Java8使用Lambda表达式(流式)快速实现List转map 、分组、过滤等操作

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。

1 数据准备

1.1 定义1个Fruit对象

package com.wkf.workrecord.work;

import org.junit.Test;

import java.math.BigDecimal;
import java.util.ArrayList;
import java.util.List;

/**
 * @author wuKeFan
 * @date 2023-03-09 15:07:04
 */
public class PublicTest {

    @Test
    public void test() {
        List<Fruit> appleList = new ArrayList<>();//存放apple对象集合
        Fruit apple1 =  new Fruit(1,"苹果1",new BigDecimal("3.25"),10);
        Fruit apple2 = new Fruit(1,"苹果2",new BigDecimal("1.35"),20);
        Fruit banana =  new Fruit(2,"香蕉",new BigDecimal("2.89"),30);
        Fruit lychee =  new Fruit(3,"荔枝",new BigDecimal("9.99"),40);
        appleList.add(apple1);
        appleList.add(banana);
        appleList.add(apple2);
        appleList.add(lychee);
    }

}

1.2 添加一些测试数据

        List<Fruit> fruitList = new ArrayList<>();//存放水果对象集合
        Fruit apple1 =  new Fruit(1,"苹果1",new BigDecimal("3.25"),10);
        Fruit apple2 = new Fruit(1,"苹果2",new BigDecimal("1.35"),20);
        Fruit banana =  new Fruit(2,"香蕉",new BigDecimal("2.89"),30);
        Fruit lychee =  new Fruit(3,"荔枝",new BigDecimal("9.99"),40);
        fruitList.add(apple1);
        fruitList.add(banana);
        fruitList.add(apple2);
        fruitList.add(lychee);

2 正文开始

2.1 分组

Map<Integer, List<Fruit>> groupBy = fruitList.stream().collect(Collectors.groupingBy(Fruit::getId));
System.err.println("groupBy:"+groupBy);

输出:

groupBy:{1=[Fruit(id=1, name=苹果1, money=3.25, num=10), Fruit(id=1, name=苹果2, money=1.35, num=20)], 2=[Fruit(id=2, name=香蕉, money=2.89, num=30)], 3=[Fruit(id=3, name=荔枝, money=9.99, num=40)]}

2.2 List转Map

id为key，apple对象为value，可以这么做：

/**
 * 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, Fruit> appleMap = fruitList.stream().collect(Collectors.toMap(Fruit::getId, a -> a,(k1,k2)->k1));
System.out.println(appleMap);

输出:

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

2.3 过滤Filter

从集合中过滤出来符合条件的元素：

//过滤出符合条件的数据
List<Fruit> filterList = fruitList.stream().filter(a -> a.getName().equals("香蕉")).collect(Collectors.toList());
System.err.println("filterList:"+filterList);

输出:

filterList:[Fruit(id=2, name=香蕉, money=2.89, num=30)]

2.4 求和

将集合中的数据按照某个属性求和:

//计算 总金额
BigDecimal totalMoney = appleList.stream().map(Apple::getMoney).reduce(BigDecimal.ZERO, BigDecimal::add);
System.err.println("totalMoney:"+totalMoney);  //totalMoney:17.48

2.5 查找流中最大 最小值

Collectors.maxBy 和 Collectors.minBy 来计算流中的最大或最小值

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);

2.6 去重

import static java.util.Comparator.comparingLong;
import static java.util.stream.Collectors.collectingAndThen;
import static java.util.stream.Collectors.toCollection;

List<Fruit> unique = fruitList.stream().collect(
    collectingAndThen(
    toCollection(() -> new TreeSet<>(comparingLong(Fruit::getId))), ArrayList::new)
);

3 下表展示 Collectors 类的静态工厂方法