List<String> items =
        Arrays.asList("apple", "apple", "banana",
                "apple", "orange", "banana", "papaya");
// 分组
Map<String, List<String>> result1 = items.stream().collect(
        Collectors.groupingBy(
                Function.identity()
//{papaya=[papaya], orange=[orange], banana=[banana, banana], apple=[apple, apple, apple]}
System.out.println(result1);
// 分组计数
Map<String, Long> result2 = items.stream().collect(
        Collectors.groupingBy(
                Function.identity(), Collectors.counting()
// {papaya=1, orange=1, banana=2, apple=3}
System.out.println(result2);
Map<String, Long> finalMap = new LinkedHashMap<>();
//分组, 计数和排序
result2.entrySet().stream()
        .sorted(Map.Entry.<String, Long>comparingByValue().reversed())
        .forEachOrdered(e -> finalMap.put(e.getKey(), e.getValue()));
// {apple=3, banana=2, papaya=1, orange=1}
System.out.println(finalMap);

集合按照多个属性分组

1.多个属性拼接出一个组合属性

public static void main(String[] args) {
    User user1 = new User("zhangsan", "beijing", 10);
    User user2 = new User("zhangsan", "beijing", 20);
    User user3 = new User("lisi", "shanghai", 30);
    List<User> list = new ArrayList<User>();
    list.add(user1);
    list.add(user2);
    list.add(user3);
    Map<String, List<User>> collect = list.stream().collect(Collectors.groupingBy(e -> fetchGroupKey(e)));
    //{zhangsan#beijing=[User{age=10, name='zhangsan', address='beijing'}, User{age=20, name='zhangsan', address='beijing'}], 
    // lisi#shanghai=[User{age=30, name='lisi', address='shanghai'}]}
    System.out.println(collect);
private static String fetchGroupKey(User user){
    return user.getName() +"#"+ user.getAddress();

2.嵌套调用groupBy

User user1 = new User("zhangsan", "beijing", 10);
User user2 = new User("zhangsan", "beijing", 20);
User user3 = new User("lisi", "shanghai", 30);
List<User> list = new ArrayList<User>();
list.add(user1);
list.add(user2);
list.add(user3);
Map<String, Map<String, List<User>>> collect
        = list.stream().collect(
                Collectors.groupingBy(
                        User::getAddress, Collectors.groupingBy(User::getName)
System.out.println(collect);

3. 使用Arrays.asList

我有一个与Web访问记录相关的域对象列表。这些域对象可以扩展到数千个。
我没有资源或需求将它们以原始格式存储在数据库中,因此我希望预先计算聚合并将聚合的数据放在数据库中。
我需要聚合在5分钟窗口中传输的总字节数,如下面的sql查询

select 
  round(request_timestamp, '5') as window, --round timestamp to the nearest 5 minute
  http_result_code, 
  transaction_time, 
  sum(bytes_transferred)
from web_records
group by 
    round(request_timestamp, '5'), 
    http_result_code, 
    transaction_time


在java 8中,我当前的第一次尝试是这样的,我知道这个解决方案类似于Group by multiple field names in java 8

Map<Date, Map<String, Map<String, Map<String, Map<String, Integer>>>>>>> aggregatedData =
webRecords
    .stream()
    .collect(Collectors.groupingBy(WebRecord::getFiveMinuteWindow,
               Collectors.groupingBy(WebRecord::getCdn,
                 Collectors.groupingBy(WebRecord::getIsp,
                   Collectors.groupingBy(WebRecord::getResultCode,
                       Collectors.groupingBy(WebRecord::getTxnTime,
                         Collectors.reducing(0,
                                             WebRecord::getReqBytes(),
                                             Integer::sum)))))));


这是可行的,但它是丑陋的,所有这些嵌套的地图是一个噩梦!要将地图“展平”或“展开”成行,我必须这样做

for (Date window : aggregatedData.keySet()) {
  for (String cdn : aggregatedData.get(window).keySet()) {
    for (String isp : aggregatedData.get(window).get(cdn).keySet()) {
      for (String resultCode : aggregatedData.get(window).get(cdn).get(isp).keySet()) {
        for (String txnTime : aggregatedData.get(window).get(cdn).get(isp).get(resultCode).keySet()) {
           Integer bytesTransferred = aggregatedData.get(window).get(cdn).get(distId).get(isp).get(resultCode).get(txnTime);
           AggregatedRow row = new AggregatedRow(window, cdn, distId...


