java 将list 转为map

#将list转为map  【注意：要对list 进行判空】

Map<Integer, User> collect = users.stream().collect(Collectors.toMap(User::getUserId, User -> User);

#将list转为map并且key去重

Map<Integer, User> collect = users.stream().collect(Collectors.toMap(User::getUserId, User -> User,(oldValue, newValue) -> newValue));

#Collectors.toMap() 多字段拼接key，并且去重

 3.1：去掉重复key 【(key1, key2) -> key2)】, key 使用多字段拼接。返回 Map<String, UserTalk> userMap1类型数据
 　　　　Map<String, UserTalk> userMap1 = userTalkList.stream().collect(Collectors.toMap(userTalk -> userTalk.getSubjectId() + "" + userTalk.getGradeId() +
         "" + userTalk.getUnitId() + "" + userTalk.getLectureType(), UserTalk -> UserTalk, (key1, key2) -> key2));

 #如果集合list Objects == null，那么需要配合 Optional 对object进行判空

        Map<String, BigDecimal> collect = Optional.ofNullable(users).orElseGet(() -> {
            System.out.println("如果是空的，就会执行该处代码，返回一个值");
            return new ArrayList<>();
        }).stream().collect(Collectors.toMap(User::getBirthday, User::getAccumulatedAmount));

#如果value是null，需要使用如下Stream流

 HashMap<Object, Object> collect1 = objects.stream().collect(HashMap::new, (k, v) -> k.put(v.getRoleName(), v.getSysId()), HashMap::putAll);
 System.out.println(collect1);

 #key 重复时，将前面的value 和后面的value拼接起来按照既定的规则

        List<User> users = new ArrayList<>();
        User user5 = new User();
        user5.setUserId(35);
        user5.setNum(0);
        user5.setUserName("王一博");
        user5.setPhone("17319088796");
        users.add(user5);

        User user = new User();
        user.setUserId(1);
        user.setNum(0);
        user.setUserName("王二博");
        user.setPhone("17319088769");
        users.add(user);

        User user2 = new User();
        user2.setUserId(2);
        user2.setNum(0);
        user2.setUserName("王三博");
        user2.setPhone("17319088789");
        users.add(user2);

        User user3 = new User();
        user3.setUserId(3);
        user3.setNum(0);
        user3.setUserName("王三博");
        user3.setPhone("16519088789");
        users.add(user3);

        User user4 = new User();
        user4.setUserId(3);
        user4.setNum(0);
        user4.setUserName("王five博");
        user4.setPhone("16088789");
        users.add(user4);

        //当key 和 value 都是 String时
        Map<String, String> collect = users.stream().collect(Collectors.toMap(User::getUserName, User::getPhone, (key1, key2) -> key1 + "," + key2));
        //测试结果result： {王二博=17319088769, 王三博=17319088789,16519088789, 王five博=16088789, 王一博=17319088796}

        //当key 是Integer , value 都是 String时
        Map<Integer, String> collect1 = users.stream().collect(Collectors.toMap(User::getUserId, User::getPhone, (key1, key2) -> key1 + "," + key2));
        //测试结果result：{1=17319088769, 2=17319088789, 3=16519088789,16088789, 35=17319088796}

        //当key 是Integer , value 都是 Integer时getAge[https://stackoverflow.com/questions/55681921/convert-mapinteger-string-to-mapstring-string]
        Map<String, String> collect3 = users.stream().collect(Collectors.toMap(entry -> String.valueOf(entry.getUserId()), entry -> String.valueOf(entry.getPhone()), (key1, key2) -> key1 + "-" + key2));
        //测试结果result：{1=17319088769, 2=17319088789, 35=17319088796, 3=16519088789-16088789}

 #key 重复时，将数据组成集合【组成何种形式的集合任意发挥】

        Map<Integer, List<String>> map = users.stream().collect(Collectors.toMap(User::getUserId,
                p -> {
                    List<String> getPhoneList = new ArrayList<>();
                    getPhoneList.add(p.getPhone());
                    return getPhoneList;
                },

                (List<String> value1, List<String> value2) -> {
                    value1.addAll(value2);
                    return value1;
                }
        ));
        //测试结果result: {1=[17319088769], 2=[17319088789], 3=[16519088789, 16088789], 35=[17319088796]}