如何按Java 8分组并从父级收集ID [英] How to group by java 8 and collect ids from parent

问题描述

我有A类

Company {
  String name;
  Logo logo;
}

Logo {
  int color; //can have values=1 (green),2 (red),3 (blue) ...
  String name;
  String address;
}

需要的输出:对于每种类型1,2,3，按颜色对Logo的所有实例进行分组.对于每个这样的组，什么是A.id 给我公司以其颜色徽标. 例如.哪些公司的徽标为红色?

Output needed : for each type 1,2,3 Group all instances of Logo by color. For each such group what were A.id Give me companies by their color logos. E.g. which companies have logo red?

我尝试关注

Input
List<Company> company = {//initialization}

company.stream().map(e -> e.getLogo())
          .collect(Collectors.groupingBy(e -> {Logo b = new Logo(); 
                                               b.setType(e.getType(); 
                                               return b;}, Collectors.counting()))

这将产生一个徽标图并进行计数.如何获得公司名称?

This produces a map of Logo and count How do I get names of Company?

推荐答案

如果要查找所有logo为color的公司，则应将其分组为:

If you were to look for all companies given a color of their logo, you shall group as:

Map<Integer, List<Company>> collect = company.stream()
        .collect(Collectors.groupingBy(e -> e.getLogo().getColor()));

以防万一，只是这些值的计数是无光泽的，然后应使用Collectors.counting，例如:

Just in case, just the count of such values matte, you should then use Collectors.counting such as:

Map<Integer, Long> count = company.stream()
        .collect(Collectors.groupingBy(e -> e.getLogo().getColor(),
                Collectors.counting()));

简而言之，如果您希望值本身为Company类型，请不要map流.

In short, do not map the stream if you want the values to be of type Company itself.

编辑:基于注释，如果您打算将Company转换为其name，则可以将其分组后进行映射.使用Collectors.mapping如:

Edit: Based on comments, if the idea is to convert Company to its name you can map once grouped. Using Collectors.mapping such as :

Map<Integer, List<String>> collect = company.stream()
        .collect(Collectors.groupingBy(e -> e.getLogo().getColor(), 
                        Collectors.mapping(Company::getName, Collectors.toList())));

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