根据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

我正在尝试根据某些属性从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以简单的方式完成吗

can we do it in a simple way using java 8

List<Employee> employee

我们可以根据 id 员工的财产。我见过帖子从字符串的arraylist中删除重复的字符串。

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

你可以从获取一个流列出并输入 TreeSet ，从中提供一个唯一比较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.

然后，如果你真的需要一个列表，你可以把这个集合放回一个ArrayList。

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));

给出示例：

List<Employee> employee = Arrays.asList(new Employee(1, "John"), new Employee(1, "Bob"), new Employee(2, "Alice"));

它将输出：

[Employee{id=1, name='John'}, Employee{id=2, name='Alice'}]

---

另一个想法可能是使用包装员工的包装器，并使用基于其id的equals和hashcode方法： / p>

---

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());
    }
}

然后你包装每个实例，调用 distinct（），打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，我认为你可以做到这一点通过提供将进行比较的函数来包装泛型：

---

In fact, I think you can make this wrapper generic by providing a function that will do the comparison:

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));
    }
}

，映射将为：

.map(e -> new Wrapper<>(e, Employee::getId))

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