Java8集合合并列表
我正在尝试使用员工姓名合并集合 我有一个Java8集合合并列表,java,collections,java-8,Java,Collections,Java 8,我正在尝试使用员工姓名合并集合 我有一个MainDTO,它有列表 员工拥有字符串名称和列表地址 地址是一个字符串 MainDTO -> List<Employee> empList; Employee-> String name, List<String> 输出将是: ( ("emp1",[("KY"),"("NY"),("MN"),"("FL"),("TN"),"("AZ")]), ("em2",[("KY"),"("NY")]) ) 使用java
MainDTO
,它有列表
员工
拥有字符串名称
和列表地址
地址
是一个字符串
MainDTO -> List<Employee> empList;
Employee-> String name, List<String>
输出将是:
( ("emp1",[("KY"),"("NY"),("MN"),"("FL"),("TN"),"("AZ")]),
("em2",[("KY"),"("NY")])
)
使用java 8或java 7对这些数据进行排序的任何最佳方法。这是解决问题的一个可能的解决方案-如果要避免重复值,请使用Set;否则,请使用列表:
public static void main(String[]args){
final List<Map.Entry<String, List<String>>> inputData = new ArrayList<>();
inputData.add(new AbstractMap.SimpleEntry<String,
List<String>>("emp1", Arrays.asList("KY","NY")));
inputData.add(new AbstractMap.SimpleEntry<String,
List<String>>("emp2", Arrays.asList("KY","NY")));
inputData.add(new AbstractMap.SimpleEntry<String,
List<String>>("emp1", Arrays.asList("MN","FL")));
inputData.add(new AbstractMap.SimpleEntry<String,
List<String>>("emp1", Arrays.asList("MN","FL")));
//If you do not care about duplicates - use List
final Map<String,List<String>> possibleWithDuplicatesData =
inputData.stream()
.collect(Collectors.toMap(Map.Entry::getKey,
entry -> new ArrayList<String>()));
inputData.stream()
.filter(entry -> possibleWithDuplicatesData.containsKey(entry.getKey()))
.forEach(entry->
possibleWithDuplicatesData.get(entry.getKey()).addAll(entry.getValue()));
//If you want to avoid duplicates - use Set
final Map<String,Set<String>> noDuplicatesData =
inputData.stream()
.collect(Collectors.toMap(Map.Entry::getKey,
entry -> new HashSet<String>()));
inputData.stream()
.filter(entry -> noDuplicatesData.containsKey(entry.getKey()))
.forEach(entry->
noDuplicatesData.get(entry.getKey()).addAll(entry.getValue()));
}
publicstaticvoidmain(字符串[]args){
最终列表inputData=new ArrayList();
add(新的AbstractMap.SimpleEntry(“emp1”,Arrays.asList(“KY”,“NY”));
add(新的AbstractMap.SimpleEntry(“emp2”,Arrays.asList(“KY”,“NY”));
add(新的AbstractMap.SimpleEntry(“emp1”,Arrays.asList(“MN”,“FL”));
add(新的AbstractMap.SimpleEntry(“emp1”,Arrays.asList(“MN”,“FL”));
//如果您不关心重复项,请使用列表
最终地图可能重复数据=
inputData.stream()
.collect(Collectors.toMap(Map.Entry::getKey、,
条目->新建ArrayList());
inputData.stream()
.filter(entry->possibleWithDuplicatesData.containsKey(entry.getKey()))
.forEach(条目->
可能是重复数据.get(entry.getKey()).addAll(entry.getValue());
//如果要避免重复,请使用Set
最终地图节点重复数据=
inputData.stream()
.collect(Collectors.toMap(Map.Entry::getKey、,
条目->新建哈希集());
inputData.stream()
.filter(entry->noDuplicatesData.containsKey(entry.getKey()))
.forEach(条目->
noDuplicatesData.get(entry.getKey()).addAll(entry.getValue());
}
试试这个:
private static class Address {
private String address;
public Address(String address) {
this.address = address;
}
public String getAddress() {
return address;
}
public void setAddress(String address) {
this.address = address;
}
@Override
public String toString() {
return "Address{" +
"address='" + address + '\'' +
'}';
}
}
private static class Employee {
private String name;
private List<Address> addresses;
public Employee(String name, List<Address> addresses) {
this.name = name;
this.addresses = addresses;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public List<Address> getAddresses() {
return addresses;
}
public void setAddresses(List<Address> addresses) {
this.addresses = addresses;
}
@Override
public String toString() {
return "Employee{" +
"name='" + name + '\'' +
", addresses=" + addresses +
