[Lamada] Lamada 集合操作
阿新 • • 發佈:2020-07-14
分組
//List 以ID分組 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='蘋果1', money=3.25, num=10}, Apple{id=1, name='蘋果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 如果集合物件有重複的key,會報錯Duplicate key .... * apple1,apple12的id都為1。 * 可以用 (k1,k2)->k1 來設定,如果有重複的key,則保留key1,捨棄key2 */ Map<Integer, Apple> appleMap = appleList.stream().collect(Collectors.toMap(Apple::getId, a -> a,(k1,k2)->k1));
過濾
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;
// 根據id去重
List<Person> unique = appleList.stream().collect(
collectingAndThen(
toCollection(() -> new TreeSet<>(comparingLong(Apple::getId))), ArrayList::new)
)