问题
How to filter a Map<String, List<Employee>> using Java 8 Filter?
I have to filter only when any of employee in the list having a field value Gender = "M".
Input: Map<String,List<Employee>>
Output: Map<String,List<Employee>>
Filter criteria: Employee.genter = "M"
Also i have to filter out the key in the output map (or filter map [new map we get after filter]) if the List<> is empty on the map value
回答1:
To filter out entries where a list contains an employee who is not of the "M" gender:
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));
To filter out employees who are not of the "M" gender:
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())));
To filter out entries where a list doesn't contain any "M" employee.
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));
Let's have 2 entries in the map:
"1" -> ["M", "M", "M"]
"2" -> ["M", "F", "M"]
The results for them will be:
r1 = {1=[M, M, M], 2=[M, M]}
r2 = {1=[M, M, M]}
r3 = {1=[M, M, M], 2=[M, F, M]}
回答2:
In Java 8 you can convert a Map.entrySet() into a stream, follow by a filter() and collect() it. Example taken from 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));
And for your case it would look like this, because you also need to find out if there is a match in a List of object of class Employee.
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()));
回答3:
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())));
回答4:
You may do it like so,
Map<String, List<Employee>> resultMap = input.entrySet().stream()
.collect(Collectors.toMap(Map.Entry::getKey,
e -> e.getValue().stream().filter(emp -> emp.getGender().equals("M")).collect(Collectors.toList())));
You can use Collectors.toMap with the same key, and derive the new value using the existing one.
回答5:
If you want to filter the map entries if any Employee in the list of the entry has gender = M, use the following code:
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));
And, If you want to filter out all the Employees with gender M from each list, use the following code:
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())));
回答6:
Filter only map entries that have only male Employes:
@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());
}
回答7:
For instance:
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()));
回答8:
Returning map of employee
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));
}
来源：https://stackoverflow.com/questions/51629897/java-8-filter-mapstring-listemployee