文章来源于：http://ddrv.cn/a/280939
详细分享，移步：https://blog.csdn.net/weixin_41552767/article/details/107690660
java8 lambda小试牛刀，利用Stream把list转map,并将两个list的数据对象合并起来
public static void main(String[] args) {

// 集合1

List lists = new ArrayList<>();

SkillUpgrade s = new SkillUpgrade();

s.setLv(1);

s.setAppearNum(100);

lists.add(s);

SkillUpgrade s2 = new SkillUpgrade();

s2.setLv(2);

s2.setAppearNum(200);

lists.add(s2);

// 集合1

List listx = new ArrayList<>();

SkillUpgrade x = new SkillUpgrade();

x.setLv(1);

x.setSelectNum(1100);

listx.add(x);

SkillUpgrade x2 = new SkillUpgrade();

x2.setLv(2);

x2.setSelectNum(1200);

listx.add(x2);

// 把list转map,{k=lv,vaule=并为自身}  . SkillUpgrade->SkillUpgrade或Function.identity()

Map<Integer, SkillUpgrade> map = listx.stream()

.collect(Collectors.toMap(SkillUpgrade::getLv, SkillUpgrade -> SkillUpgrade));

System.out.println(“map:=”+map);

// 合并

lists.forEach(n -> {

// 如果等级一致

if (map.containsKey(n.getLv())) {

SkillUpgrade obj = map.get(n.getLv());

// 把数量复制过去

n.setSelectNum(obj.getSelectNum());

}

});

System.out.println(“lists:=”+lists);

// 重复问题

Map<Integer, SkillUpgrade> keyRedo = listx.stream()

.collect(Collectors.toMap(SkillUpgrade::getLv, Function.identity(), (key1, key2) -> key2));

// 方式二：指定实例的map

Map<Integer, SkillUpgrade> linkedHashMap = listx.stream().collect(Collectors.toMap(SkillUpgrade::getLv,

SkillUpgrade -> SkillUpgrade, (key1, key2) -> key2, LinkedHashMap::new));

}
/** 
 * output:map:={1=SkillUpgrade [skillId=null, skillName=null, lv=1, persNum=null, selectNum=1100, appearNum=null], 2=SkillUpgrade [skillId=null, skillName=null, lv=2, persNum=null, selectNum=1200, appearNum=null]} 
 *        lists:=[SkillUpgrade [skillId=null, skillName=null, lv=1, persNum=null, selectNum=1100, appearNum=100], SkillUpgrade [skillId=null, skillName=null, lv=2, persNum=null, selectNum=1200, appearNum=200]] 
 */  

输出结果：
map:={1=SkillUpgrade [skillId=null, skillName=null, lv=1, persNum=null, selectNum=1100, appearNum=null], 2=SkillUpgrade [skillId=null, skillName=null, lv=2, persNum=null, selectNum=1200, appearNum=null]}

lists:=[SkillUpgrade [skillId=null, skillName=null, lv=1, persNum=null, selectNum=1100, appearNum=100], SkillUpgrade [skillId=null, skillName=null, lv=2, persNum=null, selectNum=1200, appearNum=200]]

java8 lambda小试牛刀，利用Stream把list转map,并将两个list的数据对象合并起来



public static void main(String[] args) {
		// 集合1
		List<SkillUpgrade> lists = new ArrayList<>();
		SkillUpgrade s = new SkillUpgrade();
		s.setLv(1);
		s.setAppearNum(100);
		lists.add(s);
		SkillUpgrade s2 = new SkillUpgrade();
		s2.setLv(2);
		s2.setAppearNum(200);
		lists.add(s2);
		// 集合1
		List<SkillUpgrade> listx = new ArrayList<>();
		SkillUpgrade x = new SkillUpgrade();
		x.setLv(1);
		x.setSelectNum(1100);
		listx.add(x);
		SkillUpgrade x2 = new SkillUpgrade();
		x2.setLv(2);
		x2.setSelectNum(1200);
		listx.add(x2);
		// 把list转map,{k=lv,vaule=并为自身}  . SkillUpgrade->SkillUpgrade或Function.identity()
		Map<Integer, SkillUpgrade> map = listx.stream()
				.collect(Collectors.toMap(SkillUpgrade::getLv, SkillUpgrade -> SkillUpgrade));
		System.out.println("map:="+map);
		// 合并
		lists.forEach(n -> {
			// 如果等级一致
			if (map.containsKey(n.getLv())) {
				SkillUpgrade obj = map.get(n.getLv());
				// 把数量复制过去
				n.setSelectNum(obj.getSelectNum());
			}
		});
		System.out.println("lists:="+lists);
		// 重复问题
		Map<Integer, SkillUpgrade> keyRedo = listx.stream()
				.collect(Collectors.toMap(SkillUpgrade::getLv, Function.identity(), (key1, key2) -> key2));
		// 方式二：指定实例的map
		Map<Integer, SkillUpgrade> linkedHashMap = listx.stream().collect(Collectors.toMap(SkillUpgrade::getLv,
				SkillUpgrade -> SkillUpgrade, (key1, key2) -> key2, LinkedHashMap::new));
	}
	
