通过生成映射到其集合值中元素数量的每个键的新映射来报告多图 [英] Report on a multimap by producing a new map of each key mapped to the count of elements in its collection value

问题描述

想象一个多重地图跟踪人员，每个人都有分配的任务列表.

Imagine a multimap tracking persons, each of whom has a list of assigned tasks.

Map< Person , List< Task > >  personToTasks = 
    Map.of(
        new Person( "Alice" ) , List.of( new Task( "a1" ), new Task( "a2") ) , 
        new Person( "Bob" )   , List.of( new Task( "b1" ) ) , 
        new Person( "Carol" ) , List.of( new Task( "c1" ), new Task( "c2"), new Task( "c3") ) 
    )
;

如何使用流获取新地图，将每个Person映射到Integer，并在其分配的任务列表中找到项目数?

How can I use streams to get a new map, mapping each Person to an Integer with the count of items found in their list of assigned tasks?

如何获得与该硬编码地图等效的结果:

How to get a result equivalent to this hard-coded map:

Map< Person , Integer >  personToTaskCount = 
    Map.of(
        new Person( "Alice" ) , 2 , 
        new Person( "Bob" )   , 1 , 
        new Person( "Carol" ) , 3 
    )
;

我一直在尝试以下排列:

I have been trying permutations of:

Map < Person, Integer > personToTaskCount = 
        personToTasks.keySet().stream().collect
        (
            Collectors.mapping
            (
                Map.Entry :: getKey ,
                ???
            )
        )
;

推荐答案

您在正确的轨道上，这是一个可行的解决方案:

You are on the right track, here's a possible solution:

Map<Person, Integer> personToTaskCount = personToTasks
    .entrySet()
    .stream()
    .collect(Collectors.toMap(e -> e.getKey(), e -> e.getValue().size()));

The Map#entrySet method returns a Set of the elements in a map, a set of Map.Entry objects. A Map.Entry holds the key-value pairing of each entry in the map. We can ask that object for the key and for the value.

在您的示例情况下，询问每个Map.Entry的值会得到一个List< Task >.然后，我们可以查询该列表的大小，包含的元素数量.这个大小数字是我们想要的新地图Map < Person, Integer >的值.

In your example case, asking for the value of each Map.Entry gets us a List< Task >. We can then interrogate that list for its size, the number of elements it contains. This size number is the value we want for your new map, Map < Person, Integer >.

这是上面代码的扩展版本，以使这些Map.Entry对象的作用更加明显.

Here is an expanded version of the code above, to make more obvious these Map.Entry objects in action.

Map < Person, List < Task > > personToTasks =
        Map.of(
                new Person( "Alice" ) , List.of( new Task( "a1" ) , new Task( "a2" ) ) ,
                new Person( "Bob" ) , List.of( new Task( "b1" ) ) ,
                new Person( "Carol" ) , List.of( new Task( "c1" ) , new Task( "c2" ) , new Task( "c3" ) )
        );


Map < Person, Integer > personToTaskCount =
        personToTasks
                .entrySet()
                .stream()
                .collect(
                        Collectors.toMap(
                                ( Map.Entry < Person, List < Task > > e ) -> { return e.getKey(); } ,
                                ( Map.Entry < Person, List < Task > > e ) -> { return e.getValue().size(); }
                        )
                );

将结果转储到控制台.

System.out.println( "personToTasks = " + personToTasks );
System.out.println( "personToTaskCount = " + personToTaskCount );

运行时.

personToTasks = {Person [name = Alice] = [Task [title = a1]，Task [title = a2]]，Person [name = Carol] = [Task [title = c1]，Task [title = c2] ，Task [title = c3]]，Person [name = Bob] = [Task [title = b1]]}

personToTasks = {Person[name=Alice]=[Task[title=a1], Task[title=a2]], Person[name=Carol]=[Task[title=c1], Task[title=c2], Task[title=c3]], Person[name=Bob]=[Task[title=b1]]}

personToTaskCount = {人物[名称=鲍勃] = 1，人物[名称=爱丽丝] = 2，人物[名称=卡罗尔] = 3}

personToTaskCount = {Person[name=Bob]=1, Person[name=Alice]=2, Person[name=Carol]=3}

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