Java 8 Lambda - 两个列表的交集

Question

我试图找到intersection两个基于某些条件并执行一些步骤的列表.找不到办法(在学习阶段):)

Double totalAmount = 0.00d;
Double discount = 0.00d;


List<OrderLineEntry> orderLineEntryList = orderEntry.getOrderReleases().stream()
    .flatMap(orderReleaseEntry -> orderReleaseEntry.getOrderLines().stream())
    .filter(orderLineEntry -> orderLineEntry.getStatus().equals("PP") || orderLineEntry.getStatus().equals("PD"))
    .collect(Collectors.toList());

for (OrderLineEntry orderLineEntry : orderLineEntryList) {
    for (SplitLineEntry splitLineEntry : splitReleaseEntry.getLineEntries()) {
        if (splitLineEntry.getOrderLineId().equals(orderLineEntry.getId()) && splitLineEntry.getStatusCode() != "PX") {
            totalAmount += orderLineEntry.getFinalAmount();
            couponDiscount += orderLineEntry.getCouponDiscount() == null ? 0.00d : orderLineEntry.getCouponDiscount();
        }
    }
}

如您所见,逻辑很简单

根据一些过滤器从订单获取所有项目list并与另一个过滤list并做一些事情.

Answer 1

最简单的方法是:

List<T> intersect = list1.stream()
                         .filter(list2::contains)
                         .collect(Collectors.toList());

Answer 2

我需要在假设list1.id == list2.fk_id上比较它们

首先建立一组fk_id;

Set<Integer> orderLineEntrSet = orderEntry.getOrderReleases().stream()
    .flatMap(orderReleaseEntry ->
orderReleaseEntry.getOrderLines().stream())
    .filter(orderLineEntry -> { 
            String s = orderLineEntry.getStatus(); 
            return "PP".equals(s) || "PD".equals(s); 
    })
    .map(e -> e.getId())
    .collect(Collectors.toSet());

double[] totalAmount = { 0.0 };
double[] couponDiscount = { 0.0 };
orderLineEntryList.stream()
    .flatMap(sre -> sre.getLineEntries().stream())
    .filter(ole -> orderLineEntrySet.contains(ole.getOrderLineId())
    .filter(ole -> !"PX".equals(ole.getStatusCode()))
    .forEach(ole -> {
            totalAmount[0] += ole.getFinalAmount();
            if (ole.getCouponDiscount() != null)
                couponDiscount[0] += ole.getCouponDiscount();
        });

您可以使用reduce函数避免使用对数组对象的引用.例如,看看如何实现Collectors.averagingDouble.但我发现这更复杂.

注意:这是O(N),使用一组id而不是使用匹配的id列表,它们是O(N ^ 2)