在 Map 中分组键和值

Question

我有一个 Map，它以 String 为键，以 String[] 为值。

Map<String,String[]> map = new HashMap<>();
map.put("String1", new String[]{ "abc", "dfe", "new"});
map.put("String8", new String[]{"xyz","hji","new"});
map.put("String2", new String[]{"abc","dfe","old"});
map.put("String3", new String[]{"abc","dfe","past"});
map.put("String5", new String[]{"xyz","hji","ancient"});
map.put("String6",new String[]{"xyz","hji","past"});
map.put("String4", new String[]{"abc","dfe","ancient"});
map.put("String7", new String[]{"xyz","hji","old"});

我想使用值打印地图分组。
值是字符串数组。
String 数组的前两个元素将用于分组。
代码必须在不通过任何过滤器的情况下找到相似的值并按值分组。
最终输出将如下所示：

String1:[{ "abc", "dfe", "new"}]
String2:[{"abc","dfe","old"}]
String3:[{"abc","dfe","past"}]
String4:[{"abc","dfe","ancient"}]
String5:[{"xyz","hji","ancient"}]
String6:[{"xyz","hji","past"}]
String7:[{"xyz","hji","old"}]
String8:[{"xyz","hji","new"}]

如何在 Java 8 中实现这一点？

Answer 1

根据您的输出，您似乎只想根据键进行排序。 所以你可以这样做。 创建一个SortedMap并在构造函数中将当前地图添加到它。

SortedMap<String,String[]> smap = new TreeMap<>(map);
smap.forEach((k,v)-> System.out.println(k + ":" + Arrays.toString(v)));

印刷

String1:[abc, dfe, new]
String2:[abc, dfe, old]
String3:[abc, dfe, past]
String4:[abc, dfe, ancient]
String5:[xyz, hji, ancient]
String6:[xyz, hji, past]
String7:[xyz, hji, old]
String8:[xyz, hji, new]

Answer 2

你可以像这样进行双重分组：

Map<String, String[]> map = new HashMap<>();
map.put("String1", new String[]{"abc", "dfe", "new"});
map.put("String8", new String[]{"xyz", "hji", "new"});
map.put("String2", new String[]{"abc", "dfe", "old"});
map.put("String3", new String[]{"abc", "dfe", "past"});
map.put("String5", new String[]{"xyz", "hji", "ancient"});
map.put("String6", new String[]{"xyz", "hji", "past"});
map.put("String4", new String[]{"abc", "dfe", "ancient"});
map.put("String7", new String[]{"xyz", "hji", "old"});


Map<String, Map<String, List<Map.Entry<String, String[]>>>> collect = map.entrySet()
        .stream()
        .collect(Collectors.groupingBy(x -> x.getValue()[0], Collectors.groupingBy(x -> x.getValue()[1])));

Map<String, String[]> collect1 = collect.values()
        .stream()
        .flatMap(x -> x.values().stream())
        .flatMap(Collection::stream)
        .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));

System.out.println(collect1);

Answer 3

如果您只想根据键进行排序，那么您可以使用 Java 8 来完成：

        Map<String, String[]> map = new HashMap<>();
        map.put("String1", new String[]{"abc", "dfe", "new"});
        map.put("String8", new String[]{"xyz", "hji", "new"});
        map.put("String2", new String[]{"abc", "dfe", "old"});
        map.put("String3", new String[]{"abc", "dfe", "past"});
        map.put("String5", new String[]{"xyz", "hji", "ancient"});
        map.put("String6", new String[]{"xyz", "hji", "past"});
        map.put("String4", new String[]{"abc", "dfe", "ancient"});
        map.put("String7", new String[]{"xyz", "hji", "old"});

        map = map.entrySet().stream().sorted(comparingByKey()).collect(toMap(Map.Entry::getKey, Map.Entry::getValue, (e1, e2) -> e2, LinkedHashMap::new));

        map.forEach((s1, strings) -> System.out.println(s1 + " : " + Arrays.toString(strings)));

输出将是：

String1 : [abc, dfe, new]
String2 : [abc, dfe, old]
String3 : [abc, dfe, past]
String4 : [abc, dfe, ancient]
String5 : [xyz, hji, ancient]
String6 : [xyz, hji, past]
String7 : [xyz, hji, old]
String8 : [xyz, hji, new]

Answer 4

同样正如您提到的，您想根据数组的第 2 个元素（值）对地图进行排序，然后您可以尝试比较器：

从值分组数组的第 2 个元素然后排序：

    Map<String, String[]> map = new HashMap<>();
    map.put("String1", new String[]{"abc", "dfe", "new"});
    map.put("String8", new String[]{"xyz", "hji", "new"});
    map.put("String2", new String[]{"abc", "dfe", "old"});
    map.put("String3", new String[]{"abc", "dfe", "past"});
    map.put("String5", new String[]{"xyz", "hji", "ancient"});
    map.put("String6", new String[]{"xyz", "hji", "past"});
    map.put("String4", new String[]{"abc", "dfe", "ancient"});
    map.put("String7", new String[]{"xyz", "hji", "old"});

    // Compare based on 1st 2 element oof Array (Map Value)
    Comparator<String[]> byArrayValue = (String[] o1, String[] o2) -> o1[0].concat(o1[1]).compareTo(o2[0].concat(o2[1]));


    // Sort based on 1st 2 element of array
    map = map.entrySet().stream()
            .sorted(Map.Entry.<String, String[]>comparingByValue(byArrayValue))
            .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (e1, e2) -> e1, LinkedHashMap::new));


    map.forEach((s1, strings) -> System.out.println(s1 + " : " + String.join(",", strings)));

输出将基于值数组的第 2 个元素：

String4 : [abc, dfe, ancient]
String3 : [abc, dfe, past]
String2 : [abc, dfe, old]
String1 : [abc, dfe, new]
String8 : [xyz, hji, new]
String7 : [xyz, hji, old]
String6 : [xyz, hji, past]
String5 : [xyz, hji, ancient]