如你所见,这是相当混乱和难以维持。
有谁知道更好的方法吗?任何帮助都将不胜感激。
我想知道是否有更好的方法来展开嵌套的映射,或者是否有一个库允许您对集合进行分组。

您应该为地图创建自定义密钥。最简单的方法是使用Arrays.asList

Function<WebRecord, List<Object>> keyExtractor = wr ->
    Arrays.<Object>asList(wr.getFiveMinuteWindow(), wr.getCdn(), wr.getIsp(),
             wr.getResultCode(), wr.getTxnTime());
Map<List<Object>, Integer> aggregatedData = webRecords.stream().collect(
      Collectors.groupingBy(keyExtractor, Collectors.summingInt(WebRecord::getReqBytes)));


在这种情况下,键是按固定顺序列出的5个元素。不是很面向对象,但很简单。或者,您可以定义自己的表示自定义键的类型,并创建适当的hashCode/equals实现。

参考链接:

Java8的groupingBy实现集合的分组,类似Mysql的group by分组功能,注意得到的是一个map对集合按照单个属性分组、分组计数、排序List&lt;String&gt; items = Arrays.asList("apple", "apple", "banana", "apple", "orange", "ba...
List<TbOrganinfo> organinfo = (List<TbOrganinfo>)JSONObject.parseObject(oil).get("data"); //单个key Map map = organinfo.stream().collect(Collectors.toMap(TbOrganinfo::getOrganName,Function.identity())); //两个属性组合作为key Map map = organinfo.s...
SELECT m.className,m.course (SELECT AVG(s.score) score,c.className,s.course FROM score s , class c ON s.name = c.name GROUP BY c.className,s.course) m inner JOIN (SELECT MAX(t.score) score,t.className (SELECT AVG(s.score) AS score,c..
Collectors.groupingBy根据一个或多个属性对集合的项目进行分组 数据准备: public Product(Long id, Integer num, BigDecimal price, String name, String category) { this.id = id; this.num = num; this.price = price; this.name = name; this.category = category; Product prod1 = ne
Java 8引入了Stream API,有许多新方法,其有一个对于分组和聚合操作非常有用,那就是groupingBy()方法。它可以将一个流分组成一个Map,其Entry的key是分组的条件,value是分组的结果,通常是一个List或其他集合。groupingBy()方法的另一个形式是groupingByConcurrent(),它返回一个并发Map,对于并发访问更加友好。 利用groupingBy()方法进行字段分组求和操作的示例如下: 假设有一个Person类,其包含属性:name, age和salary。现在我们需要根据name和age两个字段进行分组,并求出每组的salary总和。可以使用groupingBy()方法加上summingDouble()方法来实现: List<Person> persons = Arrays.asList( new Person("Tom", 20, 5000), new Person("Tom", 21, 4000), new Person("Jerry", 22, 6000), new Person("Jerry", 23, 5500), new Person("Kate", 24, 7000), new Person("Kate", 25, 8000) Map<String, Map<Integer, Double>> result = persons.stream() .collect(Collectors.groupingBy(Person::getName, Collectors.groupingBy(Person::getAge, Collectors.summingDouble(Person::getSalary)))); 这里的personList是一个包含了6个Person对象的List,我们希望将其相同name和age的对象分组,求得salary的总和。在groupingBy()方法,第一个参数是分组条件,这里是Person::getName,第二个参数是分组的结果,这里是一个嵌套的groupingBy()方法,用于再次按照age进行分组,结果是一个Map<Integer, Double>。最后,我们使用summingDouble()方法对salary字段进行求和,得到各个分组的salary总和。这里的result是一个Map<String, Map<Integer, Double>>类型的对象,其key是name,value是以age为key,salary总和为value的子Map,就是我们需要的结果。 这样,我们就利用Java 8Stream API和groupingBy()方法进行了多字段分组求和操作,代码简洁,可读性强,非常方便。
博主,为什么我按照你的方法升级报错[code=html] Additionally, a control plane component may have crashed or exited when started by the container runtime. To troubleshoot, list all containers using your preferred container runtimes CLI. Here is one example how you may list all running Kubernetes containers by using crictl: - 'crictl --runtime-endpoint unix:///var/run/containerd/containerd.sock ps -a | grep kube | grep -v pause' Once you have found the failing container, you can inspect its logs with: - 'crictl --runtime-endpoint unix:///var/run/containerd/containerd.sock logs CONTAINERID' error execution phase wait-control-plane: couldn't initialize a Kubernetes cluster To see the stack trace of this error execute with --v=5 or higher [/code] Centos: 磁盘空间分配,将home空间分配给root Zeillion: 简直是坑,千万别用。 基于Docker快速搭建 PostgreSQL 高可用方案 今天就努力: 你这是把docker hub中的复制过来了? Docker-in-Docker: Jenkins CI 内部如何运行 docker 你的名字有点酸: 执行docker命令报错bash: /usr/bin/docker: cannot execute: required file not found,大佬这个是什么情况呢? iPerf3 命令通用选项 m0_74911788: 你好,为什么我刚刷机没安装iperf 工具也能直接使用iperf 命令?