	/**
	 * output:map:={1=SkillUpgrade [skillId=null, skillName=null, lv=1, persNum=null, selectNum=1100, appearNum=null], 2=SkillUpgrade [skillId=null, skillName=null, lv=2, persNum=null, selectNum=1200, appearNum=null]}
     *        lists:=[SkillUpgrade [skillId=null, skillName=null, lv=1, persNum=null, selectNum=1100, appearNum=100], SkillUpgrade [skillId=null, skillName=null, lv=2, persNum=null, selectNum=1200, appearNum=200]]
	 */




输出结果：

map:={1=SkillUpgrade [skillId=null, skillName=null, lv=1, persNum=null, selectNum=1100, appearNum=null], 2=SkillUpgrade [skillId=null, skillName=null, lv=2, persNum=null, selectNum=1200, appearNum=null]}

lists:=[SkillUpgrade [skillId=null, skillName=null, lv=1, persNum=null, selectNum=1100, appearNum=100], SkillUpgrade [skillId=null, skillName=null, lv=2, persNum=null, selectNum=1200, appearNum=200]]

java8 lambda小试牛刀，利用Stream把list转map,并将两个list的数据对象合并起来
publicstaticvoidmain(String[] args) {
// 集合1
List lists = newArrayList<>();
SkillUpgrade s = newSkillUpgrade();
s.setLv(1);
s.setAppearNum(100);
lists.add(s);
SkillUpgrade s2 = newSkillUpgrade();
s2.setLv(2);
s2.setAppearNum(200);
lists.add(s2);
// 集合1
List listx = newArrayList<>();
SkillUpgrade x = newSkillUpgrade();
x.setLv(1);
x.setSelectNum(1100);
listx.add(x);
SkillUpgrade x2 = newSkillUpgrade();
x2.setLv(2);
x2.setSelectNum(1200);
listx.add(x2);
// 把list转map,{k=lv,vaule=并为自身}  . SkillUpgrade->SkillUpgrade或Function.identity()
Map map = listx.stream()
.collect(Collectors.toMap(SkillUpgrade::getLv, SkillUpgrade -> SkillUpgrade));
System.out.println(“map:=”+map);
// 合并
lists.forEach(n -> {
// 如果等级一致
if(map.containsKey(n.getLv())) {
SkillUpgrade obj = map.get(n.getLv());
// 把数量复制过去
n.setSelectNum(obj.getSelectNum());
}
});
System.out.println(“lists:=”+lists);
// 重复问题
Map keyRedo = listx.stream()
.collect(Collectors.toMap(SkillUpgrade::getLv, Function.identity(), (key1, key2) -> key2));
// 方式二：指定实例的map
Map linkedHashMap = listx.stream().collect(Collectors.toMap(SkillUpgrade::getLv,
SkillUpgrade -> SkillUpgrade, (key1, key2) -> key2, LinkedHashMap::new));
}
/**
* output:map:={1=SkillUpgrade [skillId=null, skillName=null, lv=1, persNum=null, selectNum=1100, appearNum=null], 2=SkillUpgrade [skillId=null, skillName=null, lv=2, persNum=null, selectNum=1200, appearNum=null]}
*        lists:=[SkillUpgrade [skillId=null, skillName=null, lv=1, persNum=null, selectNum=1100, appearNum=100], SkillUpgrade [skillId=null, skillName=null, lv=2, persNum=null, selectNum=1200, appearNum=200]]
*/
输出结果：
map:={1=SkillUpgrade [skillId=null, skillName=null, lv=1, persNum=null, selectNum=1100, appearNum=null], 2=SkillUpgrade [skillId=null, skillName=null, lv=2, persNum=null, selectNum=1200, appearNum=null]}
lists:=[SkillUpgrade [skillId=null, skillName=null, lv=1, persNum=null, selectNum=1100, appearNum=100], SkillUpgrade [skillId=null, skillName=null, lv=2, persNum=null, selectNum=1200, appearNum=200]]

List<User>中，User有id和money两个字段，如果id相同，就把User合并，并且将money相加，java怎么处理？？



假如 id 是 String类型

User tmpUser = null;
Map<String, User> map = new HashMap<String, User>();//合并后的User存在map中
for(User oUser userList) {
     tmpUser = map.get(oUser.getId());
     if (tmpUser != null) {
             tmpUser.setMoney(tmpUser.getMoney() + oUser.getMoney);  //money合并
     } else {
          map.put(oUser.getId(), oUser);
    }
}