Question

我有一个Map地图，需要使用lambda表达式基于另一个Map进行过滤

我试图在地图上进行过滤，并找到基于另一张地图的所有匹配项，但它似乎不起作用。似乎值未正确过滤。有什么方法可以执行流和映射并将过滤逻辑放置在那里？有人可以帮忙吗

 public static void main(String []args){
    System.out.println("Hello World");
    Map<String,List<String>> items = new HashMap<>();
    List<String> ut1=new ArrayList<>();
    ut1.add("S");
    ut1.add("C");
    List<String> ut2=new ArrayList<>();
    ut2.add("M");
    List<String> ut3=new ArrayList<>();
    ut3.add("M");
    ut3.add("C");
    items .put("1010016",ut1);
    items .put("1010019",ut2);
    items .put("1010012",ut3);
    System.out.println("Map"+items);

    Map<String,Map<String,String>> sKey = new HashMap<>();
    Map<String,String> utKey1 = new HashMap<>();
    utKey1.put("S","1001");
    utKey1.put("M","1002");
    utKey1.put("C","1003");

    Map<String,String> utKey2 = new HashMap<>();
    utKey2.put("S","1004");

    Map<String,String> utKey3 = new HashMap<>();
    utKey3.put("S","1005");
    utKey3.put("M","1006");

    Map<String,String> utKey4 = new HashMap<>();
    utKey4.put("S","1007");
    utKey4.put("M","1008");
    utKey4.put("C","1009");

    sKey.put("1010016",utKey1);
    sKey.put("1010019",utKey2);
    sKey.put("1010012",utKey3);
    sKey.put("1010011",utKey4);

    System.out.println("Map2"+sKey);

    Map<String,Map<String,String>> map3 =
        sKey.entrySet().stream()
            .filter(x -> 
                items.containsKey(x.getKey())
                && x.getValue().entrySet().stream().allMatch(y -> 
                    items.entrySet().stream().anyMatch(list ->
                        list.getValue().contains(y.getKey()))))
            .collect(Collectors.toMap(Entry::getKey, Entry::getValue));
    System.out.println("Map3"+map3);
 }

过滤后的地图返回为：

Map3 {1010012 = {S = 1005，M = 1006}，1010016 = {S = 1001，C = 1003，M = 1002}，1010019 = {S = 1004}}

但是实际结果应该是：

Map3 {1010012 = {M = 1006}，1010016 = {S = 1001，C = 1003}}

Answer 1

我宁愿说这是一种解决方法，可以使用流实现预期的输出。

Map<String, Map<String, String>> result = 
    sKey.entrySet().stream()
    .filter(detail -> items.keySet().contains(detail.getKey()) && 
            !Collections.disjoint(detail.getValue().keySet(), items.get(detail.getKey())))
    .collect(HashMap::new, 
            (m,v) -> m.put(v.getKey(), v.getValue().entrySet().stream()
                .filter(detail -> items.get(v.getKey()).contains(detail.getKey()))
                .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue))), 
            HashMap::putAll);

输出

{1010012={M=1006}, 1010016={S=1001, C=1003}}

如何使用Lamda表达式过滤基于其他地图的地图

1 个答案: