需求描述：电商订单数据处理，根据下面的list1和list2各十个订单
统计出同时被两个人购买的商品列表（取交集）统计出两个人购买商品的差集统计出全部被购买商品的去重并集统计两个人的分别购买订单的平均价格统计两个人的分别购买订单的总价格准备类对象：
public class VideoOrder { private String tradeNO; private int money; private String title; public VideoOrder() { } public VideoOrder(String tradeNO, String title, int money) { this.tradeNO = tradeNO; this.money = money; this.title = title; } public String getTradeNO() { return tradeNO; } public void setTradeNO(String tradeNO) { this.tradeNO = tradeNO; } public int getMoney() { return money; } public void setMoney(int money) { this.money = money; } public String getTitle() { return title; } public void setTitle(String title) { this.title = title; } @Override public String toString() { return "VideoOrder{" + "title='" + title + ", money=" + money + '\'' + '}'; } }先重写了videoOrder父类object类中的equals和hashcode方法
@Override public boolean equals(Object obj) { if (obj instanceof VideoOrder){ return this.title.equals(((VideoOrder) obj).getTitle()); } return super.equals(obj); } @Override public int hashCode() { return this.title.hashCode(); } // 数据源是相同的 就不再贴上了 减少点篇幅 public static void main(String[] args) { // 总价 35 List<VideoOrder> videoOrders1 = Arrays.asList(new VideoOrder("20190242812", "springboot教程", 3), new VideoOrder("20194350812", "微服务SpringCloud", 5), new VideoOrder("20190814232", "Redis教程", 9), new VideoOrder("20190523812", "⽹⻚开发教程", 9), new VideoOrder("201932324", "百万并发实战Netty", 9)); // 总价 54 List<VideoOrder> videoOrders2 = Arrays.asList(new VideoOrder("2019024285312", "springboot教程", 3), new VideoOrder("2019081453232", "Redis教程", 9), new VideoOrder("20190522338312", "⽹⻚开发教程", 9), new VideoOrder("2019435230812", "Jmeter压⼒测试", 5), new VideoOrder("2019323542411", "Git+Jenkins持续集成", 7), new VideoOrder("2019323542424", "Idea全套教程", 21)); // 交集 List<VideoOrder> result1 = videoOrders1.stream().filter(obj -> videoOrders2.contains(obj)).collect(Collectors.toList()); System.out.println("交集："+result1); // 差集 List<VideoOrder> result2forDiff = videoOrders1.stream().filter(obj -> !videoOrders2.contains(obj)).collect(Collectors.toList()); List<VideoOrder> result3forDiff = videoOrders2.stream().filter(obj -> !videoOrders1.contains(obj)).collect(Collectors.toList()); System.out.println("订单1的差集："+result2forDiff); System.out.println("订单2的差集："+result3forDiff); // 去重并集 // 先拿到所有的并集 为不影响原集合所以不能直接用addAll() List<VideoOrder> midList = videoOrders1.parallelStream().collect(Collectors.toList()); midList.addAll(videoOrders2); List<VideoOrder> resultForAll = midList.stream().distinct().collect(Collectors.toList()); System.out.println("订单去重并集："+resultForAll); // 订单平均价格 Double order1Average = videoOrders1.stream().collect(Collectors.averagingInt(VideoOrder::getMoney)); System.out.println("订单1平均价格："+order1Average); Double order2Average = videoOrders2.stream().collect(Collectors.averagingInt(VideoOrder::getMoney)); System.out.println("订单2平均价格："+order2Average); // 订单总价 int order1Sum = videoOrders1.stream().collect(Collectors.summingInt(VideoOrder::getMoney)).intValue(); System.out.println("订单1总价格："+order1Sum); int order2Sum = videoOrders2.stream().collect(Collectors.summingInt(VideoOrder::getMoney)).intValue(); System.out.println("订单1总价格："+order2Sum); } 运行结果： 交集：[VideoOrder{title='springboot教程, money=3'}, VideoOrder{title='Redis教程, money=9'}, VideoOrder{title='⽹⻚开发教程, money=9'}] 订单1的差集：[VideoOrder{title='微服务SpringCloud, money=5'}, VideoOrder{title='百万并发实战Netty, money=9'}] 订单2的差集：[VideoOrder{title='Jmeter压⼒测试, money=5'}, VideoOrder{title='Git+Jenkins持续集成, money=7'}, VideoOrder{title='Idea全套教程, money=21'}] 订单去重并集：[VideoOrder{title='springboot教程, money=3'}, VideoOrder{title='微服务SpringCloud, money=5'}, VideoOrder{title='Redis教程, money=9'}, VideoOrder{title='⽹⻚开发教程, money=9'}, VideoOrder{title='百万并发实战Netty, money=9'}, VideoOrder{title='Jmeter压⼒测试, money=5'}, VideoOrder{title='Git+Jenkins持续集成, money=7'}, VideoOrder{title='Idea全套教程, money=21'}] 订单1平均价格：7.0 订单2平均价格：9.0 订单1总价格：35 订单1总价格：54总结：取交、并、差集这种重写父类equals方法的写法挺好用的，简单，比我自己写的憨憨的方法好多了；统计均价和总价的方式各有千秋吧，我自己写的那种会计算统计最大最小平均多种结果，如果只需要其中一种的话可能会有点浪费资源，如果统计多种结果的话用summarizing的方法刚好。
