Java 8 过滤器映射<String,List<Employee>>

Question

如何使用 Java 8 过滤器过滤 Map<String, List<Employee>>？

只有当列表中的任何员工具有字段值Gender = "M" 时，我才需要过滤。

输入：Map<String,List<Employee>>
输出：Map<String,List<Employee>>
过滤条件：Employee.genter = "M"

如果列表在映射值上为空，我还必须过滤掉输出映射（或过滤映射[过滤后得到的新映射]）中的键

Answer 1

员工返回地图

public static Map<Integer, Employee> evaluatemapEmployee()
    {
        //return Dao.getselectedEmployee().entrySet().stream().collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));
        //return Dao.getselectedEmployee().entrySet().stream().filter(emp->emp.getValue().getsalary()>8000).collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));
        //return Dao.getselectedEmployee().entrySet().stream().filter(emp->emp.getValue().getEmpname().matches("om")).collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));

        //return Dao.getselectedEmployee().entrySet().stream().filter(emp->emp.getValue().getEmpid()==103).collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));
        return Dao.getselectedEmployee().entrySet().stream().filter(emp->emp.getValue().getEmpname().matches("kush")).collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));
    }

Answer 2

过滤掉列表中包含不属于"M" 性别的员工的条目：

Map<String, List<Employee>> r2 = map.entrySet().stream()
    .filter(i -> i.getValue().stream().allMatch(e-> "M".equals(e.gender)))
    .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));

---

过滤掉不属于"M" 性别的员工：

Map<String, List<Employee>> r1 = map.entrySet().stream()
    .filter(i -> !i.getValue().isEmpty())
    .collect(Collectors.toMap(Map.Entry::getKey,
        i -> i.getValue().stream()
              .filter(e -> "M".equals(e.gender)).collect(Collectors.toList())));

---

过滤列表中不包含任何"M" 员工的条目。

Map<String, List<Employee>> r3 = map.entrySet().stream()
    .filter(i -> i.getValue().stream().anyMatch(e -> "M".equals(e.gender)))
    .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));

---

让我们在地图中有 2 个条目：

"1" -> ["M", "M", "M"]
"2" -> ["M", "F", "M"]

他们的结果将是：

r1 = {1=[M, M, M], 2=[M, M]}
r2 = {1=[M, M, M]}
r3 = {1=[M, M, M], 2=[M, F, M]}

Answer 3

例如：

Map<String, List<Employee>> result = originalMap.entrySet().stream()
    .filter(es -> es.getValue().stream().anyMatch(emp -> emp.getGender().equals("M")))
    .collect(Collectors.toMap(e -> e.getKey(), e -> e.getValue()));

Answer 4

仅过滤只有男性员工的地图条目：

@Test
public void filterOnlyMales(){
        String gender = "M";
        Map<String, List<Employee>> maleEmployees = map.entrySet()
                 .stream()
                 /*Filter only keys with male Employes*/
                 .filter(entry -> entry.getValue().stream()
                                  .anyMatch(empl -> gender.equals(empl.getGender())))
                 .collect(Collectors.toMap(
                          Map.Entry::getKey,
                          p -> filterMalesOnly(gender, p));

    }

private List<Employee> filterMalesOnly(String gender,
                                       Map.Entry<String, List<Employee>> p) {
    return p.getValue()
          .stream()
          .filter(empl -> gender.equals(empl.getGender()))
          .collect(
                  Collectors.toList());
}

Answer 5

如果条目列表中的任何 Employee 的性别 = M，如果要过滤映射条目，请使用以下代码：

    Map<String,List<Employee>> result = employeeMap.entrySet()
                                .stream()
                                .filter(e -> e.getValue()
                                            .stream()
                                            .anyMatch(employee -> employee.getGender().equalsIgnoreCase("M")))
                                .collect(Collectors.toMap(Entry::getKey,Entry::getValue));

而且，如果您想从每个列表中过滤掉所有性别为M 的员工，请使用以下代码：

Map<String,List<Employee>> result = employeeMap.entrySet()
                       .stream()
                       .collect(Collectors.toMap(Entry::getKey,
                           e -> e.getValue().stream()
                           .filter(employee -> employee.getGender().equalsIgnoreCase("M"))
                           .collect(Collectors.toList())));

Answer 6

Map<String, List<Employee>> result = yourMap.entrySet()
            .stream()
            .flatMap(ent -> ent.getValue().stream().map(emp -> new SimpleEntry<>(ent.getKey(), emp)))
            .filter(ent -> "M".equalsIgnoreCase(ent.getValue().getGender()))
            .collect(Collectors.groupingBy(
                    Entry::getKey,
                    Collectors.mapping(Entry::getValue, Collectors.toList())));

Answer 7

在 Java 8 中，您可以将 Map.entrySet() 转换为流，然后是 filter() 和 collect()。示例取自 here。

    Map<Integer, String> map = new HashMap<>();
    map.put(1, "linode.com");
    map.put(2, "heroku.com");

    //Map -> Stream -> Filter -> String
    String result = map.entrySet().stream()
        .filter(x -> "something".equals(x.getValue()))
        .map(x->x.getValue())
        .collect(Collectors.joining());

    //Map -> Stream -> Filter -> MAP
    Map<Integer, String> collect = map.entrySet().stream()
        .filter(x -> x.getKey() == 2)
        .collect(Collectors.toMap(x -> x.getKey(), x -> x.getValue()));

    // or like this
    Map<Integer, String> collect = map.entrySet().stream()
        .filter(x -> x.getKey() == 3)
        .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));

对于您的情况，它看起来像这样，因为您还需要确定在 List 的类 Employee 的对象中是否存在匹配项。

Map<String, List<Employee>> collect = map.entrySet().stream()
            .filter(x -> x.getValue().stream()
        .anyMatch(employee -> employee.Gender.equals("M")))
            .collect(Collectors.toMap(x -> x.getKey(), x -> x.getValue()));