时间: 2021-01-07 02:38:41

标签:

javascript

Java数据聚合问题

public class Demo {public static void main(String[] args) throws IOException {

HashMap m1 = new HashMap<>();

m1.put("month","2020-06");

m1.put("plan_working_hours",272.5);

m1.put("month_total_working_hours",120);

m1.put("project_code","ToB-APP");

m1.put("this_years_total_working_hours",287);

m1.put("project_name","B端APP框架");

m1.put("total_working_hours",307);

HashMap m2 = new HashMap<>();

m2.put("month","2020-07");

m2.put("plan_working_hours",272.5);

m2.put("month_total_working_hours",93);

m2.put("project_code","ToB-APP");

m2.put("this_years_total_working_hours",287);

m2.put("project_name","B端APP框架");

m2.put("total_working_hours",307);

HashMap m3 = new HashMap<>();

m3.put("month","2020-11");

m3.put("plan_working_hours",104);

m3.put("month_total_working_hours",16);

m3.put("project_code","1001");

m3.put("this_years_total_working_hours",67);

m3.put("project_name","C端底座");

m3.put("total_working_hours",150);

HashMap m4 = new HashMap<>();

m4.put("month","2020-12");

m4.put("plan_working_hours",104);

m4.put("month_total_working_hours",51);

m4.put("project_code","1001");

m4.put("this_years_total_working_hours",67);

m4.put("project_name","C端底座");

m4.put("total_working_hours",150);

ArrayListmaps = new ArrayList<>();

maps.add(m1);

maps.add(m2);

maps.add(m3);

maps.add(m4);

//todo 做数据聚合(

// 将project_name值相同的数据中的month和month_total_working_hours聚合到一个list中

// 每个project_name只有一个map数据)

/**

* 格式如下

*

* {

* "monthList": ["2020-06":120 ,"2020-07":93]

* "plan_working_hours": 272.5,

* "project_code": "ToB-APP",

* "this_years_total_working_hours": 287,

* "project_name": "B端APP框架",

* "total_working_hours": 307

* },

*/

}

}

最佳答案

Map one = Maps.newHashMap();

one.put("one",1);

one.put("two",2);

Map two = Maps.newHashMap();

two.put("one",1);

two.put("two",2);

List> list = Lists.newArrayList();

list.add(one);

list.add(two);

Set keySet = one.keySet();

Map> result = list.stream().reduce(new HashMap<>(), (s, o) -> {

keySet.forEach(a -> {

if (s.containsKey(a)) {

List value = s.get(a);

value.add(o.get(a));

} else {

List value = Lists.newArrayList();

value.add(o.get(a));

s.put(a, value);

}

});

return s;

}, (s, o) -> s);

System.out.println(result);