基于 Java 8 中的属性从对象列表中删除重复项 [英] Remove duplicates from a list of objects based on property in Java 8
问题描述
我正在尝试从基于某些属性的对象列表中删除重复项.
I am trying to remove duplicates from a List of objects based on some property.
我们可以使用 java 8 以简单的方式做到这一点吗
can we do it in a simple way using java 8
List<Employee> employee
我们可以根据员工的 id
属性从中删除重复项吗?我看到过从字符串数组列表中删除重复字符串的帖子.
Can we remove duplicates from it based on id
property of employee. I have seen posts removing duplicate strings form arraylist of string.
推荐答案
您可以从 List
中获取一个流并将其放入您提供的 TreeSet
唯一比较 id 的自定义比较器.
You can get a stream from the List
and put in in the TreeSet
from which you provide a custom comparator that compares id uniquely.
如果你真的需要一个列表,你可以把这个集合放回一个 ArrayList 中.
Then if you really need a list you can put then back this collection into an ArrayList.
import static java.util.Comparator.comparingInt;
import static java.util.stream.Collectors.collectingAndThen;
import static java.util.stream.Collectors.toCollection;
...
List<Employee> unique = employee.stream()
.collect(collectingAndThen(toCollection(() -> new TreeSet<>(comparingInt(Employee::getId))),
ArrayList::new));
举个例子:
List<Employee> employee = Arrays.asList(new Employee(1, "John"), new Employee(1, "Bob"), new Employee(2, "Alice"));
它会输出:
[Employee{id=1, name='John'}, Employee{id=2, name='Alice'}]
另一个想法可能是使用一个包装器来包装员工,并根据其 id 使用 equals 和 hashcode 方法:
Another idea could be to use a wrapper that wraps an employee and have the equals and hashcode method based with its id:
class WrapperEmployee {
private Employee e;
public WrapperEmployee(Employee e) {
this.e = e;
}
public Employee unwrap() {
return this.e;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
WrapperEmployee that = (WrapperEmployee) o;
return Objects.equals(e.getId(), that.e.getId());
}
@Override
public int hashCode() {
return Objects.hash(e.getId());
}
}
然后包装每个实例,调用 distinct()
,解开它们并将结果收集到列表中.
Then you wrap each instance, call distinct()
, unwrap them and collect the result in a list.
List<Employee> unique = employee.stream()
.map(WrapperEmployee::new)
.distinct()
.map(WrapperEmployee::unwrap)
.collect(Collectors.toList());
事实上,我认为您可以通过提供一个进行比较的函数来使这个包装器通用:
In fact, I think you can make this wrapper generic by providing a function that will do the comparison:
public class Wrapper<T, U> {
private T t;
private Function<T, U> equalityFunction;
public Wrapper(T t, Function<T, U> equalityFunction) {
this.t = t;
this.equalityFunction = equalityFunction;
}
public T unwrap() {
return this.t;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
@SuppressWarnings("unchecked")
Wrapper<T, U> that = (Wrapper<T, U>) o;
return Objects.equals(equalityFunction.apply(this.t), that.equalityFunction.apply(that.t));
}
@Override
public int hashCode() {
return Objects.hash(equalityFunction.apply(this.t));
}
}
并且映射将是:
.map(e -> new Wrapper<>(e, Employee::getId))
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