b=[2], c=[1], d=[2], e=[2]

Answer 1

You can do this by flat mapping each value of the map.

In this code, only the values of the map are kept (since you are not interested in the keys). Each value, which is a List<String> is flat mapped so that the Stream pipeline, which is Stream<List<String>> becomes Stream<String> . Finally, the Stream is grouped by ( groupingBy ) the identity function and the reduction performed on the elements is just counting the number of occurences ( counting() ).

Map<String, Long> ex = 
      Directmap.values()
               .stream()
               .flatMap(Collection::stream)
               .collect(Collectors.groupingBy(v -> v, Collectors.counting()));

Note that if a value is appearing twice in the initial map, it will also counted twice. If you want to count the number of distinct values, you can add a call to distinct() before collecting the result (or use a Set instead of a List ).