'}';
}
}
private static class MainDTO {
private List<Employee> employees;
public MainDTO(List<Employee> employees) {
this.employees = employees;
}
public List<Employee> getEmployees() {
return employees;
}
public void setEmployees(List<Employee> employees) {
this.employees = employees;
}
@Override
public String toString() {
return "MainDTO{" +
"employees=" + employees +
'}';
}
}
public static void main(String[] args) {
List<Employee> employees = new ArrayList<>();
employees.add(new Employee("emp1", Arrays.asList(new Address("KY"), new Address("NY"))));
employees.add(new Employee("emp2", Arrays.asList(new Address("KY"), new Address("NY"))));
employees.add(new Employee("emp1", Arrays.asList(new Address("MN"), new Address("FL"))));
employees.add(new Employee("emp1", Arrays.asList(new Address("TN"), new Address("AZ"))));
MainDTO dto = new MainDTO(employees);
List<Employee> merged = dto.getEmployees().stream()
.collect(Collectors.toMap(Employee::getName,
e -> e,
(l, r) ->
new Employee(l.getName(),
Stream.concat(l.getAddresses().stream(), r.getAddresses().stream())
.collect(Collectors.toList())),
LinkedHashMap::new))
.values()
.stream()
.collect(Collectors.toList());
System.out.println(merged);
}
如果Java 9是一个选项,您可以使用:
Map addressesByEmployeeName=empList
.stream()
.收集(分组)(
Employee::getName,flatMapping(
e->e.getAddresses().stream(),
toList());
但是我有一种奇怪的感觉,Java 9不是一个选项。很好,因为它很好地利用了新的JDK 9flatMapping
收集器。如果jdk9不是一个选项,那么可以在jdk8中执行类似的操作,尽管它需要更多的参与。它在流中执行平面映射操作,生成一个成对的员工姓名和地址流。我将使用AbstractMap.SimpleEntry
,而不是创建一个特殊的pair类
其基本思想是将员工列表流化,并将其平面映射为成对的(姓名、地址)流。完成后,将它们收集到地图中,按名称分组,并将地址收集到每个组的列表中。下面是执行此操作的代码:
import static java.util.AbstractMap.SimpleEntry;
import static java.util.Map.Entry;
import static java.util.stream.Collectors.*;
Map<String, List<String>> mergedMap =
empList.stream()
.flatMap(emp -> emp.getAddresses().stream().map(
addr -> new SimpleEntry<>(emp.getName(), addr)))
.collect(groupingBy(Entry::getKey, mapping(Entry::getValue, toList())));
创建地图条目对象并从中提取数据有些麻烦,但并不可怕。如果您愿意,可以将这里的一些习惯用法提取到实用方法中,以使事情变得更清晰。如果您使用具有多重映射的库,例如 由于这种模式有些常见,我们可以向库中添加
RichIterable.toMultimap()
。那么这个例子可以归结为一行
MutableListMultimap<String, String> addressesByEmployeeName =
empList.toMultimap(Employee::getName, Employee::getAddresses);
MutableListMultimap地址ByEmployeeName=
toMultimap(Employee::getName,Employee::getAddresses);
注意:我是Eclipse集合的提交者。很好地使用了新的JDK 9特性+1.在JDK 8中实现这一点是可能的,尽管更复杂;看见
Map<String, List<String>> addressesByEmployeeName = empList
.stream()
.collect(groupingBy(
Employee::getName, flatMapping(
e -> e.getAddresses().stream(),
toList())));
import static java.util.AbstractMap.SimpleEntry;
import static java.util.Map.Entry;
import static java.util.stream.Collectors.*;
Map<String, List<String>> mergedMap =
empList.stream()
.flatMap(emp -> emp.getAddresses().stream().map(
addr -> new SimpleEntry<>(emp.getName(), addr)))
.collect(groupingBy(Entry::getKey, mapping(Entry::getValue, toList())));
List<Employee> mergedEmps =
mergedMap.entrySet().stream()
.map(entry -> new Employee(entry.getKey(), entry.getValue()))
.collect(toList());
ListIterable<Employee> empList = ...;
MutableSortedSetMultimap<String, String> addressesByEmployeeName =
new TreeSortedSetMultimap<>();
empList.each(e -> e.getAddresses().each(a -> addressesByEmployeeName.put(e.getName(), a)));
ListAdapter.adapt(empList).each(
e -> ListAdapter.adapt(e.getAddresses()).each(
a -> addressesByEmployeeName.put(e.getName(), a)));
MutableListMultimap<String, String> addressesByEmployeeName =
empList.toMultimap(Employee::getName, Employee::getAddresses);