https://blog.csdn.net/lu930124/article/details/77595585

利用java8新特性，可以用简洁高效的代码来实现一些数据处理。

定义1个Apple对象：

public class Apple {private Integer id;private String name;private BigDecimal money;private Integer num;public Apple(Integer id, String name, BigDecimal money, Integer num) {this.id = id;this.name = name;this.money = money;this.num = num;    }}

添加一些测试数据：

List<Apple> appleList = new ArrayList<>();Apple apple1 =  new Apple(1,"苹果1",new BigDecimal("3.25"),10);Apple apple12 = new Apple(1,"苹果2",new BigDecimal("1.35"),20);Apple apple2 =  new Apple(2,"香蕉",new BigDecimal("2.89"),30);Apple apple3 =  new Apple(3,"荔枝",new BigDecimal("9.99"),40);appleList.add(apple1);appleList.add(apple12);appleList.add(apple2);appleList.add(apple3);

1、分组

List里面的对象元素，以某个属性来分组，例如，以id分组，将id相同的放在一起：

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}]}

2、List转Map

id为key，apple对象为value，可以这么做：

Map<Integer, Apple> appleMap = appleList.stream().collect(Collectors.toMap(Apple::getId, a -> a,(k1,k2)->k1));

打印appleMap

{1=Apple{id=1, name='苹果1', money=3.25, num=10}, 2=Apple{id=2, name='香蕉', money=2.89, num=30}, 3=Apple{id=3, name='荔枝', money=9.99, num=40}}

3、过滤Filter

从集合中过滤出来符合条件的元素：

List<Apple> filterList = appleList.stream().filter(a -> a.getName().equals("香蕉")).collect(Collectors.toList());System.err.println("filterList:"+filterList);[Apple{id=2, name='香蕉', money=2.89, num=30}]

4.求和