Java分组依据:对多个字段求和 [英] Java groupingBy: sum multiple fields

问题描述

此问题是帖子的扩展:通过返回多个返回值来进行Java 8分组字段.

This question is an extension to the post: Java 8 groupingby with returning multiple field.

对于相同的问题，如何返回Customer的列表?例如，它应该返回:

For the same problem, how do you return a list of Customer? For example, it should return:

Customer("A",4500,6500)
Customer("B",3000,3500)
Customer("C",4000,4500)

如果您想将Map<String, Customer>作为结果集+1，

推荐答案

@Pankaj Singhal的帖子是正确的主意.但是，我会将合并逻辑提取到其自己的函数中，例如在Customer类中，您将具有这样的功能:

@Pankaj Singhal's post is the right idea if you want a Map<String, Customer> as the result set +1. However, I would extract the merging logic into its own function e.g. in the Customer class you would have a function as such:

public static Customer merge(Customer first, Customer second) {
        first.setTotal(first.getTotal() + second.getTotal());
        first.setBalance(first.getBalance() + second.getBalance());
        return first;
}

然后，流查询将变为:

Map<String, Customer> retObj = 
           listCust.stream()
                   .collect(Collectors.toMap(Customer::getName, Function.identity(), Customer::merge)); 

• listCust.stream()创建流对象，即Stream<Customer>.
• Collector .
• Function.identity() ，如果映射的键包含重复项，合并功能Customer::merge用于解决冲突.
• listCust.stream() creates a stream object i.e. Stream<Customer>.
• collect performs a mutable reduction operation on the elements of this stream using the provided Collector.
• The result of toMap is the provided collector, the toMap method extracts the keys Customer::getName and values Function.identity() and if the mapped keys contain duplicates, the merge function Customer::merge is used to resolve collisions.

将合并逻辑提取到其自身的功能中，我看到了三个好处:

There are three benefits I see with extracting the merging logic into its own function:

• 代码更紧凑.
• 代码更具可读性.
• 合并的复杂性与流隔离开了.

但是，如果您打算检索Collection<Customer>:

if however, your intention is to retrieve a Collection<Customer>:

Collection<Customer> result = listCust.stream()
                .collect(Collectors.toMap(Customer::getName,
                        Function.identity(),
                        Customer::merge))
                .values();

或List<Customer>作为结果集，那么您要做的就是调用values()并将其结果传递给ArrayList构造函数:

or List<Customer> as the result set then all you have to do is call values() and pass the result of that to the ArrayList constructor:

List<Customer> result = new ArrayList<>(listCust.stream()
                .collect(Collectors.toMap(Customer::getName,
                        Function.identity(),
                        Customer::merge))
                .values());

更新:

如果您不想改变源中的对象，则只需修改merge函数，如下所示:

if you don't want to mutate the objects in the source then simply modify the merge function as follows:

public static Customer merge(Customer first, Customer second) {
        Customer customer = new Customer(first.getName(), first.getTotal(), first.getBalance());
        customer.setTotal(customer.getTotal() + second.getTotal());
        customer.setBalance(customer.getBalance() + second.getBalance());
        return customer;
}

一切都保持不变.

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