java - 将对象值列表转换为组

Question

我有如下一段代码

OrderCriteria o1 = new OrderCriteria(1, 1, 101, 201);
OrderCriteria o2 = new OrderCriteria(1, 1, 102, 202);
OrderCriteria o4 = new OrderCriteria(1, 1, 102, 201);
OrderCriteria o5 = new OrderCriteria(2, 2, 501, 601);
OrderCriteria o6 = new OrderCriteria(2, 2, 501, 602);
OrderCriteria o7 = new OrderCriteria(2, 2, 502, 601);
OrderCriteria o8 = new OrderCriteria(2, 2, 502, 602);
OrderCriteria o9 = new OrderCriteria(2, 2, 503, 603);

OrderCriteria 如下所示:

public class OrderCriteria {
    private final long orderId;
    private final long orderCatalogId;
    private final long procedureId;
    private final long diagnosisId;

    public OrderCriteria(long orderId, long orderCatalogId, long procedureId, long diagnosisId) {
        this.orderId = orderId;
        this.orderCatalogId = orderCatalogId;
        this.procedureId = procedureId;
        this.diagnosisId = diagnosisId;
    }

    // Getters
}

我想要的是获取按订单 ID 分组的程序列表和诊断列表。所以它应该返回:

{1, {101, 102}, {201, 202}}
{2, {501, 502, 503}, {601, 602, 603}}

这意味着 ID 为 1 的订单具有过程 ID 101、102 和诊断 ID 201、202 等。我尝试使用 google guava 表但无法提出任何有效的解决方案。

Answer 1

首先，您需要一个新结构来保存分组数据:

class OrderCriteriaGroup {
    final Set<Long> procedures = new HashSet<>();
    final Set<Long> diagnoses = new HashSet<>();

    void add(OrderCriteria o) {
        procedures.add(o.getProcedureId());
        diagnoses.add(o.getDiagnosisId());
    }

    OrderCriteriaGroup merge(OrderCriteriaGroup g) {
        procedures.addAll(g.procedures);
        diagnoses.addAll(g.diagnoses);
        return this;
    }
}

add() 和 merge() 是帮助我们流式传输和收集数据的便捷方法，如下所示:

Map<Long, OrderCriteriaGroup> grouped = criteriaList.stream()
        .collect(Collectors.groupingBy(OrderCriteria::getOrderId,
                Collector.of(
                        OrderCriteriaGroup::new,
                        OrderCriteriaGroup::add,
                        OrderCriteriaGroup::merge)));