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  • [Lamada] Lamada 集合操作

    分组

    
    //List 以ID分组 Map<Integer,List<Apple>>
    Map<Integer, List<Apple>> groupBy = appleList.stream().collect(Collectors.groupingBy(Apple::getId));
    System.err.println("groupBy:"+groupBy);
    {1=[Apple{id=1, name='苹果1', money=3.25, num=10}, Apple{id=1, name='苹果2', money=1.35, num=20}], 2=[Apple{id=2, name='香蕉', money=2.89, num=30}], 3=[Apple{id=3, name='荔枝', money=9.99, num=40}]}
    
    

    List ---> Map

    /**
     * List -> Map
     * 需要注意的是:
     * toMap 如果集合对象有重复的key,会报错Duplicate key ....
     *  apple1,apple12的id都为1。
     *  可以用 (k1,k2)->k1 来设置,如果有重复的key,则保留key1,舍弃key2
     */
    Map<Integer, Apple> appleMap = appleList.stream().collect(Collectors.toMap(Apple::getId, a -> a,(k1,k2)->k1));
    
    

    过滤

    List<Apple> filterList = appleList.stream().filter(a -> a.getName().equals("香蕉")).collect(Collectors.toList())
    

    求和

    
    BigDecimal totalMoney = appleList.stream().map(Apple::getMoney).reduce(BigDecimal.ZERO, BigDecimal::add);
    
    

    最大值和最小值

    
    Optional<Dish> maxDish = Dish.menu.stream().
          collect(Collectors.maxBy(Comparator.comparing(Dish::getCalories)));
    maxDish.ifPresent(System.out::println);
     
    Optional<Dish> minDish = Dish.menu.stream().
          collect(Collectors.minBy(Comparator.comparing(Dish::getCalories)));
    minDish.ifPresent(System.out::println);
    
    

    去重

    import static java.util.Comparator.comparingLong;
    import static java.util.stream.Collectors.collectingAndThen;
    import static java.util.stream.Collectors.toCollection;
     
    // 根据id去重
         List<Person> unique = appleList.stream().collect(
                    collectingAndThen(
                            toCollection(() -> new TreeSet<>(comparingLong(Apple::getId))), ArrayList::new)
            )
    
    
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  • 原文地址:https://www.cnblogs.com/anycc/p/13299005.html
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