java 8如何在多个属性上获得不同的列表

java 8 how to get distinct list on more than one property(java 8如何在多个属性上获得不同的列表)

问题描述

如何从对象列表中获取不同的(基于两个属性的不同)列表.例如，让有属性名称和价格的对象列表.现在我如何才能获得具有不同名称或价格的列表.
假设

How can one get the distinct (distinct based on two property) list from a list of objects. for example let there are list of objects with property name and price. Now how can I get a list with distinct name or price.
suppose

list<xyz> l1 = getlist(); // getlist will return the list.

现在让 l1 具有以下属性(名称、价格):-
n1, p1
n1, p2
n2, p1
n2, p3

Now let l1 has the following properties(name, price) :-
n1, p1
n1, p2
n2, p1
n2, p3

现在过滤后的列表应该是-
n1, p1
n2, p3

Now after the filter the list should be-
n1, p1
n2, p3

我试过这样解决 -

public List<xyz> getFilteredList(List<xyz> l1) {

        return l1
                .stream()
                .filter(distinctByKey(xyz::getName))
                .filter(distinctByKey(xyz::getPrice))
                .collect(Collectors.toList());
    }

    private static <T> Predicate<T> distinctByKey(Function<? super T, Object> keyExtractor) {
        Map<Object,Boolean> seen = new ConcurrentHashMap<>();
        return t -> seen.putIfAbsent(keyExtractor.apply(t), Boolean.TRUE) == null;
    }

现在的问题是，当我对名称进行过滤时，列表返回将是 -
n1, p1
n2, p1

Now the problem is when i did filter on name the list return would be -
n1, p1
n2, p1

然后它会对返回的价格运行过滤器 -
n1, p1

and then it would have run filter on price which return -
n1, p1

这不是预期的结果.

推荐答案

我会选择这样的东西，它相当简单灵活，并且基于您的示例:

I'd go for something like this, which is fairly simple and flexible, and builds on your example:

public static <T> List<T> distinctList(List<T> list, Function<? super T, ?>... keyExtractors) {

    return list
        .stream()
        .filter(distinctByKeys(keyExtractors))
        .collect(Collectors.toList());
}

private static <T> Predicate<T> distinctByKeys(Function<? super T, ?>... keyExtractors) {

    final Map<List<?>, Boolean> seen = new ConcurrentHashMap<>();

    return t -> {

        final List<?> keys = Arrays.stream(keyExtractors)
            .map(ke -> ke.apply(t))
            .collect(Collectors.toList());

        return seen.putIfAbsent(keys, Boolean.TRUE) == null;

    };

}

然后可以通过以下方式调用它:

This can then be called in the following manner:

final List<Xyz> distinct = distinctList(list, Xyz::getName, Xyz::getPrice)

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