Java 8 Lambda - Intersection of Two Lists

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你的背包
你的背包 2020-12-05 06:11

I am trying to find intersection of two lists based on some condition and doing some steps. Couldn\'t find a way to do it (in learning stage) :)



        
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  • 2020-12-05 07:03

    The simplest approach is this:

    List<T> intersect = list1.stream()
                             .filter(list2::contains)
                             .collect(Collectors.toList());
    
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  • 2020-12-05 07:03

    List<T> intersect = list1.stream().filter(set1::contains).collect(Collectors.toList());

    This will work as long as T is a String, Integer, Float, etc. where the equals and HashCode methods are straightforward. But if the T is a custom object, we need to implement HashCode & equals

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  • 2020-12-05 07:05

    I need to compare them on assume list1.id == list2.fk_id

    First build up a set of the 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();
            });
    

    You can avoid using a reference to an array object by using reduce function. e.g. see how Collectors.averagingDouble is implemented. But I find this more complicated.

    Note: this is O(N) by using a set of ids rather than using a list of matching ids which would be O(N^2)

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