如何在Java 8 groupingby函数形成的结果中添加标签? [英] How do you add labels to the result formed by Java 8 groupingby function?

问题描述

我目前正在上课，

public class Person {
    private String country;
    private String age;
    private String name;
}

以此Person类的列表作为参数，

Taking a List of this Person class as a argument,

List<Person>

我使用 Java 8 group by(集合)函数将它们按以下数据结构进行了分组:

I managed to group them in the following data structure using Java 8 group by (collection) function :

Map<String, Map<String, Set<String>>> 

示例:

 USA={
     21=
       [
         John,
         Peter.
         Andrew
       ],
      22=
       [
         Eric,
         Mark
       ]

     ]
  },
 France = {
      etc....

下面是我的功能:

  public static Map<String, Map<String, Set<String>>> getNestedMap(List<Person> persons) {
    return persons.stream().collect(
    groupingBy(Person::getCountry,
      groupingBy(Person::getAge,
          mapping(Person::getName, toSet())

      )));
  }

但是，我希望我的数据结构看起来像这样，每个级别都有标签. Java 8分组依据(收集)有什么方法可以帮助实现这一目标?还是有更好的方法?

However, I wanted my data structure to look like this, with labels for each level. Is there a way Java 8 group by (collection) can help achieve this? or is there better way?

 Country = USA,
 AgeList = {
     Age = 21,
     People =
       [
         [Name = John],
         [Name = Peter],
         [Name = Andrew]
       ],
      Age = 22,
      People =
       [
         [Name = Eric],
         [Name = Mark]
       ]

     ]
  },
 Country = France, 
 AgeList = {
       etc....

推荐答案

您从输入(一系列Person实例)和工具(.groupingBy())开始，而不是输入和所需的输出.首先确定然后，以确定哪些工具是将输入转换为所需输出的最合适方法.

You've started with an input (a series of Person instances) and a tool (.groupingBy()), rather than an input and a desired output. Identify that first, then determine which tool(s) are the most appropriate way to transform the input into the desired output.

例如，您可能想要以Country对象结尾，该对象包含名称和Age对象列表，每个对象都包含年龄和一组People对象.考虑到这种结果结构，您可以将单个.groupingBy()传递给按国家/地区分组，然后将结果列表传递给Country(String name, List<Person> people)构造函数，然后依次由另一个.groupingBy()传递给按年龄分组，并且调用Age(int age, List<Person>)构造函数.

For example, you might want to end up with a Country object, that contains a name and a list of Age objects, each of which contains an age and a set of People objects. With that resulting structure in mind you could you could do a single .groupingBy() pass to group by country, and pass the resulting lists into a Country(String name, List<Person> people) constructor, which then in turn does another .groupingBy() pass to group by age, and invokes an Age(int age, List<Person>) constructor.

如果您的目标是然后将此结构序列化为JSON形式的字符串，则现在您已经可以在Country.toString()中轻松地这样做了，因为您已经在所需的结构中存储了数据.

If your goal is to then serialize this structure into a JSON-ish string, you can easily do so in Country.toString(), now that you have the data in the structure you need.

将结构性关注点(例如，从List<Foo>转换为复杂的东西，例如Map<Bar, Map<Baz, Set<Foo>>>)与表示性关注点(例如，然后渲染该复杂结构的字符串表示形式)分开始终是一个好主意.与一次性完成所有步骤相比，您将发现分别解决这两个步骤通常要容易得多且容易维护.

It's always a good idea to separate structural concerns (like transforming from a List<Foo> to something complex like a Map<Bar, Map<Baz, Set<Foo>>>) from representational concerns (like then rendering a string representation of that complex structure). You'll find solving the two steps separately is often significantly easier - and easier to maintain - than doing it all in one fell swoop.

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